ACR EA Unit 3 · 2017. 8. 24. · Version 2.0 ElEmEntary algEbra UnIt 3. ... Unit 3 Post-Assessment 1..... 195 Unit 3 Post-Assessment 2 ... and a stack of student activity sheets

®

AccuplacerCollege-ReAdy MATH Version 2.0

ElEmEntary algEbra UnIt 3

The classroom teacher may reproduce materials in this book for classroom use only.The reproduction of any part for an entire school or school system is strictly prohibited.

No part of this publication may be transmitted, stored, or recorded in any formwithout written permission from the publisher.

1 2 3 4 5 6 7 8 9 10

ISBN 978-0-8251-6773-7

Copyright © 2011

J. Weston Walch, Publisher

Portland, ME 04103

www.walch.com

Printed in the United States of America

EDUCATIONWALCH

Accuplacer College-Ready Mathematics: Elementary Algebra© 2011 Walch Educationiii

Unit 3 • Equations, inEqualitiEs, and word problEms

Table of Contents

iii

Teacher’s Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TG1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TG1

Station Activities Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TG3

Unit 3 Pre-Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Lesson 1: Translating Written Phrases into Algebraic Expressions . . . . . . . . . . . . . . . 7

Lesson 2: Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29

Lesson 3: Solving Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .52

Lesson 4: Solving Quadratic Equations by Factoring . . . . . . . . . . . . . . . . . . . . . . . . . .74

Lesson 5: Graphing Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .97

Lesson 6: Solving Word Problems Involving Angle Measures . . . . . . . . . . . . . . . . .141

Lesson 7: Solving Word Problems Using the Pythagorean Theorem . . . . . . . . . . . .162

Unit 3 Mixed Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .191

Unit 3 Post-Assessment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .195

Unit 3 Post-Assessment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .200

Station Activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .205

Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .239

067737F

Accuplacer College-Ready Mathematics: Elementary Algebra© 2011 Walch EducationtG1


Teacher’s GuideIntroductionThe Accuplacer College-Ready (ACR) Instructional Materials are a complete set of resources developed to support students, and their teachers, in preparation for the Accuplacer® test used by many colleges and universities to inform placement decisions. This program was developed around the content and format of the test, with input from educators familiar with mathematics and with the Accuplacer®. The ACR Materials comprise a full course, providing more than enough activities and resources for a semester of instruction that develops and reinforces the mathematics needed for the Accuplacer® and for success in future math courses. This work was funded in part by a Davis Family Foundation Grant, awarded to the Maine International Center for Digital Learning. The materials were piloted by teachers in MELMAC Learning Foundations College-Ready grant sites. They were revised and refined based on pilot feedback and are presented here as Version 2.0.

The ACR program recognizes the importance of tailoring instructional experiences to address students’ identified needs. The materials are designed in a mix-and-match model, allowing teachers to select appropriate lessons, activities, and components.

Each unit in the Accuplacer College-Ready materials contains the following elements:

• Unit Pre-Assessment • For each lesson in the unit:

• Lesson Pre-Assessment• Essential Questions• Words to Know• Warm-Up Option 1 with Debrief• Warm-Up Option 2 with Debrief• Focus Problem with Debrief• Additional Examples• Guided Practice• Independent Practice • Progress Assessment• Resource List• Answer Key

• Unit Mixed Review• Unit Post-Assessment 1• Unit Post-Assessment 2• Station Activities• Unit Glossary

Unit 3 • Equations, inEqualitiEs, and word problEmsTeacher’s Guide

Accuplacer College-Ready Mathematics: Elementary Algebra tG2

© 2011 Walch Education

All of the program materials are completely reproducible and may be printed or projected to support instruction. The complete set of materials encompasses six units, focused on six of the Accuplacer® content strands. Each unit provides overview materials as well as a series of lessons.

This unit includes the following lessons:

Elementary Algebra Unit 3: Equations, Inequalities, and Word Problems

Lesson 1: Translating Written Phrases into Algebraic Expressions

Lesson 2: Solving Linear Equations

Lesson 3: Solving Linear Inequalities

Lesson 4: Solving Quadratic Equations by Factoring

Lesson 5: Graphing Linear Equations

Lesson 6: Solving Word Problems Involving Angle Measures

Lesson 7: Solving Word Problems Using Pythagorean Theorem

Structure of the Units

Unit and Lesson Pre-Assessments provide teachers with information to guide their instructional decisions and against which to document progress. Essential Questions serve to focus teaching and learning on important concepts. Words to Know (and the Unit Glossary) allow for direct vocabulary instruction, shown to increase mathematics learning and achievement. Two Warm-Up Options (with Debriefs) offer teachers choices for engaging students’ prior knowledge and for differentiating instruction. A Focus Problem (with Debrief) and Additional Examples facilitate problem-based learning, letting teachers present concepts and skills in meaningful, real-world contexts. Guided Practice and Independent Practice problem sets give students opportunities to hone their skills, and Progress Assessments gauge their progress after each lesson. The Resource Lists include links to online resources, with synopses to help teachers make decisions about their use. (Note: Functionality of some online resources may be improved by switching to another browser, such as Safari if using a Mac or Firefox if using a PC.) A Mixed Review and two Unit Assessments with Answer Keys complete each unit. Finally, Station Activities engage small groups of students in a series of problem-solving activities, with suggested debrief prompts.



Station Activities GuideEach unit includes a collection of station-based activities to provide students with opportunities to practice and apply the mathematical skills and concepts they are learning. You may use these activities in addition to the instructional lessons, or, especially if the pre-test or other formative assessment results suggest it, instead of direct instruction in areas where students have the basic concepts but need practice. The debriefing discussions after each set of activities provide an important opportunity to help students reflect on their experiences and synthesize their thinking. It also provides an additional opportunity for ongoing, informal assessment to guide instructional planning.

Implementation Guide

The following guidelines will help you prepare for and use the activity sets in this section.

Setting Up the Stations

Each activity set consists of four stations. Set up each station at a desk, or at several desks pushed together, with enough chairs for a small group of students. Place a card with the number of the station on the desk. Each station should also contain the materials specified in the teacher’s notes, and a stack of student activity sheets (one copy per student). Place the required materials (as listed) at each station.

When a group of students arrives at a station, each student should take one of the activity sheets to record the group’s work. Although students should work together to develop one set of answers for the entire group, each student should record the answers on his or her own activity sheet. This helps keep students engaged in the activity and gives each student a record of the activity for future reference.

Forming Groups of Students

All activity sets consist of four stations. You might divide the class into four groups by having students count off from 1 to 4. If you have a large class and want to have students working in small groups, you might set up two identical sets of stations, labeled A and B. In this way, the class can be divided into eight groups, with each group of students rotating through the “A” stations or “B” stations.


Accuplacer College-Ready Mathematics: Elementary Algebra tG4


Assigning Roles to Students

Students often work most productively in groups when each student has an assigned role. You may want to assign roles to students when they are assigned to groups and change the roles occasionally. Some possible roles are as follows:

• Reader—reads the steps of the activity aloud

• Facilitator—makes sure that each student in the group has a chance to speak and pose questions; also makes sure that each student agrees on each answer before it is written down

• Materials Manager—handles the materials at the station and makes sure the materials are put back in place at the end of the activity

• Timekeeper—tracks the group’s progress to ensure that the activity is completed in the allotted time

• Spokesperson—speaks for the group during the debriefing session after the activities

Timing the Activities

The activities in this section are designed to take approximately 10 minutes per station. Therefore, you might plan on having groups change stations every 10 minutes, with a two-minute interval for moving from one station to the next. It is helpful to give students a “5-minute warning” before it is time to change stations.

Since each activity set consists of four stations, the above time frame means that it will take about 50 minutes for groups to work through all stations.

Guidelines for Students

Before starting the first activity set, you may want to review the following “ground rules” with students. You might also post the rules in the classroom.

• All students in a group should agree on each answer before it is written down. If there is a disagreement within the group, discuss it with one another.

• You can ask your teacher a question only if everyone in the group has the same question.

• If you finish early, work together to write problems of your own that are similar to the ones on the activity sheet.

• Leave the station exactly as you found it. All materials should be in the same place and in the same condition as when you arrived.



Debriefing the Activities

After each group has rotated through every station, bring students together for a brief class discussion. At this time, you might have the groups’ spokespersons pose any questions they had about the activities. Before responding, ask if students in other groups encountered the same difficulty or if they have a response to the question. The class discussion is also a good time to reinforce the essential ideas of the activities. The questions that are provided in the teacher’s notes for each activity set can serve as a guide to initiating this type of discussion.

You may want to collect the student activity sheets before beginning the class discussion. However, it can be beneficial to collect the sheets afterward so that students can refer to them during the discussion. This also gives students a chance to revisit and refine their work based on the debriefing session. If you run out of time to hold class discussions, you might want to have students journal about their experiences and follow up with a class discussion the next day.

Unit 3 • Equations, inEqualitiEs, and word problEmsUnit Pre-Assessment

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Accuplacer College-Ready Mathematics: Elementary Algebra© 2011 Walch Education1

Assessment

Circle the letter of the correct answer.

1. Which of the following most accurately translates the phrase “50% of a number less 5” into an algebraic expression?

a. 0.5x + 5 c. 5x + 0.5

b. 0.5x – 5 d. 5x – 0.5

2. Shelly purchased x DVDs at $5.99 per DVD from an online retailer, and received a 20% discount. She also paid $7.99 for shipping costs. Which of the following most accurately translates the situation into an algebraic expression?

a. 5.99x – 0.2(5.99x) + 7.99 c. 7.99x – 0.2(5.99x) + 5.99

b. 5.99x + 0.2(5.99x) + 7.99 d. 7.99x + 0.2(5.99x) + 5.99

3. Solve the equation 5y – 25 = 40.

a. 13 c. 3

b. 15 d. 65

4. Jack and 2 of his friends each bought the same kind of pizza for lunch. The total cost for all 3 pizzas was $24.30, which included 8% tax. Which of the following equations best represents the situation? Let x represent the cost of each pizza.

a. 24.3 = 3x + 0.08 c. 24.3 = 3x – 3(0.08x)

b. 3x = 24.3 + 3(0.08x) d. 24.3 = 3x + 3(0.08x)

5. Solve the inequality − + >x3

6 12 for x.

a. x > 18 c. x < –18

b. x < 18 d. x > –18

continued


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Accuplacer College-Ready Mathematics: Elementary Algebra 2


Assessment

6. A van can seat 10 passengers. Which of the following number lines most accurately represents this situation?

a. 0 2 4 6 8 10 12 14 16 18 20

x

b. 0 2 4 6 8 10 12 14 16 18 20

x

c. 0 2 4 6 8 10 12 14 16 18 20

x

d. 0 2 4 6 8 10 12 14 16 18 20

x

7. Solve the equation x2 + 8x – 20 = 0.

a. x = –2 and x = –10 c. x = –2 and x = 10

b. x = 2 and x = 10 d. x = 2 and x = –10

8. Solve the equation –x2 –10x – 16 = 0.

a. x = –2 and x = –8 c. x = –2 and x = 8

b. x = 2 and x = 8 d. x = 2 and x = –8

9. Solve the equation 2x2 – 14x – 36 = 0.

a. x = –2 and x = –9 c. x = –2 and x = 9

b. x = 2 and x = –9 d. x = 2 and x = 9

continued


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Assessment

10. Which graph best represents the equation y = –x + 1?

a.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

c.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

b.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

d.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

continued


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Assessment

11. Which graph best represents a linear equation with points (0, 2) and (1, 3)?

a.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

c.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

b.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

d.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

12. Two angles are supplementary. The larger angle is 60° more than twice the smaller angle. Which of the following are measures of the angles?

a. 10° and 80° c. 60° and 120°

b. 40° and 140° d. 20° and 160° continued


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Assessment

13. Two angles are complementary. One of the angles measures 20° more than the other. Which of the following are measures of the angles?

a. 35° and 55° c. 10° and 80°

b. 20° and 70° d. 40° and 50°

14. A right triangle has one leg that measures 7 meters. The other leg measures 24 meters. What is the measure of the hypotenuse?

a. 7 m c. 25 m

b. 16 m d. 26 m

15. A right triangle has a hypotenuse of 17 feet with one leg measuring 8 feet. What is the measure of the other leg?

a. 8 ft c. 13 ft

b. 12 ft d. 15 ft




Unit Pre-Assessment Answer Key 1. b (Lesson 1) 9. c (Lesson 4) 2. a (Lesson 1) 10. d (Lesson 5) 3. a (Lesson 2) 11. c (Lesson 5) 4. d (Lesson 1) 12. b (Lesson 6) 5. c (Lesson 3) 13. a (Lesson 6) 6. b (Lesson 3) 14. c (Lesson 7) 7. d (Lesson 4) 15. d (Lesson 7) 8. a (Lesson 4)

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 1: Translating Written Phrases into Algebraic Expressions

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Assessment

Lesson Pre-AssessmentTranslate each phrase into an algebraic expression.

1. the sum of 5 and x

2. 5 less than x

3. the quotient of 3 times x plus 7 and x

4. Shelia drinks twice as much water each day as Farimah does. Let x represent how much water Farimah drinks each day. Write an algebraic expression to describe how much water Shelia drinks.

5. Write an algebraic expression to represent the cost of 4 football game tickets, t, with a service charge of $10.




Instruction

Lesson 1: Translating Written Phrases into Algebraic Expressions

Essential Questions 1. How do you translate verbal phrases into algebraic expressions?

2. Why do we use variables?

WORDS TO KNOW

algebraic expression an expression that has one or more variables

coefficient the number multiplied by a variable in an algebraic expression

constant a fixed value that does not change, such as a number

expression a mathematical statement that includes numbers, operations, and/or variables to represent a number or quantity

variable a letter used to represent a value that can change or vary


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Warm-Up Option 1Translate each phrase into an algebraic expression.

1. the sum of 5 and 6

2. 5 less than 20

3. 8 multiplied by 3

4. half of 24

5. one-third of 27


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Warm-Up Option 2Ella purchased 2 DVDs and 3 CDs from Tyler’s Electronics at the prices listed below. After taxes, her total cost increased by $5.60.

Tyler’s ElectronicsCDs $4.99DVDs $19.99

1. How can you write the cost of 2 DVDs as an algebraic expression?

2. How can you write the cost of 2 DVDs and 3 CDs as an algebraic expression?

3. How can you write the cost of 2 DVDs and 3 CDs, increased by $5.60 for taxes, as an algebraic expression?



Instruction

Warm-Up Option 1: Debrief 1. The sum of 5 and 6 means to add 5 and 6.

The expression 5 + 6 represents the sum of 5 and 6.

2. 5 less than 20 means subtract 5 from 20.

The expression 20 – 5 represents 5 less than 20.

• It is not uncommon for students to confuse the operation “less than” (<) with the verb phrase “is less than.”

• Another common error is that students recognize “less than” as subtraction, but write “5 – 20” rather than “20 – 5”. Encourage students to think about what 5 less than 20 really is: 15.

• Concretize the material whenever possible to ensure student understanding.

3. 8 multiplied by 3 means multiply 8 by 3.

The expression 8 × 3 means 8 multiplied by 3.

4. Half of 24 means to divide 24 by 2.

The expression 242

means half of 24. The expression 12

24( ) also means half of 24. Either

expression is correct.

5. One-third of 27 means to divide 27 by 3.

The expression 273

means one-third of 27. The expression 13

27( ) also means one-third of 27.

Either expression is correct.




Instruction

Warm-Up Option 2: Debrief 1. How can you write the cost of the 2 DVDs as an algebraic expression?

The cost of the 2 DVDs can be written as an expression of addition (19.99 + 19.99) or as an expression of multiplication (2 × 19.99 or 2(19.99)).

• It will become less confusing for students if they begin to eliminate the “×” from their work as it is often confused with a variable. Encourage them to use parentheses ( ) instead.

2. How can you write the cost of 2 DVDs and 3 CDs as an algebraic expression?

The cost of the 2 DVDs and 3 CDs can be written as either the expression 2(19.99) + 3(4.99) or the expression 2 × 19.99 + 3 × 4.99.

3. How can you write the cost of 2 DVDs and 3 CDs, increased by $5.60 for taxes?

• “Increased by $5.60 for taxes” means “to add 5.60.”

2(19.99) + 3(4.99) + 5.60 would represent the cost of 2 DVDs and 3 CDs increased by $5.60.


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Focus ProblemYou and two friends had dinner at a Spanish tapas restaurant that charged $6 per tapas, or appetizer. Many tapas were ordered. The total bill was $58.32, which included taxes of $4.32.

1. What was the cost of the tapas without including the taxes?

2. What numeric expression can be used to translate the total bill without including the taxes?

3. What algebraic expression can be used to represent the number of tapas ordered multiplied by $6?

4. What algebraic expression can be used to represent the number of tapas ordered multiplied by $6, including taxes?




Instruction

Focus Problem DebriefIntroduction

To introduce translating verbal phrases into algebraic expressions, ask students to translate words they know: the sum of (addition), more than (addition), less than (subtraction), multiplied by (multiplication), times (multiplication), divided by (division), and half of (divide by 2).

• Write words for adding, subtracting, multiplying, and dividing on the board. Provide examples for each, along with examples of expressions with variables.

OperationWords that mean the

same thingExamples

Examples with variables

Addition plus, more than, the sum of, the total of, increased by, added to

the sum of a number and 6

a number plus 4

x + 6

x + 4Subtraction less than, less,

minus, the difference of, decreased by, subtracted from

five less than a number

the difference of a number and 6

x – 5

x – 6

Multiplication multiply, times, the product of, twice, double, triple, multiplied by

double a number

the product of 5 and a number

2x

5x

Division divided by, divide into, the quotient of, half of, one-third of, the ratio of

the quotient of a number and 4

one-third of a number

x4

x3

• Ask students to give examples of the words variable and constant using real-life situations. For example, “The cost of gasoline is variable from day to day; it changes daily. The number of eggs in a dozen is constant; there are always 12.”



Instruction

• Explain that these words have the same meanings in mathematics.

• A variable is a letter used to represent a value that can change or vary.

• A constant is a value that doesn’t change.

• Mathematical key words, like those shown in the chart, can be used to translate written phrases to an algebraic expression, or a mathematical statement that includes numbers, operations, and/or variables to represent a number or quantity.

Example

Sarah has saved $150. Each week she increases her savings by $15. Write an expression for the amount of savings Sarah will have after x weeks.

• Start with the information you know: Sarah has initially saved $150 and saves an additional $15 each week.

• “Increases” means “to add,” so one operation must be addition.

• Make a table to show what you know.

Week Total Savings0 $1501 $150 + $15 = $1652 $150 + $15 + $15

= $150 + 2($15) = $1803 $150 + $15 + $15 + $15

=$150 + 3 ($15) = $1954 $150 + $15 + $15 + $15 + $15

=$150 + 4($15) = $210x $150 + x($15)

• Point out that each week Sarah saves an additional constant amount of $15.

• Discuss the pattern from the table and why a variable was used to represent the unknown number of weeks of savings.

The total savings is the initial savings—a constant—plus the product of $15 times the number of weeks spent saving: 150 + 15x




Instruction

Focus Problem statement:

You and two friends had dinner at a Spanish tapas restaurant that charged $6 per tapas, or appetizer. Lots of tapas were ordered. The total bill was $58.32, which included taxes of $4.32.

Question 1

What was the cost of the tapas without including the taxes?

Instruction

• Determine what you know and translate what you know into an expression.

• The total bill was $58.32.

• The total bill included taxes of $4.32.

• Determine the amount of the bill without taxes.

$58.32 – $4.32

• Solve the problem.

$58.32 – $4.32 = $54.00

The cost of the tapas without including the taxes was $54.00.

Question 2

What numeric expression can be used to translate the total bill without including the taxes?

Instruction

• Start with the information that you know.

• The total bill was $58.32.

• The amount of taxes paid was $4.32.

• Determine whether you must subtract or add the taxes.

• Key word: “without” indicates subtraction



Instruction

• Translate “total bill without including the taxes” into a numerical expression:

total bill without taxes$58.32 ─– $4.32

• When writing mathematical expressions, symbols such as “$” are not included. This allows for readability and clarity.

The numeric expression used to translate the total bill without including the taxes is 58.32 – 4.32.

Question 3

What algebraic expression can be used to represent the number of tapas ordered multiplied by $6?

Instruction

• Translate “the number of tapas ordered multiplied by $6” into an algebraic expression.


• Each tapas costs $6.

• Determine any unknown values.

• The number of tapas ordered is unknown. Let n represent the unknown values.

• Write an expression that multiplies the unknown number of tapas ordered by $6.

cost of each tapas multiplied by unknown number$6 × n

$6 is the coefficient of the variable n: 6n




InstructionQuestion 4

What algebraic expression can be used to represent the number of tapas ordered multiplied by $6, including taxes?

Instruction

• Translate each phrase of the question into an algebraic expression.


• The number of tapas ordered is an unknown value.

• The cost of each tapas was $6.

• The tax on the total bill was $4.32.

• Determine any unknown values.

• Let n represent the unknown number of tapas ordered. So, the number of tapas ordered, multiplied by $6, can be represented by the expression 6n.

• Write an expression that represents the unknown tapas ordered, multiplied by $6, and includes taxes.

• The phrase “includes” means “to add.” So, add the taxes.

The algebraic expression to find the number of tapas ordered multiplied by $6, including taxes, is 6n + 4.32.

Additional ExamplesExample 1

A smartphone is on sale for 25% off its list price. The sale price of the smartphone is $149.25. What expression can be used to represent the list price of the smartphone?

Solution


• Key words: “off” indicates subtraction


• The sale price is $149.25.



Instruction

• The discount is 25% of the unknown list price.

• Let x represent the unknown list price.

• Translate “25% of unknown” into an expression.

• “Percent of” means “to multiply.”

• So, 25% of an unknown can be represented by 0.25 × x or 0.25x.

• Describe the situation: list price - discount = sale price

• Describe the list price: The list price is the sale price plus the discount.

sale price plus discount$149.25 + 0.25x

• Write an algebraic expression to represent the list price using what you know.

149.25 + 0.25x

The expression to represent the list price of the smartphone is 149.25 + 0.25x.

Example 2

At rest, the average human heart beats at 13

of the rate of the average monkey’s heart. What

expression can be used to represent the heart rate of a human?

Solution


• Key word: “13

of” indicates division by three or, alternatively, multiplication by 13

.


• A human’s heart beats at 13

of the rate of a monkey’s heart.

• Let x represent the unknown heart rate of a monkey.

• Translate “13

of unknown” into an expression.




Instruction

13

of an unknown can be represented by 13

× x or x3

.

• Describe the situation: human’s heart rate = 13

(monkey’s heart rate)

one-third of a monkey’s heart rate

13

─× x

• Write an algebraic expression to represent a human’s heart rate using what you know: 13

× x or 13

x or x3

The expression used to represent the heart rate of a human is 13

x or x3

.

Example 3

Helen purchased 3 books from an online book store and received a 20% discount. The shipping cost was $10 and was not discounted. Write an expression that can be used to represent the total cost Helen paid for the 3 books plus the shipping cost.

Solution


• Key words: “plus” indicates addition and “discount” indicates subtraction


• Helen purchased 3 books.

• The discount is 20% of the unknown price per book.

• There was an additional $10 cost for shipping.

• Let x represent the unknown price.

• Translate “20% of unknown” into an expression.



Instruction

• “Percent of” indicates multiplication. So, 20% of an unknown can be represented by 0.20 × x or 0.20x as the discount.

• Since there are 3 books, the discount can be multiplied by 3 to give 0.20(3x).

• Describe the situation: 3 books at an unknown price – discount + shipping = total cost

3 books at an unknown book price

subtract discount add shipping

3x – 0.20(3x) + $10

• Write an algebraic expression to represent the total cost using what you know.

3x – 0.20(3x) + 10

The expression used to represent the total cost Helen paid for the 3 books plus shipping costs is 3x – 0.20(3x) + 10.


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Guided PracticeTranslate each phrase into an algebraic expression.

1. 5 more than a number

2. a number decreased by 7

3. a number multiplied by 9

4. a number discounted by 30%

5. one-fourth of a number

6. a number increased by 29

7. 2 times a number plus 5

8. half of a number less 6

9. 40% of a number plus 15

10. a number plus 8% of a number

continued


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For problems 11–16, write algebraic expressions for each situation.

11. An airplane traveled at a speed of 340 miles per hour. Write an algebraic expression to represent the number of miles the airplane traveled in x hours.

12. It costs William $12 in tolls to drive to work each day, plus $2.50 a gallon for gas. He uses 2 gallons of gas each day. Write an algebraic expression to represent the total cost of tolls and gas for an unknown number of days.

13. Colin bought 2 theater tickets and paid a service charge of 5% for buying them from a broker. Write an algebraic expression to represent the total cost of the tickets. Let x represent the cost of each ticket.

14. Nadia and some friends went to a movie. Their total cost was $30.24, which included taxes of $2.24. Write an algebraic expression to represent the price of each movie ticket, not including taxes. Let x represent the number of Nadia’s friends that went to the movies.

15. The average heart rate of a cow is about 14

the heart rate of a chicken. Write an algebraic

expression to represent the heart rate of a cow. Let x represent the heart rate of a chicken.

16. Eddie purchased 4 packages of light bulbs and received a 15% discount. He also paid $4.85 in taxes on his purchase. Write an algebraic expression to represent the total amount Eddie paid. Let x represent the number of light bulbs purchased.


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Independent Practice Translate each phrase into an algebraic expression.

1. 7 less than a number

2. the sum of a number and 8

3. twice a number

4. a number discounted by 10%

5. one-third of a number

6. 2 more than a number

7. 5 times a number plus 10

8. half of a number less 8

9. 60% of a number less 5

10. 2 times a number plus 8% of a number

continued


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For problems 11–16, write algebraic expressions for each situation.

11. A cyclist traveled at a speed of 12 miles per hour. Write an algebraic expression to represent the number of miles the cyclist traveled in x hours.

12. It costs Shelena $45 to have her hair styled and cut each month, plus she gives her stylist an 8% tip. Write an algebraic expression to represent the total cost of having her hair cut and styled and leaving a tip for an unknown number of months.

13. Gavin agrees to buy a 6-month package deal of monthly gym passes, and in turn receives a 15% discount. Write an algebraic expression to represent the total cost of the monthly passes with the discount, if x represents the cost of each monthly pass.

14. Emily and Olivia purchased several identical pairs of shoes. Their total cost was $125.28, which included taxes of $9.28. Write an algebraic expression to represent the price of each pair of shoes before taxes. Let x represent the number of pairs of shoes purchased.

15. The maximum miles per hour for driving downtown are about one-third the maximum for driving on the highway. Write an algebraic expression to represent the maximum driving speed for driving downtown. Let x represent the maximum driving speed on the highway.

16. Andrew purchased 10 cans of tennis balls from an online store and received a 25% discount. Shipping cost $5.99. Write an algebraic expression to represent the total cost of the tennis balls with the shipping cost, if x represents the cost of each can.


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Assessment

Progress AssessmentTranslate each phrase into an algebraic expression.

1. double a number plus 8% of a number

2. a number discounted by 25% plus 5

3. a number times 3

Translate each situation into an algebraic expression. Show your work.

4. Ryanne purchased x gallons of milk and x boxes of orange juice from Happy Foods at the prices listed below. Her total cost increased by 8% for taxes. Write an algebraic expression to represent the total purchase price, including taxes.

Happy Foods Pricing

Gallon of milk $3.50

Carton of orange juice $3.75

5. Logan took tennis lessons at a rate of $45 per hour. Write an algebraic expression to represent the amount Logan paid for x lessons.

6. Ethan runs at double the rate that Julio walks. Julio walks at a rate of 4 miles per hour. Write an algebraic expression to represent the number of miles Ethan runs in x hours.



Instruction

Resource List• Math-Play.com. “Algebraic Expressions Millionaire Game.”

http://www.math-play.com/math-millionaire.html

“Algebraic Expressions Millionaire Game” can be played alone or in two teams. For each question, players have to identify the correct mathematical expression that models a given word expression.

• Quia. “Algebraic Symbolism Matching Game.”

www.quia.com/mc/319817.html

In this matching game, players pair each statement with its algebraic interpretation. There are 40 matches to the provided game.




Lesson 1 Answer KeyLesson Pre-Assessment, p. 7 1. 5 + x

2. x – 5

3. 3 7x

x+

4. 2x

5. 4t + $10

Guided Practice, p. 22 1. x + 5

2. x – 7

3. 9x

4. x – 0.30x or 0.7x

5. x4

6. x + 29

7. 2x + 5

8. x2

6−

9. 0.40x + 15

10. x + 0.08x

11. 340x

12. 12x + 2.5(2x) or 17x

13. 2x + 0.05(2x) or 2.1x

14. $ . – $ .3 24 2 2401

( )+x

15. x4

16. 4x – 0.15(4x) + $4.85

Independent Practice, p. 24 1. x – 7

2. x + 8

3. 2x

4. x – 0.1x or 0.9x

5. x3

6. x + 2

7. 5x + 10

8. x2

8−

9. 0.6x – 5

10. 2x +.08x

11. 12x

12. ($45 + 0.08($45))x

13. 6x – 0.15(6x) or 5.1x

14. 125 28 9 28. .−x

15. x3

16. 10x – 0.25(10x) + 5.99 or 7.5x + 5.99

Progress Assessment, p. 26 1. 2x + 0.08x 2. x – 0.25x + 5 3. 3x 4. 3.5x + 3.75x + 0.08(3.5x + 3.75x) 5. 45x 6. 8x

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 2: Solving Linear Equations

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Assessment

Lesson Pre-AssessmentSolve each equation. Show your work.

1. 2x + 5 = 25

2. 14

2 5x + =

3. 5x – 17 = 18

For problems 4 and 5, write algebraic equations, solve them, and show your work.

4. Jessica runs twice as far each day as Jean does. Jean runs 2.5 miles a day. How many miles a day does Jessica run?

5. James bought a plane ticket to New York City and used a coupon for 15% off the ticket price. The cost of his ticket was $253.30. What was the price of the ticket without the discount?

Accuplacer College-Ready Mathematics: Elementary Algebra30



Instruction

Lesson 2: Solving Linear EquationsEssential Questions 1. How does solving equations relate to the real world?

2. How can linear equations be represented physically?

3. Why is it important to solve linear equations?

WORDS TO KNOW

addition property of equality if the same number is added to both sides of an equation, the two sides remain equal

equation a mathematical sentence that uses an equal sign (=) to show that two quantities are equal

inverse operation pairs of opposite operations that undo each other; addition and subtraction are inverse operations; and multiplication and division are inverse operations

isolate steps taken to get the variable alone on one side of an equation

linear equation an equation that can be written in the form ax + b = c. The solution of a linear equation is a straight line.

solution the value or values that make an equation true

subtraction property of equality if the same number is subtracted from both sides of an equation, the two sides remain equal


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Warm-Up Option 1Write an algebraic expression for each phrase.

1. 3 times the sum of x and 5

2. x increased by 7

3. 8 plus the product of 2 and n




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Warm-Up Option 2To advertise his new robot, Taylor produced a 30-second TV commercial for $10,000. He has a medium-sized local TV station run his commercial n times during a one-month period at $1,500 per airing. What will the commercial cost Taylor if the commercial is aired 2, 3, and 4 times? Include the cost of producing the commercial with each airing.

Taylor’s Robot Commercialn $10,000 + $1,500n Total Cost2 $10,000 + $1,500(2) $13,0003 $10,000 + $1,500(3) $14,5004 $10,000 + $1,500(4) $16,000

1. What does n represent?

2. Is the number 30 as it relates to “a 30-second TV commercial” needed to solve the problem?

3. What algebraic expression could be used to show the total cost of n airings?



Instruction

Warm-Up Option 1: Debrief

1. 3 times the sum of x and 5

• The sum of x and 5 means add x and 5: (x + 5)

• 3 times the sum of x and 5 means to multiply (x + 5) by 3.

The expression 3(x + 5) represents 3 times the sum of x and 5.

2. x increased by 7

• “Increased by” means “to add.” Add 7 to x.

The expression x + 7 represents x increased by 7.

3. 8 plus the product of 2 and n

• The product of 2 and n is written as 2n.

The expression 8 + 2n means 8 plus the product of 2 and n.


• “Decreased by” indicates subtraction.

• “A number” indicates the use of a variable.

The expression n – 4 means a number decreased by 4.


• “One-third” means to divide the number by 3.

• “A number” indicates the use of a variable.

The expression n3

means one-third of an unknown number. It is also correct for students to write 13

n .




Instruction


1. What does n represent?

The variable n is used to represent the number of airings.

2. Is the number 30 as it relates to “a 30-second TV commercial” needed to solve the problem?

No. Some word problems, like this one, have more numbers than are necessary to find the answer.

• “30-second” describes the length of the commercial, but it is not needed to find the solution to the problem.

• Students may find it helpful to cross out information that is unnecessary.

3. What algebraic expression could be used to show the total cost of n airings?

The algebraic expression $10,000 + $1,500n could be used to show the total cost of n airings.


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Focus ProblemLucas purchased a refrigerator at a local appliance store. His total cost of $1,331 included sales tax at the rate of 8% and an additional delivery charge of $35.

1. What algebraic expression could be used to represent the tax?

2. What algebraic expression could be used to represent the total cost?

3. What was the original cost of the refrigerator?

4. How much tax did Lucas pay?

5. How can you check your solution?




Instruction


• Introduce the concept of an equation, emphasizing that both sides of an equation have the same value, or equal value. Linear equations are equations that can be written in the form ax + b = c. The solution of a linear equation is a straight line.

• Review how to translate words into algebraic expressions. Refer to Lesson 1 for more information.

• At this point, you may want to have students use a balance scale to represent a situation such as coins and dollars to represent the same amount.

• Show several examples with known values, such as 3 + 5 = 8, and then show examples using variables, such as x + 7 = x + 3 + 4.

• The goal of solving an equation is to find the value of the variable that makes the equation true.

• To solve the equation x + 7 = 15, isolate the variable x on one side of the balance.

• To isolate a variable, the operations which have been applied to the variable must be undone.

• Operations that are performed on one side of an equation must be performed on the other side of the equation to keep the equation balanced.

• To isolate x in the equation x + 7 = 15, subtract 7 from both sides of the balance.

• The solution to the equation x + 7 = 15 is x = 8.

• Describe the addition property of equality and the subtraction property of equality.

• Point out to students that the subtraction property of equality was used to find the value of x. To find the solution to the equation x – 7 = 15, the addition property of equality would be used to isolate the variable:

x – 7 + 7 = 15 + 7

x = 22

• Describe the inverse operations of addition and subtraction and of multiplication and division.

• Addition and subtraction are inverse operations, or have the reverse effects.

Example: Adding 7 to 15 results in 22 and subtracting 7 from 22 results in the original number of 15.



Instruction

• Similarly, multiplication and division are inverse operations, or have the reverse effects.

Example: Multiplying 4 by 6 results in 24 and dividing 24 by 6 results in the original number of 4.

• Discuss with students how the order of operations are undone. The opposite of PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction) is used to solve an equation.

Question 1

Which algebraic expression could be used to represent the tax?

Instruction

• Identify the variable in the focus problem, the unknown price of the refrigerator.

• Translate the tax into an algebraic expression. Let the variable p represent the price of the refrigerator.

• To calculate the amount for tax, find 8% of the price of the refrigerator, or 0.08 • p = 0.08p.

0.08p is an expression that could be used to represent the tax.

Question 2

Which expression could be used to represent the total cost?

Instruction

• The total cost of the refrigerator can be represented by:

price of the refrigerator

+ tax amount + delivery charge

p + 0.08p + 35

The expression p + 0.08p + 35 represents the total cost.





What was the original cost of the refrigerator?

Instruction

• Translate the expression for total cost into an equation. Lucas paid a total of $1,331 including the cost of the refrigerator, tax, and delivery.

Total cost = p + 0.08p + 35

1,331 = p + 0.08p + 35

• At this point, you may want to have students use a balance scale to represent the total cost equation.

• Students may find it helpful if a vertical line is drawn through the equal sign far enough for anticipated work. This line helps separate the two sides of the equation as well as replicate the balance scale on their paper.

• The goal of solving an equation is to find the value of the variable that makes the equation true. The variable, p, must be isolated to find the price of the refrigerator. To isolate a variable, the operations which have been applied to the variable must be undone and operations that are performed on one side of an equation must be performed on the other side of the equation to keep the equation balanced.

• To isolate the variable p in the equation 1,331 = p + 0.08p + 35, do the following:

• Combine like terms. Terms can be combined ONLY IF they have the exact variable part. The terms p and 0.08p both have the same variable part, p.

• To combine p + 0.08p, add the coefficients of p.

• Students may fail to recognize the coefficient of p is 1. Remind them 1 • p = p.

1,331 = 1.08p + 35

• Use the subtraction property of equality to solve this equation.



Instruction

• To undo the operation of adding 35 to the right side of the equation, use the inverse operation to subtract 35 from both sides of the equation.

1,296 = 1.08p

• To undo the multiplication of 1.08, use the inverse operation of multiplication, which is division.

• Divide both sides of the equation by 1.08.

• 1 2961 08

1 08

1 08,.

( . )

.=

p

The solution is p = =1 2961 08

1 200,.

, .

The price of the refrigerator was $1,200.

Question 4

How much tax did Lucas pay?

Instruction

• Since the value of p, the price of the refrigerator, was found, the tax can be calculated by substituting p into the expression 0.08p and solving.

0.08p = 0.08(1,200) = 96

• Point out the equality on both sides of the equation.

Lucas paid $96 in tax.





How can you check your solution?

Instruction

• To check the solution, have students substitute $1,200 into the equation used for the total cost. Remind students that both sides of the equation MUST be equal.

1,331 = p + 0.08p + 35

1,331 = 1,200 + (0.08)(1,200) + 35

1,331 = 1,200 + 96 + 35

1,331 = 1,331

Both sides of the equation are equal; therefore, the solution is correct.


Susan has saved $600 to purchase a new TV, but that is only 13

of the price of the TV she wants. What is the price of the TV?

Solution


Susan has $600.

$600 is only one-third of the price of the TV.

• Let x represent the unknown price.

• Describe the situation.

one-third of x equals $600

• Translate the situation into an equation.

13

600x =



Instruction

• Solve the equation.

13

600x =

• To isolate the variable, multiply (the reverse operation of division) both sides of the equation by 3.

313

3 600x

= ( )

x = 1,800

The price of the TV is $1,800.

• To check the results, substitute the solution into the equation 13

600x = and solve.

13

600

131 800 600

600 600

x =

=

=

( , )

Both sides of the equation are equal; therefore, the solution is correct.

Example 2

Isaac drove 35 miles to a football game. The odometer showed 45,843 miles when he arrived at the game. What was Isaac’s original odometer reading?

Solution


• The trip was 35 miles.

• The ending odometer read 45,843.




Instruction

• Let x represent the original odometer reading.


original reading plus x equals 45,843


x + 35 = 45,843


x + 35 = 45,843

• To isolate the variable, subtract (the reverse operation of addition) 35 from both sides of the equation.

x + 35 – 35 = 45,843 – 35

x = 45,808

The original odometer reading was 45,808 miles.

• To check the results substitute the solution into the equation x + 35 = 45,843 and solve.

x + 35 = 45,843

45,808 + 35 = 45,843

45,843 = 45,843

Both sides of the equation are equal, therefore the solution is correct.



InstructionExample 3

Ashley decreased her savings from $2,500 to $2,310 during a one-year period. By how much money did Ashley decrease her savings during the one-year period?

Solution


• Ashley started with $2,500.

• After one year, she had $2,310.

• Let s represent the decrease in savings.


original savings minus decreased amount equals current savings


2,500 – s = 2,310


2,500 – s = 2,310

• Because s is negative, add (the inverse operation of subtraction) s to both sides of the equation.

2,500 – s + s = 2,310 + s

2,500 = 2,310 + s

• To further isolate the variable, subtract 2,310 from both sides of the equation.

2,500 – 2,310 = 2,310 – 2,310 + s

s = 190

Ashley decreased her savings by $190.




Instruction

• Check the results by substituting the solution into the equation 2,500 – s = 2,310 and solving.

2,500 – s = 2,310

2,500 – 190 = 2,310

2,310 = 2,310

• Both sides of the equation are equal; therefore, the solution is correct.


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Guided PracticeSolve each equation. Show your work.

1. 2x + 2 = 12 6. 14

12 36x + =

2. 40 + x = 35 7. x + 15 = 30

3. 12 – y = 6 8. x + 0.08x = 162

4. 24 – y = 6 + 2y 9. 23

4 10x− =

5. 3x – 8 = x + 10 10. 3y + 3 = 27

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For problems 11–16, write algebraic equations for the situations described, and then solve them. Show your work.

11. Zach watches TV 3 times as much as Joel. Joel watches TV 2 hours a day. How many hours a day does Zach watch TV?

12. It costs Raquel $5 in tolls to drive to work and back each day, plus she uses 3 gallons of gas. It costs her a total of $12.50 to drive to work and back each day. How much per gallon is Raquel paying for her gas? How do you know?

13. Hayden bought 4 tickets to a football game. He paid a 5% service charge for buying them from a broker. His total cost was $105.00. What was the price of each ticket?

14. It cost Justin $100 to have cable TV installed in his house. Each month he pays an access fee plus tax of 7% of his monthly bill. After 6 months, Justin paid a total of $350.38 for his access fee, taxes, and his initial installation. What is Justin’s monthly access fee?

15. You and 3 friends divide the proceeds of a garage sale equally. The garage sale earned $412. How much money did you receive?

16. The area of Sofia’s herb garden is 18

the area of her vegetable garden. The area of her herb

garden is 6 square feet. What is the area of her vegetable garden?


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Independent Practice Solve each equation. Show your work.

1. 3y + 5 = 20 6. 17

3 9x + =

2. 16 + x = 7 7. y + 11 = 29

3. 27 – y = –3 8. x + 0.07x = 38.52

4. 33 – x = 3 + 2x 9. 25

6 12x− =

5. 5x – 9 = x + 23 10. 6y + 6 = 78

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11. Leah’s dog consumes four times as many calories a day as her cat. Her cat consumes 240 calories per day. How many calories per day does her dog consume?

12. It costs Marcus an access fee for each visit to his gym, plus it costs him $3 in gas for each trip to the gym and back. This month it cost Marcus $108 for 6 trips to his gym. How much is Marcus’s access fee per visit?

13. Rebecca bought x pairs of socks and received a 20% discount. Each pair of socks cost her $4.99. Her total cost without tax was $29.94. How many pairs of socks did Rebecca buy?

14. Amelia and two of her friends each ate the same lunch. The total cost was $55.08, which included an 8% tax. What was the price of each lunch, not including tax?

15. Alan mowed the lawn and trimmed the bushes in his yard. The amount of time he spent

trimming the bushes was 13

the amount of time it took him to mow the lawn. If it took him

1 hour and 15 minutes to mow the lawn, how long did it take him to trim the bushes?

16. The area of a football field is about 34

the size of an international soccer field. The area of a

football field, including the end zones, is 57,600 square feet. What is an approximate area of an

international soccer field?


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Assessment

Progress AssessmentFor problems 1–3, solve the equations. Show your work.

1. 3x – 2 = 10

2. 13

2 7y + =

3. 10 + x + 0.08x = 35.92


4. In 1903, the Tour de France was 2,428 kilometers long. In 2005, the length of the race was increased to 3,608 kilometers. How many kilometers longer was the Tour de France in 2005 than in 1903?

5. Tracey decreased her calorie intake per day from 2,800 calories to 1,900 calories. By how many calories a day did Tracey decrease her calorie intake?

6. Eben purchased a new pair of shoes and paid a 7% sales tax on his purchase. The total cost for his shoes was $63.13. What was the price of his shoes, without tax?




Instruction

Resource List• APlusMath. “Algebra Planet Blaster.”

www.aplusmath.com/Games/PlanetBlast/index.html

Players solve each multi-step linear equation to find the correct planet to “blast.” Incorrect answers cause players to destroy their own ship.

• Cool Math. “Algebra Crunchers.”

www.coolmath.com/algebra/algebra-practice-solving.html

Provides practice problems for solving equations, inequalities, and systems of equations. Users choose which type of problem they would like to work on.



Lesson 2 Answer KeyLesson Pre-Assessment, p. 29 1. x = 10 4. x = 5 2. x = 12 5. x = 298 3. x = 7

Guided Practice, p. 45 1. x = 5 9. x = 21 2. x = –5 10. y = 8 3. y = 6 11. 6 hours per day 4. y = 6 12. $2.50 per gallon 5. x = 9 13. $25 per ticket 6. x = 96 14. $39 per month 7. x = 15 15. $103 8. x = 150 16. 48 square feet

Independent Practice, p. 47 1. x = 5 9. x = 45 2. x = –9 10. y = 12 3. y = 30 11. 960 calories per day 4. x = 10 12. $15 5. x = 8 13. 6 pairs 6. x = 42 14. $17 per lunch 7. y = 18 15. 25 minutes 8. x = 36 16. 76,800 square feet

Progress Assessment, p. 49 1. x = 4 4. 1,180 kilometers 2. y = 15 5. 900 calories 3. x = 24 6. $59

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 3: Solving Linear Inequalities

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Assessment

Lesson Pre-AssessmentFind the solution set for each of the inequalities below. Show all work.

1. 2x – 1 < 3

2. − >x2

1

3. 3x + 1 ≥ 4

4. x + 5 ≤ 1

5. A golf club charges green fees of $30 per day and has an association fee of $45 per month. Eli has no more than $255 to spend each month for golf. How many days each month can Eli play golf?



Instruction

Lesson 3: Solving Linear InequalitiesEssential Questions 1. How is solving a linear inequality similar to solving a linear equation?

2. How are inequalities written?

WORDS TO KNOW

algebraic inequality an inequality that has one or more variables and contains at least one of the following symbols: <, >, ≤, ≥, or ≠

inequality a mathematical sentence that shows the relationship between quantities that are not equivalent

solution set the value or values that make a sentence or statement true


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Warm-Up Option 1Solve each algebraic equation.

1. 2x + 4 = 26

2. 35

6 27x+ =

3. t – 8 = 32

4. 9y – 3 = 105

5. –2n = 56


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Warm-Up Option 2James earns $15 per hour as a teller at a bank. He works h hours one week and pays 17% of his earnings in state and social security taxes. His pay for the week is $460.65. How many hours did James work?

1. What does h represent?

2. What algebraic equation could be used to represent the situation?

3. Solve the problem.




Instruction

Warm-Up Option 1: Debrief 1. 2x + 4 = 26

• Remind students that to solve an equation, you must reverse the operations which have been applied to the variable by using inverse operations.

• Isolate the variable, x, by first subtracting 4 from both sides of the equation.

2x + 4 = 26

2x + 4 – 4 = 26 – 4

2x = 22

• Divide both sides of the equation by 2.

2 22

22

22211

x

x

x

=

=

=

2. 35

6 27x+ =

• Isolate the variable by first subtracting 6 from both sides of the equation.

35

6 27

35

6 6 27 6

35

21

x

x

x

+ =

+ − = −

=

• Multiply both sides of the equation by 5.

35

21

3 105

x

x

=

=



Instruction


3 105

33

1053

35

x

x

x

=

=

=

3. t – 8 = 32

• Isolate the variable by first adding 8 to both sides of the equation.

t – 8 = 32

t – 8 + 8 = 32 + 8

t = 40

4. 9y – 3 = 105


9y – 3 = 105

9y – 3 + 3 = 105 + 3

9y = 108


9y = 108

9

91089

y=

y = 12

5. –2n = 56

• Isolate the variable by dividing both sides of the equation by –2.

–2n = 56−−

=−

22

562

n

n = –28




Instruction


1. What does h represent?

The variable h represents the number of hours James worked in one week.


• James earns $15 per hour, which translates to 15h.

• 17% of his earnings goes to taxes: 17% percent of 15h translates to 0.17(15h).

• James is paying 17% of his earnings, which translates to 15h – 0.17(15h).

• His pay for the week is $460.65.

The equation to represent this situation is 15h – 0.17(15h) = 460.65.

3. Solve the problem by isolating the variable.

• Simplify the left side of the equation by multiplying 0.17 by 15h.

15h – 0.17(15h) = 460.65

15h – 2.55h = 460.65

• Combine like terms.

15h – 2.55h = 460.65

12.45h = 460.65

• Divide both sides of the equation by the product.

12.45h = 460.65

h = 460.65/12.45

h = 37

James worked 37 hours.


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Focus ProblemFree checking accounts are offered at a local bank for accounts with balances of at least $200. Mr. Ferris has a balance of $250.15 and he writes a check for $75. How much must Mr. Ferris deposit in his account to keep his free checking account? Let the variable x represent the amount Mr. Ferris must deposit.

1. Is there more than one solution to this problem?

2. Which symbol can be used to represent “at least”?

3. What inequality can be used to represent the solution?

4. What is the solution set for the inequality?

5. How can the solution set be represented graphically?




Instruction


• The prefix in- in the word inequality means “not.” Inequalities are sentences stating that two things are not equal. Introduce this concept by first showing numerical inequalities such as 12 > 2 and 1 < 7.

• Remind students that the symbols >, <, ≥, ≤, and ≠ are used with inequalities.

• Use the table below to review the meanings of the inequality symbols and provide examples with their solution sets.

Symbol Description Example Solution Set> greater than, more than x > 3 all numbers greater than 3;

does not include 3≥ greater than or equal to, at least x ≥ 3 all numbers greater than or

equal to 3; includes 3< less than x < 3 all numbers less than 3; does

not include 3≤ less than or equal to, no more than x ≤ 3 all numbers less than or equal

to 3; includes 3≠ not equal to x ≠ 3 includes all numbers except 3

• Solving a linear inequality is similar to solving a linear equation. The processes used to solve inequalities are the same processes that are used to solve equations.

• Multiplying or dividing both sides of an inequality by a negative number requires reversing the inequality symbol. Students may have difficulty understanding this concept. Draw a number line on the board to show the process.

• First, show examples of the inequality 2 < 4.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

2 < 4

• Multiply both sides by –2 and the inequality becomes 2(–2) < 4(–2) or –4 < –8.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

• Ask, “Is –4 really less than –8?”



Instruction

• To make the statement true, you must reverse the inequality symbol: –4 > –8

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

–4 > –8

Question 1

Is there more than one solution to this problem?

Instruction

• There are many solutions to this problem.

Any account balance that Mr. Ferris has that is $200 or more would be a solution to the problem.

Question 2

Which symbol can be used to represent “at least”?

Instruction

• At least means “greater than or equal to.” The symbol ≥ is used to represent greater than or equal to.

Question 3

What inequality can be used to represent the solution?

Instruction

• Describe the situation and represent the situation using what you know:

current balance

minus amount of check

plus deposit is at least

$200

250.15 – 75 + d ≥ 200





What is the solution set for the inequality?

Instruction

• Use the same methods to solve the inequality as you would use to solve an equation.

250.15 – 75 + d ≥ 200

• Combine like terms.

250.15 – 75 + d ≥ 200

175.15 + d > 200

• Isolate the variable by subtracting 175.15 from both sides of the inequality.

175.15 + d > 200

175.15 – 175.15 + d > 200 – 175.15

d > 24.85

Mr. Ferris would need to deposit at least $24.85 into his checking account.

Question 5

How can the solution set be represented graphically?

Instruction

• A number line is often used to represent inequalities.

• The inequality d ≥ 24.85 can be represented by the number line shown below.

0 5 10 15 20 25 30 35 40 45 50x

The point plotted on the number line would be 24.85 and all values greater than or equal to 24.85 would be in the solution set of the inequality d ≥ 24.85.



Instruction


Solve the inequality 2x + 4 ≤ 8.

Solution

• To isolate the variable, subtract 4 from both sides of the inequality.

2x + 4 – 4 ≤ 8 – 4

2x ≤ 4

• Divide both sides of the inequality by 2, as you’ve done with equations.

2x ≤ 4

22

42

x≤

x ≤ 2

The solution set of the inequality is all numbers less than or equal to 2.

• Check the results. Substitute 2, the boundary point, back into the inequality.

2x + 4 ≤ 8

2(2) + 4 ≤ 8

4 + 4 ≤ 8

8 ≤ 8

• Also choose a number that is a part of the solution set. In this case, choose a number that is less than 2, such as 0.

2(0) + 4 ≤ 8

0 + 4 ≤ 8

4 ≤ 8

Both statements are true.




Instruction

• Graph the solution set on a number line. Notice that the point’s circle is shaded. This means that 2 is included in the solution set.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

Example 2

Solve the inequality –3x + 9 < 18.

Solution


–3x + 9 – 9 < 18 – 9

–3x < 9

• Divide both sides of the inequality by –3, as you’ve done with equations.

• However, reverse the less than symbol (<) to get a true statement. When either multiplying or dividing by a negative number, reverse the symbol of the inequality to make the statement true.

–3x + 9 – 9 < 18 – 9

–3x < 9

−−

>−

33

93

x

x > –3

The solution set of the inequality is all numbers greater than –3.

• Check the results. Substitute –3, the boundary point, back into the original inequality.

–3x + 9 < 18

–3(–3) + 9 < 18

9 + 9 < 18

18 < 18

• Also choose a number that is a part of the solution set. In this case, choose a number that is greater than –3, such as 0.



Instruction

–3(0) + 9 < 18

0 + 9 < 18

9 < 18

Both statements are true.

• Graph the solution set on a number line. Notice that the point’s circle is NOT shaded. This means that –3 is NOT included in the solution set.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

Example 3

Solve the inequality 24

3− >x

.

Solution


24

3− >x

24

2 3 2− − > −x

−>

x4

1

• Multiply both sides of the inequality by –4, as you’ve done with equations.

• Reverse the greater than symbol (>) to get a true statement.

• When either multiplying or dividing by a negative number, reverse the symbol of the inequality to make the statement true.

(– ) (– )44

1 4−

<x

x < –4

The solution set of the inequality is all numbers less than –4.




Instruction

• Check the results. Substitute –4, the boundary point, back into the original inequality.

24

3− >x

244

3− >(– )

2 – (–1) > 3

2 + 1 > 3

3 > 3

• Since the statement 3 > 3 is not true, –4 is not included in the solution set.

• Also choose a number that is a part of the solution set. In this case, choose a number that is less than –4, such as –8.

24

3− >x

284

3−−

>( )

2 – (–2) > 3

2 + 2 > 3

4 > 3

• This statement is true.

• Graph the solution set on a number line. Notice that the point’s circle is NOT shaded. This means that –4 is NOT included in the solution set.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x


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Guided PracticeFor problems 1–7, solve each inequality. Show your work.

1. 3x – 3 < 6

2. 2x + 4 ≤ 0

3. − >x3

1

4. –2x – 3 ≤ 5

5. x – 5 > 1

6. x + 2 ≤ –3

7. x – 4 < –4

For problems 8–10, solve each inequality, and then graph the solution set on a number line. Show your work.

8. x2

1 3− >

9. –2x + 2 ≤ –4

10. 22

1− ≥x

continued


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For problems 11–16, write algebraic inequalities for the situations described, and then find the solution set. Show your work.

11. A tour bus can seat 55 passengers.

12. An energy-efficient lamp can only be used with light bulbs that are 60 watts or less.

13. Camilla is saving to purchase a new pair of bowling shoes that will cost at least $39. She has already saved $19. What is the least amount of money she needs to save for the shoes?

14. Suppose you earn $20 per hour working part time at a tax office. You want to earn at least $1,800 this month, before taxes. How many hours must you work?

15. Hiram earned a score of 83 on his semester algebra test. He needs to have a total of at least 180 points from his semester and final tests to receive an A for his grade. What score must Hiram earn on his final test to ensure his A?

16. Claire purchases DVDs from an online entertainment store. Each DVD that she orders costs $15 and shipping for her order is $10. Claire can spend no more than $100. How many DVDs can Claire purchase?


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Independent Practice For problems 1–7, solve each inequality. Show your work.

1. 2x – 5 < 9

2. x + 2 ≤ 4

3. − <x2

3

4. –3x – 3 ≥ 6

5. x – 2 > 2

6. x + 3 ≤ –1

7. x – 1 < –1

For problems 8–10, solve each inequality. Show your work. Graph the solution set on a number line.

8. x3

2 2− <

9. –3x + 4 ≤ –5

10. 32

2− ≥x

continued


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11. An amusement park ride can hold 8 passengers.

12. An auditorium can seat 250 people or fewer.

13. Jeff is saving to purchase a new basketball that will cost at least $88. He has already saved $32. At least how much more does he need to save for the basketball?

14. Suppose you earn $15 per hour working part time at a hardware store. This month, you want to earn at least $950. How many hours must you work?

15. Mackenzie earned a score of 79 on her semester biology test. She needs to have a total of at least 160 points from her semester and final tests to receive a B for her grade. What score must Mackenzie earn on her final test to ensure her B?

16. Arianna purchases computer games from an online store. Each game that she orders costs $22 and shipping for her order is $9. Arianna can spend no more than $75. How many computer games can Arianna purchase?


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Assessment

Progress AssessmentFor problems 1–3, solve each inequality, then graph the solution set on a number line. Show your work.

1. 3x – 2 > 10

2. –2x – 2 ≤ 4

3. x3

2<


4. Miguel has $700 in a savings account. He wants to have at least $200 in savings. Each week he withdraws $25 for expenses. For how many weeks can Miguel withdraw money from his account before he reaches his limit of $200 in savings?

5. A cab company charges a $2.50 flat rate in addition to $1.25 per mile. Antonio has no more than $15 to spend on his ride. How many miles can Antonio travel?

6. Abby earns $18 per hour working a part time job. She needs to save at least $200 to purchase a new DVD player. She has already saved $38. How many hours must Abby work?




Instruction

Resource List• AAAMath. “Solving an Inequality.”

www.aaamath.com/g816d_x1.htm

Choose from three modes of play: Countdown, Give Me Time, or 20 Questions. Each mode asks players to find solutions to inequalities. Time is kept and score is recorded.

• Math-Play.com. “Inequality Game.”

www.math-play.com/Inequality-Game.html

In this multiple-choice game, players choose the correct answer to solve the inequality. If an incorrect answer is chosen, hints are provided.



Lesson 3 Answer KeyLesson Pre-Assessment, p. 52 1. x < 2 4. x ≤ –4 2. x < –2 5. x ≤ 7 3. x ≥ 1

Guided Practice, p. 67 1. x < 3 2. x ≤ –2 3. x < –3 4. x ≥ –4 5. x > 6 6. x ≤ –5 7. x < 0 8. x > 8

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

9. x ≥ 3

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

10. x ≤ 2

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

11. x ≤ 55 12. x ≤ 60 13. 19 + x ≥ 39; x ≥ 20 14. 20x ≥ 1,800; x ≥ 90 15. 83 + x ≥ 180; x ≥ 97 16. 15x + 10 ≤ 100; x ≤ 6

Independent Practice, p. 69 1. x < 7 2. x ≤ 2 3. x > –6 4. x ≤ –3 5. x > 4 6. x ≤ –4 7. x < 0 8. x < 12

–20 –18 –16 –14 –12 –10 –8 –6 –4 –2 0 2 4 6 8 10 12 14 16 18 20x

9. x ≥ 3

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

10. x ≤ 2

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

11. x ≤ 8 12. x ≤ 250 13. 32 + x ≥ 88; x ≥ 56 14. 15x ≥ 950; x ≥ 66 1/3 15. 79 + x ≥ 160; x ≥ 81 16. 22x + 9 ≤ 75; x ≤ 3

Progress Assessment, p. 71 1. x > 4

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

2. x ≥ –3

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

3. x < 6

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

4. 700 – 25x ≥ 200; x ≤ 20 5. 2.50 + 1.25x ≤ 15; x ≤ 10 6. 18x + 38 ≥ 200; x ≥ 9

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 4: Solving Quadratic Equations by Factoring

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Assessment

Lesson Pre-AssessmentFor problems 1–5, solve each quadratic equation. Show your work.

1. x2 – 13x + 12 = 0

2. x2 + 10x + 24 = 0

3. 2x2 – 16x + 32 = 0

4. –x2 + 121 = 0

5. The length of a rectangular patio is 6 times its width. The area of the patio is 96 square meters. What are the dimensions of the patio?



Instruction

Lesson 4: Solving Quadratic Equations by Factoring

Essential Questions 1. How does the graph of a quadratic equation compare to that of a linear equation?

2. What are the connections among the solutions of quadratic equations, the zeros of their related functions, and the horizontal intercepts of the graph of the function?

WORDS TO KNOW

binomial an algebraic expression with two unlike terms which is the sum of two monomials

monomial an expression that contains only one term, such as 4x or 6bc

quadratic equation an equation of degree 2, with two solutions at most

zero factor property If ab = 0, then a = 0, b = 0, or both a = 0 and b = 0.


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Warm-Up Option 1Solve each linear equation.

1. 2x + 1 = 25

2. x5

1 2+ =

3. x – 8 = 16

4. –x – 3 = 4

5. –2x = 24


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Warm-Up Option 2Read the statement that follows. Then answer the questions.

Harold purchased five 12-foot-long boards of lumber. The total bill before taxes was $22.50. What was the price of each board?

1. What is the unknown value?


3. What was the price of each board?




Instruction

Warm-Up Option 1: DebriefWhen solving for a given variable, be sure that students reverse the order of operations by first eliminating addition and subtraction within the equation.

1. 2x + 1 = 25


2x + 1 – 1 = 25 – 1

2x = 24

• Then divide both sides of the equation by 2.

22

242

x=

x = 12

2. x5

1 2+ =


x

x5

1 1 2 1

51

+ − = −

=

• Then multiply both sides of the equation by 5.

x

x5

5 1 5

5

( ) ( )=

=

3. x – 8 = 16


x – 8 + 8 = 16 + 8

x = 24



Instruction

4. –x – 3 = 4


–x – 3 + 3 = 4 + 3

–x = 7

• Then multiply both sides of the equation by –1.

–x(–1) = 7(–1)

x = –7

5. –2x = 24

• Isolate the variable by dividing both sides of the equation by –2.

−−

=−

22

242

x

x = –12

Warm-Up Option 2: DebriefHarold purchased five 12-foot-long boards of lumber. The total bill before taxes was $22.50. What was the price of each board?

1. What is the unknown value?

The unknown value is the price of one of the boards.


• First discuss the situation.

• It is not necessary to know the length of each board or the type of board in order to find the solution to this problem.

• Students may find it helpful to cross out irrelevant information before attempting to answer the question.




Instruction

• Harold purchased 5 boards at an unknown price per board for a total cost of $22.50.

5x = 22.50

3. What was the price of each board?

5 22 50

522 505

4 5

x

x

x

=

=

=

.

.

.

The price of each board was $4.50.


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Focus ProblemThe width of a rectangular garden is 8 feet less than its length. The area of the garden is 20 square feet.

garden w

w + 8

1. What equation can be used to find the area of the garden?

2. What information do you know?

3. What equation can be used to find the length and width of the garden?

4. What are the length and width of the garden?




Instruction


A quadratic equation is an equation that can be written as ax2 + bx + c = 0, where a ≠ 0.

The solution of a quadratic equation is the value of x when you set the equation equal to zero using the zero factor property.

• One method of solving a quadratic equation is by factoring.

Example

Solve the quadratic equation x2 + 9x + 14 = 0 by factoring.

• First, write the quadratic equation in standard form: ax2 + bx + c = 0, where a ≠ 0.

• x2 + 9x + 14 = 0 is already in standard form, where a = 1, b = 9, and c = 14.

• Find factor pairs that equal c, which in this case is 14 from the standard form; that is, find two numbers that have a product of c.

• Create a factor table to display the factor pairs.

Factors of 14 Sum of Factors–1 • –14 –15–2 • –7 –91 • 14 152 • 7 9

14 • 1 15

• From the factor pairs found, find the pair with a sum of b, which in this case is 9. There may not be a pair with a sum of b. If the pair doesn’t exist, another method would need to be used to find the solution, such as graphing.

• The factor pair of 2 and 7 has a product of c and a sum of b.

• Make two binomials from the factor pairs found.

(x + 2)(x + 7)



Instruction

• Solve each binomial using the zero factor property; that is, solve each binomial for zero.

(x + 2)(x + 7) = 0

x + 2 = 0 x + 7 = 0

x = –2 x = –7

The two solutions to the equation x2 + 9x + 14 = 0 are x = –2 and x = –7.

Focus Problem statement:

The width of a rectangular garden is 8 feet less than its length. The area of the garden is 20 square feet.

garden w

w + 8

Question 1

What equation can be used to find the area of the garden?

Instruction

The area formula for a rectangle can be used: A = lw, where l is the length of the garden and w is the width of the garden.

Question 2

What information do you know?

Instruction

• The width of the garden is 8 feet less than the length.

width = w

length = w + 8

• The area of the garden is 20 square feet.

A = 20 square feet





What equation can be used to find the length and width of the garden?

Instruction

• Substitute the known information into the area formula.

A = lw

20 = (w + 8)(w)

20 = w 2 + 8w

Question 4

What are the width and length of the garden?

Instruction

Since the equation is a quadratic equation, one method of solving the equation is to use factoring.

• Write the quadratic equation in standard form: ax2 + bx + c = 0, where a ≠ 0.

20 = w 2 + 8w

20 – 20 = w2 + 8w – 20

0 = w2 + 8w – 20, where a = 1, b = 8, and c = –20

• The equation can also be written as w2 + 8w – 20 = 0, where a = 1, b = 8, and c = –20.

• It is often helpful to write the general form of the equation (ax2 + bx + c) on the board with the equation w2 + 8w – 20 = 0 written directly under it. Point out each piece of the equation including coefficients, variables, and exponents.

• Find factor pairs that equal c from the general form, two numbers that have a product of –20.



Instruction


Factors of –20 Sum of Factors–1 • 20 19–2 • 10 8–4 • 5 1

1 • –20 –192 • –10 –84 • –5 –1

• From the factor pairs found, find the pair with a sum of b, which is 8 in this quadratic equation. The only factor pairs with a sum of 8 are –2 and 10.


(x – 2)(x + 10)

• Solve each binomial using the zero factor property.

(x – 2)(x + 10) = 0

x – 2 = 0 x + 10 = 0

x = 2 x = –10

• Since the width of the garden can’t be a negative number, eliminate the solution x = –10.

If the width of the garden is 2, then the length, w + 8, is 10.


Solve the quadratic equation x2 + 2x + 1 = 0 by factoring.

Solution

• The equation is already in standard form, where a = 1, b = 2, and c = 1.

• Find factor pairs that equal c from the general form, two numbers that have a product of 1.




Instruction


Factors of 1 Sum of Factors1 • 1 2

–1 • –1 –2

• From the factor pairs found, find the pair with a sum of b, which is 2 in this quadratic equation. The only factor pairs with a sum of 2 are 1 and 1.


(x + 1)(x + 1)


(x + 1)(x + 1) = 0

x + 1 = 0 x + 1 = 0

x = –1 x = –1

The solution to the quadratic equation x2 + 2x + 1 = 0 is x = –1.

Example 2

Solve the quadratic equation x2 + 3x = 0 by factoring.

Solution

• The equation x2 + 3x = 0 written in the standard form is x2 + 3x + 0 = 0, where a = 1, b = 3, and c = 0.

• Find factor pairs that equal c from the standard form, two numbers that have a product of 0.

• Create a factor table to display the factor pairs. Because the product of 0 and any number is 0, find factor pairs whose sum is b, 3.

Factors of 0 Sum of Factors0 • 3 3

0 • –3 –3



Instruction

• From the factor pairs found, find the pair with a sum of b, which is 3 in this quadratic equation. The only factor pairs with a sum of 3 are 0 and 3.


(x + 0)(x + 3)


(x + 0)(x + 3) = 0

x + 0 = 0 x + 3 = 0

x = 0 x = –3

The two solutions to the quadratic equation x2 + 3x = 0 are x = 0 and x = –3.

Example 3

Solve the quadratic equation –x2 – 2x + 3 = 0 by factoring.

Solution

• The equation is already in standard form. However, when a < 0, multiply both sides of the equation by –1 to avoid a negative leading coefficient.

–1(–x2 – 2x + 3) = –1(0)

x2 + 2x – 3 = 0

• Be sure students multiply all three terms by –1, not just the leading coefficient.

• Find factor pairs that equal c from the standard form, two numbers that have a product of –3.


Factors of –3 Sum of Factors–1 • 3 21 • –3 –2




Instruction

• From the factor pairs found, find the pair with a sum of b, which is 2 in this quadratic equation. The only factor pairs with a sum of 2 are –1 and 3.


(x – 1)(x + 3)


(x – 1)(x + 3) = 0

x – 1 = 0 x + 3 = 0

x = 1 x = –3

The two solutions to the quadratic equation –x2 – 2x + 3 = 0 are x = 1 and x = –3.

Example 4

Solve the quadratic equation 2x2 – 5x – 12 = 0 by factoring.

Solution

• The equation is already in standard form. However, when a > 1, divide both sides of the equation by a to follow the method used in previous examples.

• In this equation, a = 2.

2 5 12 0

2 5 122

02

52

6 0

2

2

2

x x

x x

x x

− − =

− −=

− − =

• Find factor pairs that equal c from the standard form, two numbers that have a product of –6.



Instruction

• Create a factor table to display the factor pairs. Think about factor pairs with a product of –6

and a sum of −52

.

Factors of – 6 Sum of Factors1 • –6 –5 –1 • 6 5

–4 • 32

−52

• From the factor pairs found, find the pair with a sum of b, which is −52

in this quadratic

equation. The only factor pairs with a sum of −52

are –4 and 32

.


(x – 4)(x + 32

)


(x – 4)(x + 32

) = 0

x – 4 = 0 x + 32

= 0

x = 4 x = −32

So, there are two solutions to the quadratic equation 2x2 – 5x – 12 = 0:

x = 4 and x = −32


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Guided PracticeFor problems 1–10, solve each quadratic equation. Show your work.

1. x2 – x – 20 = 0

2. –x2+ 6x – 8 = 0

3. 4x2 + 4x – 24 = 0

4. x2 – 9 = 0

5. x2 – 9x – 36 = 0

6. x2 + 3x – 4 = 0

7. x2 – 4x – 5 = 0

8. x2 + 3x – 10 = 0

9. x2 – 25 = 0

10. 2x2 + 4x – 6 = 0

continued


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For problems 11–16, write a quadratic equation to represent each situation. Then solve your equation. Show your work.

11. The product of two positive consecutive integers is 56. What are the integers?

12. The length of a rectangle is 6 more than its width. The area of the rectangle is 91 square meters. What are the width and length of the rectangle?

13. A square garden measuring 12 meters by 12 meters is scheduled to have a pedestrian pathway x meters wide installed around it, increasing the total area to 324 square meters. What is the width of the pathway?

12 m

x

x

xx

14. A ball is dropped from a height of 36 feet. The quadratic equation d = –t2 + 36 provides the distance, d, of the ball, after t seconds. After how many seconds does the ball hit the ground? Hint: The ball hits the ground at d = 0.

15. Two positive consecutive odd integers have a product of 99. What are the integers? Hint: If x is an odd integer, the next consecutive odd integer is x + 2.

16. The length of a rectangle is 2 times its width. The area of the rectangle is 72 square inches. What are the dimensions of the rectangle?


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Independent Practice For problems 1–10, solve each quadratic equation. Show your work.

1. x2 – 3x – 10 = 0

2. –x2 – 6x – 8 = 0

3. 6x2 + 6x – 36 = 0

4. x2 – 49 = 0

5. x2 – 12x + 36 = 0

6. x2 – 3x – 4 = 0

7. x2 + 4x – 5 = 0

8. x2 + 7x + 10 = 0

9. x2 – 1 = 0

10. 3x2 + 15x + 12 = 0

continued


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For problems 11–16, write a quadratic equation to represent each situation. Then solve your equation. Show your work.

11. The product of two consecutive positive integers is 90. What are the integers? Hint: If x is an integer, the next consecutive integer is x + 1.

12. The length of a rectangle is 2 more than its width. The area of the rectangle is 24 square feet. What are the width and length of the rectangle?

13. A square castle at the Children’s Museum measuring 18 feet by 18 feet is going to have a moat x meters wide installed all around it, increasing the total area to 576 square feet. What is the width of the moat?

18 ft

x

x

xx

14. A mirror has a height that is 12

its width. A new frame is being put around it that will increase

its area to 1,250 square inches. What will be the new dimensions of the mirror, after the frame is

added?

15. Two positive consecutive odd integers have a product of 63. What are the integers? Hint: If x is an odd integer, the next consecutive odd integer is x + 2.

16. The length of a mural on the side of a building is 3 times its width. The area of the mural is 75 square inches. What are the dimensions of the mural?


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Assessment

Progress AssessmentSolve each quadratic equation. Show your work.

1. x2 – 6x – 27 = 0

2. x2 + 12x + 36 = 0

3. 3x2 + 9x – 12 = 0

4. x2 – 144 = 0

5. –x2 – 25x – 24 = 0

6. The length of a rectangular herb garden is 5 times its width. If the area of the garden is 500 square feet, what are its dimensions?



Instruction

Resource List• Quia. “Rags to Riches.”

www.quia.com/rr/36611.html

Players practice factoring quadratics in this “Who Wants to Be a Millionaire?”-style game. Up to three hints are provided for the duration of the game.

• Purple Math. “Solving Quadratic Equations: Solving by Factoring.”

www.purplemath.com/modules/solvquad.htm

This Web site provides additional overview and examples of solving quadratic equations by factoring.




Lesson 4 Answer KeyLesson Pre-Assessment, p. 74 1. x = 12 and x = 1 2. x = –6 and x = –4 3. x = 4 4. x = 11 and x = –11 5. width = 4 meters; length = 24 meters

Guided Practice, p. 90 1. x = –4 and x = 5 9. x = –5 and x = 5 2. x = 4 and x = 2 10. x = –3 and x = 1 3. x = –3 and x = 2 11. 8 and 7 4. x = –3 and x = 3 12. w = 7 units; l = 13 units 5. x = –3 and x = 12 13. 3 meters 6. x = –4 and x = 1 14. 6 seconds 7. x = –1 and x = 5 15. 9 and 11 8. x = –5 and x = 2 16. w = 6 in.; l = 12 in.

Independent Practice, p. 92 1. x = –2 and x = 5 9. x = –1 and x = 1 2. x = –4 and x = –2 10. x = –1 and x = –4 3. x = –3 and x = 2 11. 9 and 10 4. x = –7 and x = 7 12. w = 4 ft; l = 6 ft 5. x = 6 13. 3 feet 6. x = –1 and x = 4 14. width = 50 in.; height = 25 in. 7. x = –5 and x = –2 15. 7 and 9 8. x = –5 and x = 2 16. w = 5 in.; l = 15 in.

Progress Assessment, p. 94 1. x = –3 and x = 9 4. x = –12 and x = 12 2. x = –6 5. x = –24 and x = –1 3. x = –4 and x = 1 6. width = 10 ft; length = 50 ft

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 5: Graphing Linear Equations

namE:


Assessment

Lesson Pre-AssessmentFor problems 1–3, graph each linear equation.

1. y = x

x

y

10

8

6

4

2

0

–2

–4

–6

–8

–10

–10 –8 –6 –4 –2 2 4 6 8 10

2. y = 2x

x

y

10

8

6

4

2

0

–2

–4

–6

–8

–10

–10 –8 –6 –4 –2 2 4 6 8 10

continued


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Assessment

3. y = –2x

x

y

10

8

6

4

2

0

–2

–4

–6

–8

–10

–10 –8 –6 –4 –2 2 4 6 8 10

4. James receives a base weekly salary of $200 plus a commission of $17 for each appliance he sells. His weekly pay can be represented by the equation y = 17x + 200. Identify the slope and y-intercept. Then use the slope and y-intercept to graph the equation for x between 0 and 20 appliances.

x

y600

570

540

510480

450420

390360

330300

270240

210180

150120

9060

300

2 4 6 8 10 12 14 16 18 20

Appliances

Sala

ry

slope = _______________

y-intercept = _______________

continued


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Assessment

5. A package shipping company charges $10 plus $16 per pound to ship a box weighing x pounds. Write an equation to find the total shipping charges. Identify the slope and y-intercept, and use them to graph the equation for boxes weighing between 1 and 10 pounds.

x

y200

190

180

170

160

150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

01 2 3 4 5 6 7 8 9 10

Weight (pounds)

Ship

ping

cha

rge

slope = _______________

y-intercept = _______________

Accuplacer College-Ready Mathematics: Elementary Algebra100



Instruction

Lesson 5: Graphing Linear EquationsEssential Questions 1. What does the slope of a linear equation represent?

2. How are the x- and y-intercepts of a linear equation useful when graphing?

3. What do the points on the graph of a linear equation represent?

WORDS TO KNOW

slope of a line the ratio of vertical change to horizontal change of a line

x-intercept the point where the graph of a line intercepts, or crosses, the x-axis

y-intercept the point where the graph of a line intercepts, or crosses, the y-axis


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Warm-Up Option 1Solve each equation for y.

1. 3y + 6x = –36

2. –2y – 4x = –20

3. x – 8 + y = 16

4. x – 20 – 5y = 30

5. –2x + y = 24


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Warm-Up Option 2A cell phone company charges a $20 flat fee plus $0.05 for every minute used for calls.

1. Make a table of values from 0 to 60 minutes in 10-minute intervals that represent the total amount charged.

2. Write an algebraic equation that could be used to represent the situation.

3. What do the unknown values in your equation represent?



Instruction

Warm-Up Option 1: DebriefWhen solving for a given variable, be sure that students remember to reverse the order of operations by first eliminating addition and subtraction within the equation.

1. 3y + 6x = –36

• Isolate the y variable by first subtracting 6x from both sides of the equation.

3y = –36 – 6x

• Divide both sides of the equation by 3. Be sure students divide all terms by 3.

3

3363

63

y x=−

−

y = –12 – 2x

• Set up the equation in the form y = mx + b.

y = –2x – 12

2. –2y – 4x = –20

• Isolate the y variable by first adding 4x to both sides of the equation.

–2y = –20 + 4x

• Divide both sides of the equation by –2. Be sure students divide all terms by –2.

y = –2x + 10

3. x – 8 + y = 16

• Isolate the y variable by first subtracting x from both sides of the equation.

–8 + y = 16 – x

• Then add 8 to both sides of the equation.

y = –x + 24




Instruction

4. x – 20 – 5y = 30

• Begin to isolate the y variable by first subtracting x from both sides of the equation.

–20 – 5y = 30 – x• Add 20 to both sides of the equation and then divide all terms on both sides by –5.

y x= −15

10

5. –2x + y = 24

• Isolate the y variable by adding 2x to both sides of the equation.

y = 2x + 24

Warm-Up Option 2: DebriefA cell phone company charges a $20 flat fee plus $0.05 for every minute used for calls.

1. Make a table of values from 0 to 60 minutes in 10-minute intervals that represent the total amount charged.

Minutes Used Total Amount Charged ($)0 20 + 0(0.05) = 20.00

10 20 + 10(0.05) = 20.5020 20 + 20(0.05) = 21.0030 20 + 30(0.05) = 21.5040 20 + 40(0.05) = 22.0050 20 + 50(0.05) = 22.5060 20 + 60(0.05) = 23.00

2. Write an algebraic equation that could be used to represent the situation.

y = 0.05x + 20

3. What do the unknown values in your equation represent?

x represents the number of minutes used, and y represents the total amount charged.


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Focus ProblemYou can buy a 6-hour phone card for $5, but the fine print states that each minute you use actually costs you 1.5 minutes.

1. What equation can be used to represent the number of remaining available minutes?

2. Find the x-intercept of the line of the equation.

3. Find the y-intercept of the line of the equation.

4. What is the slope of the line?

5. Use the x- and y-intercepts to graph the line on the coordinate grid.

x

y500

450

400

350

300

250

200

150

100

50

025 50 75 100 125 150 175 200 225 250

Minutes used

Rem

aini

ng m

inut

es




Instruction


Many relationships can be represented by linear equations. The slope of a linear graph is a measure of the rate of change of one variable with respect to another variable.

• The slope of a linear equation is also defined by the ratio of the rise of the graph compared to the run. Given two points on a line, (x

1, y

1) and (x

2, y

2), the slope is the ratio of the change in the

y-values of the points (rise) to the change in the corresponding x-values of the points (run).

sloperiserun

= =−−

y yx x2 1

2 1

• The slope-intercept form of an equation of a line is often used to easily identify the slope and y-intercept, which then can be used to graph the line. The slope-intercept form of an equation is shown below, where m represents the slope of the line and b represents the y-value of the point where the line intersects the y-axis at point (0, y).

y = mx + b

• Horizontal lines have a slope of 0. They have a run but no rise. Vertical lines have no slope.

• The x-intercept of a line is the point where the line intersects the x-axis at (x, 0).

• If a point lies on a line, its coordinates make the equation true.

• The graph of a line is the collection of all points that satisfy the equation. The graph of the linear equation y = –2x + 2 is shown, with its x- and y-intercepts plotted.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5



Instruction

• Lines can be graphed by plotting points or by using the y-intercept and the slope using the point-slope form of a linear equation.

y – y1 = m(x – x

1)

Question 1

What equation can be used to represent the number of remaining available minutes?

Instruction

• The equation y = –1.5 x + 360 can be used to represent the situation written in slope-intercept form.

• A common error for students would be to write y = –1.5x + 6, where the constant is 6 hours rather than 360 minutes, the number of minutes in 6 hours (60 minutes • 6).

Question 2

Find the x-intercept of the line of the equation.

Instruction

• The x-intercept represents the point on the graph where no access time remains.

• To find the x-intercept, or where the line crosses the x-axis, substitute 0 for y to find the x-value of the x-intercept.

0 = –1.5x + 360

• Solve for x.

1.5x = 360

x = 240 when y = 0

• It is common for students to reverse the order of the coordinates. Remind students that coordinates are always written as (x, y).

The x-intercept is the point (240, 0), the point where the line intercepts the x-axis.





Find the y-intercept of the line of the equation.

Instruction

• The y-intercept represents the point on the graph where all 6 hours of access time is available.

• Since the equation is written in slope-intercept form, the y-intercept is (0, 360). The constant term in the equation is the y-value of the y-intercept.

• Students could also find the value of the y-intercept in a similar way as finding the x-intercept.

• Substitute 0 for x to find the value of the y-intercept.

y = –1.5(0) + 360

• Solve for y.

y = 360 when x = 0

The y-intercept is (0, 360), the point where the line intercepts the y-axis.

Question 4

What is the slope of the line?

Instruction

• Since the equation is written in slope-intercept form, the slope of the line is the coefficient of x. So, the slope of the line is –1.5. A line with a negative slope goes downward from left to right.

• Students could also calculate the slope using the slope formula: y yx x2 1

2 1

−−

• Students have already found two points on the line (240, 0) and (0, 360).



Instruction

• Substitute the coordinates into the slope formula.

(x1, y

1) = (240, 0) and (x

2, y

2) = (0, 360)

slope =−−

=−

−

=−

= −

y yx x2 1

2 1

360 00 24036024032

• Remind students that it does not matter which coordinate is labeled (x1, y

1) and which is

labeled (x2, y

2). The results will be the same.

• It is not uncommon for students to need to see this done both ways to believe it is true.

(x1, y

1) = (0, 360) and (x

2, y

2) = (240, 0)

slope =−−

=−

−

=−

= −

y yx x2 1

2 1

0 360240 036024032

−32

is equal to –1.5, the slope value previously identified in the equation.





Use the x- and y-intercepts to graph the line on the coordinate grid.

Instruction

• First, plot the y–intercept (0, 360).

• Next, plot the x–intercept (240, 0).

• Connect each of these points using a straightedge.

• Each point on the line is a solution to the equation y = –1.5x + 360.

• Remind students that using fractions rather than decimals for slope will make it easier to visualize the “rise over run” of the equation.

• To visualize the slope of the line, draw the slope triangle:

• A slope of −32

has a rise of –3 and a run of 2.

• From the y-intercept, (0, 360), go down 3 units and over 2 units.

• Remind students that a negative slope indicates that there is a negative rise.

• The point (2, 357) has been found.

• From (2, 357), go down 3 more units and over 2 more units to the point (4, 354) as shown in the graph below. This process would continue to the x-intercept of (240, 0).

360

359

358

357

356

355

354

353

352

351

350x

y

0 1 2 3 4



Instruction

• The final graph is shown below.

x

y500

450

400

350

300

250

200

150

100

50

025 50 75 100 125 150 175 200 225 250

Rem

aini

ng m

inut

es

Minutes used




Instruction


Graph the line y x= +14

6 .

Solution

• Find the slope of the line.

• Since the equation is written in slope-intercept form (y = mx + b), the slope of the line is 14

.

• Find the y-intercept. Let x = 0.

y x= +14

6

y = +14

0 6( )

y = 6 when x = 0

• Graph the equation using the y-intercept (0, 6) and the slope of 14

.

• First, plot (0, 6).

• The slope of 14

is a positive number indicating a rise of 1 and a run of 4.

• From this point, rise 1 unit and run 4 units to the right.

• Make a dot at the point (4, 7).

• Connect the points (0, 6) and (4, 7).

• Be sure students include arrows.

x

y

1 2 3 4 5 6 7 8 9 10

10

9

8

7

6

5

4

3

2

1

0




Graph the line y x=23

.

Solution

• Find the slope of the line.

• Since the equation is written in slope-intercept form (y = mx + b), the slope of the line is 23

.

• Find the y-intercept. Let x = 0.

y x=23

y =23

0( )

y = 0 when x = 0

• Graph the equation using the y-intercept, (0, 0) and the slope of 23

.

• First, plot (0, 0).

• The slope of 23

is a positive number indicating a rise of 2 and a run of 3.

• From this point, rise 2 units and run 3 units to the right.

• Make a dot at the point (3, 2).

• Connect the points (0, 0) and (3, 2).

• Be sure students include arrows.

x

y

1 2 3 4 5 6 7 8 9 10

10

9

8

7

6

5

4

3

2

1

0





Two points on a line are (0, 1) and (1, 3). Find the slope of the line and the equation of the line. Graph the line.

Solution

• Use the given points to find the slope.

slope =−−

=−−

= =

y yx x2 1

2 1

3 11 021

2

• Again, remind students that the naming of each point is irrelevant. The resulting slope will be the same.

• Use the point-slope form of a linear equation to find the slope-intercept equation using the slope and the point (0, 1).

y – y1 = m(x – x

1)

y – 1 = 2(x – 0)

y = 2x + 1

• It is also possible to use the point (1, 3) in the point-slope form.

y – y1 = m(x – x

1)

y – 3 = 2(x – 1)

y = 2x + 1

• The equation is now in slope-intercept form, for which the slope is 2 and the y–intercept is 1.



Instruction

• Graph the equation.

• Begin by plotting the y–intercept of (0, 1).

• The slope of the line is 2 or the fraction equivalent 21

.

• From (0, 1), rise 2 units and run 1 unit.

• Remind students to think of slope as a fraction. A common error is for students to read 2 as a rise of 2 and a run of 0, resulting in a vertical line.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

Example 4

Find the x- and y-intercepts of the line –2y = 6x – 6. Use the intercepts to graph the equation.

Solution

• Write the equation of the line in slope-intercept form.

–2y = 6x – 6

y = –3x + 3

• Find the x- and y-intercepts of the line y = –3x + 3.

• Since the equation is written in slope-intercept form, the y-intercept is (0, 3).




Instruction

• To find the x-intercept, substitute 0 for y into the equation and solve to find the value for x.

y = –3x + 3

0 = –3x + 3

x = 1

• The x-intercept is (1, 0).

• Plot the x- and y-intercepts. Draw a line extending through the two points.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5


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Guided PracticeFor problems 1–4, graph the lines using the x- and y-intercepts. Show your work.

1. y = x + 2 3. y = 13

x + 2

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

2. y = x – 1 4. y = –2x + 2

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

continued


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For problems 5–8, use the given two points on a line to find the slope and the equation of the line. Graph the line.

5. (0, 0) and (1, –1) 7. (0, 1) and (–1, –3)

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

6. (0, 0) and (2, 1) 8. (0, 5) and (1, 2)

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

continued


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For problems 9–14, write an equation in slope-intercept form that represents the situation described. Graph the line using the y-intercept and slope. Show your work.

9. A gear on a machine turns at a rate of 2 revolutions per second. Let x = time in seconds and y = number of revolutions.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

10. The formula F = 95

C + 32 represents the relationship between Celsius temperature and

Fahrenheit temperature.

x

y

50

45

40

35

30

25

20

15

10

5

0

–5

–10

–15

–20

–25

–30

–35

–40

–45

–50

–50 –45 –40 –35 –30 –25 –20 –15 –10 –5 5 10 15 20 25 30 35 40 45 50

continued


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11. A cab company charges an initial rate of $2.50 and $0.40 for each additional minute. Find total fares for 0 to 20 minutes.

x

y10

9

8

7

6

5

4

3

2

1

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

12. Matthew receives a base weekly salary of $300 plus a commission of $50 for each person he recruits. Find Matthew’s weekly salaries potential for 0 to 10 people recruited.

x

y1000

900

800

700

600

500

400

300

200

100

01 2 3 4 5 6 7 8 9 10

continued


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13. A water company charges a monthly fee of $6.70 plus a usage fee of $2.60 per 1,000 gallons used. Find total monthly charges for 0 to 15,000 gallons of water used.

x

y75

70

65

60

55

50

45

40

35

3025201510

50

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

14. Maddie borrowed $1,250 from a friend to buy a new TV. Her friend doesn’t charge any interest, and Maddie makes $40 payments each month. Show the amounts owed from $1,250 to $0.

x

y150014001300

120011001000

900

800700600

500

400300200

100

03 6 9 12 15 18 21 24 27 30 33 36 39 42 45


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Independent Practice For problems 1–4, graph the lines using the x- and y-intercepts. Show your work.

1. y = –x – 2 3. y = –x + 2

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

2. y = 12

x + 4 4. y = –2x – 3

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

continued


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For problems 5–10, use the given two points on a line to find the slope and the equation of the line. Graph the line and show your work.

5. (0, 0) and (1, –2) 7. (0, 2) and (2, –4)

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

6. (0, 0) and (4, 1) 8. (0, 4) and (1, 0)

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

continued


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For problems 9–14, write an equation in slope-intercept form that represents the situation described. Graph the line using the y-intercept and slope. Show your work.

9. A gear on a machine turns at a rate of 12

revolution per second. Let x = time in seconds and y = number of revolutions.

x

y

10

8

6

4

2

0

–2

–4

–6

–8

–10

–10 –8 –6 –4 –2 2 4 6 8 10

continued


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10. The formula C = 59

(F – 32) represents the relationship between Fahrenheit temperature and

Celsius temperature.

x

y

–50 –40 –30 –20 –10 0 10 20 30 40 50

50

40

30

20

10

–10

–20

–30

–40

–50

11. A limousine company charges an initial rate of $50.00 and $75.00 for each hour. Find the total fares for 0 to 5 hours.

x

y500

450

400

350

300

250

200

150

100

50

01 2 3 4 5

continued


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12. Angela receives a base weekly salary of $100 plus a commission of $65 for each computer she installs. Find Angela’s weekly salary potential for 0 to 10 computer installations.

x

y1000

950900850800750700650600550500450400350300250200150100

500

2 4 6 8 10 12 14 16 18 20

13. A cable company charges a monthly fee of $59.00 plus $8 for each movie-on-demand rented. Find total monthly charges that include renting from 1 to 10 on-demand movies.

x

y150140130120110100

90

807060

50

403020

10

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15

continued


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14. Garrett borrowed $500 from his aunt. She doesn’t charge any interest, and he makes $15 payments each month. Show the amounts owed from $500 to $0.

x

y500

450

400

350

300

250

200

150

100

50

05 10 15 20 25 30 35 40 45 50


namE:



Assessment

Progress AssessmentFor problems 1 and 2, graph the lines using the x- and y-intercepts. Show your work.

1. y = 3x + 4

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

2. y = 34

x + 1

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

continued


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Assessment

For problems 3–5, use the given two points on a line to find the slope and the equation of the line. Graph the line and show your work.

3. (0, –3) and (1, –4)

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

4. (0, 2) and (5, 6)

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

continued


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Assessment

5. (0, 0) and (2, –4)

x

y

10

8

6

4

2

0

–2

–4

–6

–8

–10

–10 –8 –6 –4 –2 2 4 6 8 10

6. A company rents personal watercraft for $70 per hour plus an initial $15 fee. What are the total costs to rent a watercraft for 0 to 8 hours? Write an equation to represent the situation and graph it.

x

y750

675

600

525

450

375

300

225

150

75

01 2 3 4 5 6 7 8 9 10



Instruction

Resource List• Math-Play.com. “Hoop Shoot.”

www.math-play.com/slope-intercept-game.html

This one- or two-player game includes 10 multiple-choice questions about slope and y-intercept. Correct answers result in a chance to make a 3-pointer in a game of basketball.

• Oswego City School District Regents Exam Prep Center. “Graphing Equations of Straight Lines.”

www.regentsprep.org/regents/math/algebra/AC1/EqLines2.htm

This Web site contains a thorough summary of the methods used to graph linear equations.




Lesson 5 Answer KeyLesson Pre-Assessment, p. 97 1.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

2.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

3.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

4. slope = 17; y-intercept: (0, 200)

x

y600570540510480450420390360330300270240210180150120

906030

02 4 6 8 10 12 14 16 18 20

Appliances

Sala

ry


5. y = 16x + 10; slope = 16; y-intercept = (0, 10)

x

y200190180170160150140130120110100

908070605040302010

01 2 3 4 5 6 7 8 9 10

Weight (pounds)

Ship

ping

cha

rge

Guided Practice, p. 117 1.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

2.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

3.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

4.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

5. slope = –1; y = –x

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

133



6. slope = 1/2; y = 1/2x

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

7. slope = 4; y = 4x + 1

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

8. slope = –3; y = –3x + 5

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

9. y = 2x. The two plotted points (0, 0) and (1, 2) would be in a table of values and shown on the graph.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

10. y = 9/5x + 32; points could include (0, 32) and (5, 41)

x

y

50

45

40

35

30

25

20

15

10

5

0

–5

–10

–15

–20

–25

–30

–35

–40

–45

–50

–45 –40 –35 –30 –25 –20 –15 –10 –5 0 5 10 15 20 25 30 35 40 45 50


11. y = 0.4x + 2.5; points could include (0, 2.5) and (5, 4.5)

x

y10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

12. y = 50x + 300; points could include (0, 300) and (1, 350)

x

y1000

900

800

700

600

500

400

300

200

100

0 1 2 3 4 5 6 7 8 9 10

13. y = 2.6x + 6.7; points could include (0, 6.7) and (10, 32.7)

x

y7570656055504540353025201510

5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

14. y = –40x + 1,250; points could include (0, 1,250) and (1, 1,210)

x

y150014001300120011001000

900800700600500400300200100

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45



Independent Practice, p. 122 1.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

2.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

3.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

4.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

5. y = –2x; slope = –2

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

6. y = 1/4x; slope = 1/4

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5


7. y = –3x + 2; slope = –3

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

8. y = –4x + 4; slope = –4

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

9. y = 1/2x; slope = 1/2; y-intercept: (0, 0)

x

y

10

8

6

4

2

0

–2

–4

–6

–8

–10

–10 –8 –6 –4 –2 2 4 6 8 10

10. y = 5/9(x – 32); slope = 5/9; y-intercept: (0, –17 7/9)

x

y

–50 –40 –30 –20 –10 0 10 20 30 40 50

50

40

30

20

10

–10

–20

–30

–40

–50



11. y = 75x + 50; slope = 75; y-intercept: (0, 50)

x

y500

450

400

350

300

250

200

150

100

50

01 2 3 4 5


x

y1000

950900850800750700650600550500450400350300250200150100

500

2 4 6 8 10 12 14 16 18 20


x

y150140130120110100

908070605040302010

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15

14. y = –15x + 500; slope = –15; y-intercept: (0, 500)

x

y500

450

400

350

300

250

200

150

100

50

05 10 15 20 25 30 35 40 45 50


Progress Assessment, p. 128 1.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

2.

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

3. slope = –1; y = –x – 3

x

y

–5 –4 –3 –2 –1 0 1 2 3 4 5

5

4

3

2

1

–1

–2

–3

–4

–5

4. slope = 4/5; y = 4/5x + 2

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

5. slope = –2; y = –2x

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10



6. y = 70x + 15

x

y750

675

600

525

450

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75

01 2 3 4 5 6 7 8 9 10

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 6: Solving Word Problems Involving Angle Measures

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Assessment

Lesson Pre-AssessmentFor problems 1–5, write an equation and solve each problem. Show your work.

1. Two angles are supplementary. One has a measure of 140°. What is the measure of the other angle?

2. Two angles are complementary. One has a measure of 45°. What is the measure of the other angle?

3. A triangle has one angle that measures 50° and another angle that measures 17°. What is the measure of the third angle?

4. Two angles are complementary. One angle is 30° more than the other angle. What are the measures of the two angles?

5. A telephone pole is attached to a wire support that forms a 45° angle with the pole. What is the measure of the other supplementary angle?

x

45°




Instruction

Lesson 6: Solving Word Problems Involving Angle Measures

Essential Questions 1. How are linear equations used to find angle measures?

2. How are complementary and supplementary angles related to linear equations?

WORDS TO KNOW

complementary angles two angles whose sum is 90°

supplementary angles two angles whose sum is 180°


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Warm-Up Option 1Given each equation below, solve for y. Show your work.

1. Solve y + x = 180. Let x = 90.

2. Solve y + x = 180. Let x = 30.

3. Solve y + x = 90. Let x = 45.

4. Solve y + x = 90. Let x = 20.

5. Solve y + 2x = 180. Let x = 30.


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Warm-Up Option 2After her arm surgery, Heather was put into a cast that held her arm at a 90° angle. When her cast was taken off, she was able to extend her arm by an additional 10° each week. How many weeks will it take for Heather to extend her arm to 180°?


2. What do you need to know?


4. How many weeks will it take for Heather to extend her arm to 180°?

5. Check the results by making a table of values.



Instruction

Warm-Up Option 1: DebriefWhen solving for a given variable, be sure that students remember to reverse the order of operations by first eliminating addition and subtraction within the equation.

1. Solve y + x = 180. Let x = 90.

y + (90) = 180

y = 90

2. Solve y + x = 180. Let x = 30.

y + (30) = 180

y = 150

3. Solve y + x = 90. Let x = 45.

y + (45) = 90

y = 45

4. Solve y + x = 90. Let x = 20.

y + (20) = 90

y = 70

5. Solve y + 2x = 180. Let x = 30.

y + 2(30) = 180

y = 120




Instruction



• Heather’s arm is at a 90° angle initially.

• She extends her arm another 10° each week.

• When her arm is fully extended it will be a straight line.

• A straight line measures 180°.


• The number of weeks it will take for Heather to be able to fully extend her arm


• Let w represent the number of weeks.

Starting position of Heather’s arm

PlusDegrees increased

each weekEquals

Full extension

90 + 10w = 180

4. How many weeks will it take for Heather to extend her arm to 180°?

• Solve for w.

• When solving for a given variable, reverse the order of operations by first eliminating addition and subtraction within the equation.

180 = 10w + 90

10w = 90

w = 9

It will take 9 weeks for Heather to fully extend her arm.



Instruction

5. Check the results by making a table of values.

Week ° of Arm Extension0 90°1 90° + 10° = 100°2 100° + 10° = 110°3 110° + 10° = 120°4 120° + 10° = 130°5 130° + 10° = 140°6 140° + 10° = 150°7 150° + 10° = 160°8 160° + 10° = 170°9 170° + 10° = 180°


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Focus ProblemThe lie of a hockey stick is the angle between the shaft and the blade. Anthony’s hockey stick has a lie of 137°, as shown below.

shaft

bladelie

x137˚

1. Write an algebraic equation to determine the value for the angle that is supplementary to the lie.

2. Use your equation to find the angle that is supplementary to the lie.

3. Suppose a different hockey stick has a lie 65° more than its supplement. Write an algebraic equation to find the value of the lie.

4. Use your equation to find the angle that is 65° more than its supplement.



Instruction


Many relationships can be represented by linear equations, including problems finding the sum of the measures of a triangle, complementary angles, and supplementary angles.

• The sum of the measures of a triangle is always 180°.

• Two angles are complementary angles when their sum measures 90°. The two angles do not have to share a vertex. If one angle measures 30°, then the measure of its complement is 60°.

• Two angles are supplementary angles when their sum measures 180°. The two angles do not have to share a vertex. If one angle measures 80°, then the measure of its supplement is 100°.

• Students often confuse the words supplement and complement. Use this mnemonic device to help students remember: when written alphabetically, complement comes before supplement; similarly, when 90° and 180° appear on a number line, 90° comes before 180°. So, complementary angles add up to 90° and supplementary angles add up to 180°.

• Word problems involving angle measures can be translated into linear equations and solved using problem-solving strategies. Some problem-solving strategies include:

• writing an equation

• drawing a picture, diagram, or table of values

• looking for a pattern

• systematically guessing and checking

• making a list

• acting it out

• working a simpler problem

• working backward

• using logical reasoning





Write an algebraic equation to determine the value for the angle that is supplementary to the lie.

Instruction

• One of the first steps in problem solving is to determine what you know by organizing the information.

• In this problem, you know that the lie of the hockey stick is 137°.

• You also know that the two angles, 137° and x, form supplementary angles because their sum is 180° (a straight line, as shown in the diagram).

• Translate what you know and what you need to know into an algebraic equation.

• You know that 180° is equal to the sum of the two supplementary angles.

180 = 137 + x

Question 2

Use your equation to find the angle that is supplementary to the lie.

Instruction

• To find the angle that is supplementary to the lie of 137°, solve for x.

• Isolate the variable x by subtracting 137 from both sides of the equation.

x = 43°

The angle that is supplementary to 137° is 43°.




Suppose a different hockey stick has a lie 65° more than its supplement. Write an algebraic equation to find the value of the lie.

Instruction

• Determine what you know.

• One of the angles measures 65° more than its supplement.

• The degree measure of the supplement can be represented by 180 – x.

• Determine what you need to know.

• You need to know the value of x to find one of the supplementary angles.

• Translate what you know into an algebraic equation.

measure of one of the supplements = x

measure of the other supplements = x + 65

x + x + 65 = 180

Question 4

Use your equation to find the angle that is 65° more than its supplement.

Instruction

• Solve the equation for x by first simplifying x + x.

x + x + 65 = 180

2x + 65 = 180

2x = 115

x = 57.5°

• The lie is 65° more than x.

65° + 57.5° = 122.5°

The lie measures 122.5°.




Instruction


In a pair of complementary angles, the measure of one angle is 15° greater than the measure of the other. What are the measures of the complementary angles?

Solution


• One of the angles measures 15° more than its complement.

• The degree measure of the complement can be represented by 90 – x.


• You need to know the value of x to find one of the complementary angles.


• The symbol used to indicate angle is ∠ ; m∠ refers to the measure of the angle.

m x

m x

m m

x x

∠ =∠ = +

∠ + ∠ =

+ + =

1

2 15

1 2 90

15 90

• Solve for x.

x = 37.5°

• Check the results.

m∠ =1 37 5.

m∠ =2 2 is 15° more than m∠1 , or 37.5° + 15° = 52.5°.

37.5° + 52.5° = 90°

The measures of the complementary angles are 37.5° and 52.5°.




A triangle has angle measures of 80°, 40°, and x. Find the measure of unknown angle.

Solution


• You know the value of two angles: 80° and 40°.

• You know that the sum of the angles of a triangle is 180°.


• You need to know the measure of the third angle.


• Let x represent the value of the third angle.

180° = x + 80° + 40°

180 – 120 = x + 120 – 120

x = 60°


180° = 80° + 40° + 60° = 180°

Example 3

Twice the measure of one supplementary angle is equal to 140°. What are the measures of each of the supplementary angles?

Solution


• You know the value of one supplement: 2x = 140°, so x = 70°.

• You know the sum of the two supplements is 180°.




Instruction


• You need to know the measure of the other supplementary angle.

• Translate what you know into an algebraic equation. Let x represent the value of the unknown supplement.

180 = x + 70

180 – 70 = x + 70 – 70

x = 110°


180° = 70° + 110°

The measures of the supplementary angles are 70° and 110°.


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Guided PracticeFor problems 1–4, find the complement to each angle by writing an equation and solving it. Show your work.

1. 76° 3. 7°

2. 42° 4. 15°

For problems 5–8, find the supplement to each angle by writing an equation and solving it. Show your work.

5. 45° 7. 4°

6. 90° 8. 110°

For problems 9 and 10, two angle measures of a triangle are provided. Find the missing angle of the triangle by writing an equation and solving it. Show your work.

9. 90° and 15° 10. 45° and 30°

continued


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For problems 11–16, write an equation and then solve it. Show your work.

11. The supplement of an angle measures 30° more than twice its complement. Find the measures of the angles.

12. One of two complementary angles measures 60° more than the other measures. Find the measure of each angle.

13. Two angles are supplementary. The angles have equal measure. What are their measures?

14. Two angles are complementary. The larger angle is 15° more than twice the smaller angle. What are the measures of the angles?

15. Two angles are supplementary. The angle measures have a ratio of 5 to 7. What are the measures of the angles?

16. A windshield wiper forms a straight angle when fully extended. If the wiper is opened to a 140° angle, how many more degrees does it need to be opened to be fully extended?


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Independent Practice For problems 1–4, find the complement to each angle by writing an equation and solving it. Show your work.

1. 45° 3. 65°

2. 80° 4. 30°

For problems 5–8, find the supplement to each angle by writing an equation and solving it. Show your work.

5. 75° 7. 5°

6. 120° 8. 55°


9. 40° and 60° 10. 45° and 45°

continued


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For problems 11–16, write an equation and then solve it. Show your work.

11. The supplement of an angle measures 20° more than twice its complement. Find the measures of the angles.

12. One of two complementary angles measures 30° more than the other measures. Find the measure of each angle.

13. Two angles are supplementary. One angle is twice the measure of the other. What are their measures?

14. Two angles are complementary. The larger angle is 30° more than twice the smaller angle. What are the measures of the angles?

15. Two angles are supplementary. The angle measures have a ratio of 2 to 3. What are the measures of the angles?

16. A folding fan forms a straight angle when fully extended. If the fan is opened to a 120° angle, how many more degrees does it need to be opened to be fully extended?


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Assessment

Progress AssessmentFor problems 1 and 2, find the complement to each angle by writing an equation and solving it. Show your work.

1. 70° 2. 20°

For problems 3 and 4, find the supplement to each angle by writing an equation and solving it. Show your work.

3. 10° 4. 150°


5. 60° and 30° 6. 110° and 20°

7. Two angles are supplementary. One angle is 40° larger than the other. What are the measures of the angles? Find the missing angles by writing an equation and then solving it. Show your work.




Instruction

Resource List• Math Warehouse. “Complementary Angles.”

www.mathwarehouse.com/geometry/angle/complementary-angles.php

Users answer practice problems about complementary angles.

• Math Warehouse. “Interactive Supplementary Angles.”

www.mathwarehouse.com/geometry/angle/interactive-supplementary-angles.php

This Web site allows the user to click and drag points to discover the rule for supplementary angles.

• Math123xyz. “Angles Puzzle.”

www.math123xyz.com/Nav/Geometry/Angle_Puzzle.php

For this four-part memory game, players match the pairs of angle measures that make complementary angles. After successful completion, players match supplementary angle pairs. For the next two rounds, players match complementary angles and then supplementary angles, respectively, that do not list their angle measures; they must correctly eyeball the corresponding angles.



Lesson 6 Answer KeyLesson Pre-Assessment, p. 141 1. 180 = x + 140; x = 40° 2. 90 = x + 45; x = 45° 3. 180 = x + 50 + 17; x = 113° 4. 90 = 2x + 30; x = 30° and 60° 5. 180 = x + 45; x = 135°

Guided Practice, p. 155 1. 90 = x + 76; x = 14° 2. 90 = x + 42; x = 48° 3. 90 = x + 7; x = 83° 4. 90 = x + 15; x = 75° 5. 180 = x + 45; x = 135° 6. 180 = x + 90; x = 90° 7. 180 = x + 4; x = 176° 8. 180 = x + 110; x = 70° 9. 180 = x + 90 + 15; x = 75° 10. 180 = x + 45 + 30; x = 105° 11. 180 – x = 30 + 2 (90 – x); 30°; supplement: 150° and complement: 160° 12. 90 = x + x + 60; 15° and 75° 13. 180 = 2x; 90° and 90° 14. 90 = 15 + 2x + x; 25° and 65° 15. 5x + 7x = 180; 75° and 105° 16. x + 140 = 180; 40°

Independent Practice, p. 157 1. 90 = x + 45; x = 45° 2. 90 = x + 80; x = 10° 3. 90 = x + 65; x = 25° 4. 90 = x + 30; x = 60° 5. 180 = x + 75; x = 105° 6. 180 = x + 120; x = 60° 7. 180 = x + 5; x = 175° 8. 180 = x + 55; x = 125° 9. 180 = x + 40 + 60; x = 80° 10. 180 = x + 45 + 45; x = 90° 11. 180 – x = 20 + 2 (90 – x); 20°; supplement: 160° and complement: 70° 12. 90 = x + x + 30; 30° and 60° 13. 180 = 2x + x; 60° and 120° 14. 90 = 30 + 2x + x; 20° and 70° 15. 2x + 3x = 180; 72° and 108° 16. x + 120 = 180; 60°

Progress Assessment, p. 159 1. 90 = x + 70; x = 20° 2. 90 = x + 20; x = 70° 3. 180 = x + 10; x = 170° 4. 180 = x + 150; x = 30° 5. 180 = x + 60 + 30; x = 90° 6. 180 = x + 110 + 20; x = 50° 7. 180 = x + x + 40; 70° and 110°

Unit 3 • Equations, inEqualitiEs, and word problEmsLesson 7: Solving Word Problems Using the Pythagorean Theorem

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Assessment

Lesson Pre-AssessmentA right triangle has legs a and b and hypotenuse c. For the problems that follow, use the Pythagorean theorem to find the unknown side length of each triangle. Be sure to show your work.

1. a = 3 cm and b = 4 cm

2. a = 10 m and b = 24 m

3. a = 8 ft and c = 17 ft

4. Jose travels 90 miles north and 120 miles east. If he takes the shortest diagonal route to return to his starting point, how many miles will he have to travel on his return trip?

5. A support wire is attached to a 30-foot-tall telephone pole. It meets the ground 16 feet from the base of the pole. What is the length of the wire?



Instruction

Lesson 7: Solving Word Problems Using the Pythagorean Theorem

Essential Questions 1. How does the Pythagorean theorem apply in real-world situations?

2. Does the Pythagorean theorem apply to all triangles?

WORDS TO KNOW

hypotenuse the side opposite of the right angle of any right triangle

leg of a triangle either of the two shorter sides of any right triangle

perfect square an integer that is a square of an integer

Pythagorean theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs of any right triangle; for a right triangle with legs a and b, and hypotenuse c, a2 + b2 = c2

right triangle a triangle with exactly one right (90°) angle


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Warm-Up Option 1For problems 1–4, solve each equation for x.

1. 49 = x2

2. 16 = x2

3. 25 = x2 – 144

4. x2 + 64 = 289

5. Estimate the value, to the tenths place, for x when x2 = 80.


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Warm-Up Option 2A water tank is shaped like a square prism and holds 45 cubic meters. The height of the tank is 5 meters. Use the formula V = l • w • h, for which V is volume, l is length, w is width, and h is height, to find the length of each side of the square base.




Instruction

Warm-Up Option 1: Debrief 1. 49 = x2

• When solving for a given variable, reverse the order of operations. Similar to how subtraction is the inverse operation of addition, taking the square root of a number is the inverse operation of squaring a number.

• In this equation, students are asked to find which number results in 49 when squared.

• The two solutions to this equation are 7 and –7.

7 • 7 = 49 and –7 • –7 = 49

x = 7 and x = –7

2. 16 = x2

• Again, students are asked to find which number results in 16 when squared.


4 • 4 = 16 and –4 • –4 = 16

x = 4 and x = –4

3. 25 = x2 – 144

• To solve for x, first add 144 to both sides of the equation.

169 = x2

• Now find the number that results in 169 when squared.


13 • 13 = 169 and –13 • –13 = 169

x = 13 and x = –13



Instruction

4. x2 + 64 = 289

• To solve for x, first subtract 64 from both sides of the equation.

x2 = 225

• Now find the number that results in 225 when squared.


15 • 15 = 225 and –15 • –15 = 225

x = 15 and x = –15

5. Estimate the value, to the tenths place, for x when x2 = 80.

• Unlike the previous problems, there is no integer that will result in 80 when squared.

• To find an approximate answer, first think of the squares of numbers that are close to 80.

9 • 9 = 81, which is larger than 80

8 • 8 = 64, which is less than 80

• The solution to x2 = 80 is less than 9, but greater than 8.

• The solution is also closer to 9 because the square of 9 is closer to 80 than the square of 8.

• To estimate to the tenths place, choose a number less than 9.

8.9 • 8.9 = 79.21

• Because 79.21 is almost 80 and choosing a number less than 8.9 would result in a number less than 79.21, we can estimate x to have the value of 8.9. Since a negative value of x would also result in a squared answer of 79.21, –8.9 is also a possible answer.

x ≈ 8.9 and x ≈ –8.9




Instruction

Warm-Up Option 2: Debrief• Identify the known information.

volume of tank = 45 cubic meters

height of tank = 5 meters

• Substitute the known information into the equation V = l • w • h to find the length and width of the tank.

45 = l • w • 5

• Students may not recognize that because the tank has a square base, the length and width can be written as the same variable, x.

45 = x • x • 5

45 = 5x2

• To solve for x, first divide both sides of the equation by 5.

9 = x2

• To find the value of x, take the square root of both sides. This will give us the number that will result in 9 when squared.

• 9 is a perfect square, meaning it is an integer with a square root that is also an integer.

• The square root of 9 is 3.

x = 3

• It should be noted that –3 • –3 also results in 9, but would not apply to this situation.

The length and width of the water tank is 3 meters.


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Focus ProblemA house painter places the bottom of his 20-foot ladder 12 feet from a house. How far up the house does the ladder reach?

x20 ft

12 ft




4. How far up the house does the ladder reach?

5. Suppose the same 20-foot ladder is placed so that it reaches 17 feet up the side of the house. Approximately how far out from the house was the ladder placed? Round your answer to the nearest tenth.




Instruction


• A right triangle has exactly one right angle. The longest side of a right triangle is called the hypotenuse. The hypotenuse is the side opposite the right angle. The other two sides are called legs. Both of the legs are shorter than the hypotenuse and form the sides of the right angle.

• The Pythagorean theorem is one of the most famous theorems in all of mathematics and states that for any right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. If a and b are the lengths of the legs and c is the length of the hypotenuse, then:

a2 + b2 = c2

cb

a

Question 1

What information do you know?

Instruction

• One of the first steps in problem solving is to determine what you know by organizing the information.

• The figure shown is a right triangle.

hypotenuse = 20 ft

one of the legs = 12 ft

Question 2

What do you need to know?

Instruction

• You need to know the length of the missing leg of the right triangle, which will tell you how far up the house the ladder reaches.




What algebraic equation could be used to represent the situation?

Instruction

• To represent this situation, use the Pythagorean theorem: a2 + b2 = c2.

• Substitute known values into the equation.

122 + x2 = 202

Question 4

How far up the house does the ladder reach?

Instruction

• First, simplify the equation by squaring 12 and 20.

144 + x2 = 400

• To solve the problem, isolate the variable x by reversing the order of operations.

• Subtract 144 from both sides of the equation.

x2 = 256

• To solve for x, take the square root of both sides of the equation to maintain equality.

x2 256=

x = 256

• To find the square root of 256, find the number that results in 256 when squared.

• 256 is a perfect square, meaning its square root is also an integer.

16 • 16 = 256

• It is also true that –16 • –16 = 256, but in this case, it would be impossible to have a ladder that reaches –16 feet high.

The ladder reaches 16 feet up the side of the house.





Suppose the same 20-foot ladder is placed so that it reaches 17 feet up the side of the house. Approximately how far out from the house was the ladder placed? Round your answer to the nearest tenth.

Instruction

• Draw a picture to help understand the problem.

x

20 ft17 ft

• Identify the known information.

• hypotenuse = 20 ft

• one of the legs = 17 ft

• Use the Pythagorean theorem: a2 + b2 = c2.


172 + x2 = 202

• Simplify the equation by squaring 17 and 20.

289 + x2 = 400

• To solve the problem, isolate the variable x by reversing the order of operations.

• Subtract 289 from both sides of the equation.

x2 = 111



Instruction


x2 111=

x = 111


• 111 is not a perfect square, meaning the square root of 111 is not an integer.



11 • 11 = 121, which is greater than 111


• To continue estimating to the tenths place, choose a number between 10 and 11.

• 111 is about halfway between 100 and 121, so try 10.5.

10.5 • 10.5 = 110.25

• 110.25 is less than 111, so try a number greater than 10.5 to be sure your number is accurate.

10.6 • 10.6 = 112.36

• The value for x that results in 111 when squared is approximately 10.5.

• It is also true that –10.5 • –10.5 = 110.25, but it would be impossible to have a ladder that reaches –10.5 feet high.

The ladder reaches approximately 10.5 feet up the side of the house.




Instruction


A wire is used to hold a wooden door in a fence open at a 90° angle. It is tied from the edge of the door to one of the wooden slats of the fence. The length of the door is 3 feet. The wire measures 5 feet. What is the length of the fence between the door’s hinge and the wire tied to the fence?

x

5 ft

Solution


• The situation describes a right triangle.


• one leg (length of the door) = 3 ft


• You need to know the value of x to find the length of the fence.

• What algebraic equation could be used to represent the situation?



52 = 32 + x2

• Simplify the equation by squaring 5 and 3.

25 = 9 + x2

• Isolate the variable x by reversing the order of operations. Subtract 9 from both sides of the equation.

x2 = 16



Instruction


x2 16=

x = 16


• 16 is a perfect square.

4 • 4 = 16 and –4 • –4 = 16

x = 4 and x = –4

• However, since a fence can’t have a negative length, x = –4 is not a valid solution to the problem.

The length of the fence is 4 feet.

• To check the results, use the problem-solving strategy of working backward. Substitute 4 feet and additional known values into the original equation and solve.

52 = 32 + (4)2

25 = 9 + 16

25 = 25

Example 2

A ramp was constructed to allow wheelchair access to a building. If the ramp is 25 feet long and the horizontal distance is 24 feet, what is the height of the ramp?

Solution




• one leg (horizontal distance) = 24 ft




Instruction


• You need to know the height of the ramp.




252 = 242 + x2

• Solve the problem. Start by simplifying the equation.

625 = 576 + x2

• Isolate the variable x by reversing the order of operations. Subtract 576 from both sides of the equation.

x2 = 49


x2 49=

x = 49


• 49 is a perfect square.

7 • 7 = 49 and –7 • –7 = 49

• Height cannot be a negative number.

x = 7 feet

The height of the ramp is 7 feet.


252 = 242 + (7)2

625 = 576 + 49

625 = 625




You’re walking in a park that has a flower garden in the middle of it. To avoid the garden, you walk 9 meters south and 22 meters east. If you could have walked directly through the garden, about how many meters would you have walked? Round your answer to the nearest tenth.

Solution



one leg = 9 meters

other leg = 22 meters


• You need to know the hypotenuse of the triangle.

• Draw a picture to represent the situation as part of your problem-solving process.

22 m

x9 m




x2 = 222 + 92

• Simplify the equation.

x2 = 484 + 81

x2 = 565


x = 565

• 565 is not a perfect square.





Instruction

• Students may want to create a chart of integers and their squares as a reference.

Integer Square of Integer1 12 43 94 165 25x x2


24 • 24 = 576, which is greater than 565


• To continue estimating to the tenths place, choose a number between 23 and 24.

232 = 529

242 = 576

• 565 is closer to 576 than it is to 529. Choose a number closer to 24 than 23.

23.7 • 23.7 = 561.69

• 561.69 is less than 565, so try a number greater than 23.7.

23.8 • 23.8 = 566.44; 566.44 is too large.

• The value for x that results in 565 when squared is approximately 23.7.• –23.7 squared is also approximately 565, but distance can’t be a negative number.

You would have walked approximately 23.7 meters.


23.72 ≈ 222 + 92

561.69 ≈ 484 + 81

561.69 ≈ 5653 ft


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Guided PracticeFor problems 1–5, use the given leg lengths to find the length of the hypotenuse for each right triangle. Round your answers to the nearest tenth and show your work.

1. a = 9 cm and b = 12 cm

2. a = 10 cm and b = 24 cm

3. a = 5 m and b = 2 m

4. a = 5 ft and b = 12 ft

5. a = 7 in. and b = 8 in.

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For problems 6–10, find the unknown leg of each right triangle given the lengths of one leg and the hypotenuse. Round your answer to the nearest tenth and show your work.

6. a = 8 yd and c = 17 yd

7. a = 12 m and c = 18 m

8. b = 20 yd and c = 25 yd

9. a = 10 cm and c = 14 cm

10. b = 21 ft and c = 29 ft

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For problems 11–16, write an equation for each scenario described and solve it for the unknown value, rounded to nearest tenth. Show your work.

11. A rectangular flag has a length of 8 feet and a width of 5 feet. What is the diagonal length of the flag?

12. A baseball diamond is a square with 90 feet between each base. If a catcher throws a ball from home plate to 2nd base (the diagonal of the square), how far will he throw the ball?

13. Kelly walked 16 meters east and then 30 meters north. How far is Kelly from her starting point?

Kelly’s starting point

Kelly’s ending point

30 m

16 m

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14. The dimensions of a soccer field are 100 yards long and 60 yards wide. Edward runs from one corner to the corner that’s diagonally across from it. What distance did Edward run?

15. Chloe walks 3 miles north and 4 miles west. If she takes the shortest route to return to where she started, how far does she have to walk?

16. The sides of an A-frame house begin at the foundation and meet at the top, forming a letter A or triangle. One A-frame house has sides that are 30 feet long. If the house is 36 feet wide at the base, what is the height of the house at its tallest point?

30 ft30 ft

18 ft18 ft36 ft


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Independent Practice For problems 1–5, use the given leg lengths to find the length of the hypotenuse for each right triangle. Round your answers to the nearest tenth and show your work.

1. a = 12 cm and b = 16 cm

2. a = 15 cm and b = 36 cm

3. a = 3 m and b = 6 m

4. a = 12 ft and b = 35 ft

5. a = 20 in. and b = 12 in.

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For problems 6–10, find the unknown leg of each right triangle given the lengths of one leg and the hypotenuse. Round your answer to the nearest tenth and show your work.

6. b = 12 yd and c = 15 yd

7. a = 10 m and c = 15 m

8. b = 40 yd and c = 41 yd

9. b = 14 cm and c = 23 cm

10. a = 11 ft and c = 61 ft

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For problems 11–16, write an equation and solve it for the unknown value, rounded to nearest tenth. Show your work.

11. A rectangle has a width of 13 feet and a length of 84 feet. What is the diagonal length of the rectangle?

12. A softball diamond is a square with 60 feet between each base. If a catcher throws a ball from home plate to 2nd base (the diagonal of the square), how far will she throw the ball?

13. Simon walked 15 meters south and then 8 meters west. How far is Simon from his starting point?

15 m

8 m

continued


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14. The dimensions of an Olympic-size swimming pool are 50 meters by 25 meters. Sula swims from one corner to the corner diagonally across from it. What distance did Sula swim?

15. A rectangular HDTV screen has a diagonal that is 61 inches long and a height of 30 inches. What is the screen’s length?

16. A 15-foot ladder rests against a wall. The base of the ladder is 5 feet away from the wall. How far up the wall does the ladder reach?


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Assessment

Progress AssessmentA right triangle has two legs of length a and b and a hypotenuse of length c. Use the Pythagorean theorem to find the unknown side length of each triangle described. Be sure to show your work.

1. a = 12 cm and b = 16 cm

2. a = 16 m and b = 30 m

3. a = 9 ft and b = 7 ft

4. a = 10 yd and c = 26 yd

5. An HDTV has a length of 36.7 inches and a height of 20.4 inches. What is its diagonal length?

6. A storage shed is shaped like an A-frame tent. The two sides are each 10 feet long. The base of the shed is 12 feet long. What is the height of the shed at its tallest point?

10 ft10 ft

6 ft6 ft12 ft




Instruction

Resource List• CRCTLessons. “Pythagorean Theorem Game.”

www.crctlessons.com/Pythagorean-theorem-game.html

Answer multiple-choice questions related to the Pythagorean theorem. More than one opportunity is given to find the correct answer, at which point a hint is provided.

• Math Warehouse. “Right Triangles: The Good Old Pythagorean Theorem.”

www.mathwarehouse.com/geometry/triangles/right-triangle.html

This Web site provides a summary of the Pythagorean theorem as well as additional practice.

• Quia. “Pythagorean Theorem Jeopardy.”

www.quia.com/cb/278769.html

Choose between one or two players in this Jeopardy-style game. Correct answers are given if incorrect answers are typed in.



Lesson 7 Answer KeyLesson Pre-Assessment, p. 162 1. 5 cm 4. 150 mi 2. 26 m 5. 34 ft 3. 15 ft

Guided Practice, p. 179 1. 15 cm 9. 9.8 cm 2. 26 cm 10. 20 ft 3. 5.4 m 11. 9.4 ft 4. 13 ft 12. 127.3 ft 5. 10.6 in. 13. 34 m 6. 15 yd 14. 116.6 yd 7. 13.4 m 15. 5 mi 8. 15 yd 16. 24 ft

Independent Practice, p. 183 1. 20 cm 9. 18.2 cm 2. 39 cm 10. 60 ft 3. 6.7 m 11. 85 ft 4. 37 ft 12. 84.9 ft 5. 23.3 in. 13. 17 m 6. 9 yd 14. 55.9 m 7. 11.2 m 15. 53.1 in. 8. 9 yd 16. 14.1 ft

Progress Assessment, p. 187 1. 20 cm 4. 24 yd 2. 34 m 5. 42.0 in. 3. 11.4 ft 6. 8 ft

Unit 3 • Equations, inEqualitiEs, and word problEmsMixed Review

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Solve each equation. Show your work.

1. − =−x3

52 11

2. 60 – y = 8 + 7y

3. 6x2 + 24x – 126 = 0

4. x2 – 5x – 36 = 0

5. x2 – 100 = 0

For problems 6 and 7, solve each inequality, and then graph the solution set on a number line.

6. 3x – 18 ≤ 0

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

7. − <x

43

2

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

continued


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Find the answers to the following problems. Show your work.

8. Identify the x- and y-intercepts of the line = +yx

210 .

9. Identify the x- and y-intercepts of the line =− +y x31

4.

10. Determine the slope and the equation of a line containing the points (0, 7) and (5, 22).

11. A passenger train traveled at a speed of 80 miles per hour. Write an algebraic expression to represent the number of miles the train traveled in x hours.

12. Elsa purchased 8 shirts and received a 15% discount. She also paid $6.45 in taxes on her purchase. Write an algebraic expression to represent the total amount Elsa paid.

13. Robert bought 6 concert tickets. He paid a 5% service charge for buying them online. His total cost was $472.50. What was the price of each ticket?

14. Carmen earned a score of 81 on her semester science test. She needs to have a total of 170 points from her semester and final tests to receive an A for the class. Write and solve an algebraic inequality to determine the score Carmen must earn on her final test to ensure that she gets an A.

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15. The product of two consecutive odd integers is 143. What are the integers? (Hint: If x is an odd integer, the next consecutive odd integer is x + 2.)

16. A photo service charges $25.00 a year as well as $0.03 for each photo ordered. Write an equation in slope-intercept form that represents the cost of photos for one year. Identify the slope and y-intercept of your equation.

17. Find the supplement to an angle measuring 53° by writing an equation and then solving it.

18. Find the complement to an angle measuring 39° by writing an equation and then solving it.

19. Two of the three angles of a triangle measure 14° and 57°. Write an equation and solve it to find the measure of the third angle.

20. Two angles are complementary. The larger angle is 5° more than 3 times the smaller angle. What are the measures of the angles?

21. The leg of a right triangle measures 9 meters and the hypotenuse measures 41 meters. Write and solve an equation to determine the length of the other leg.

22. A rectangular flowerbed has a length of 36 feet and a width of 77 feet. What is the diagonal length of the flowerbed?




Mixed Review Answer Key 1. x = –15 2. y = 6.5 3. x = 3 and x = –7 4. x = 9 and x = –4 5. x = 10 and x = –10 6. x < 6

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

7. x > 6

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10x

8. y-intercept: 10; x-intercept: –20 9. y-intercept: 1/4; x-intercept: 1/12 10. slope = 3; y = 3x + 7 11. 80x 12. 8x – 0.15(8x) + 6.45 = 6.8x + 6.45 13. 6(x + 0.05x) = 472.50; x = $75 14. 81 + x > 170; x > 89 15. 11 and 13 16. y = 0.03x + 25; slope: 0.03; y-intercept: 25 17. 90 = x + 53; x = 127° 18. 90 = x + 39; x = 51° 19. 180 = x + 14 + 57; x = 109° 20. 90 = x + 3x + 5; 21.25° and 68.75° 21. 40 meters 22. 85 feet

Unit 3 • Equations, inEqualitiEs, and word problEmsUnit Post-Assessment 1

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Assessment

Solve the following problems. Show your work.

1. A truck driver traveled at a speed of 65 miles per hour. Write an algebraic expression to represent the number of miles the truck driver traveled in x hours.

2. A new sushi restaurant is having a special. Write an algebraic expression to represent the cost of $2 per x plates of sushi eaten plus a $5 flat fee for unlimited beverages.

3. Hillary plays tennis twice as much as Michelle. Michelle plays tennis x hours a week. Together, the girls play a total of 9 hours. How many hours a week does Hillary play tennis? Write an equation and solve.

4. Oscar purchased a new tennis racket and paid a 7% sales tax on his purchase. The total cost of his racket was $68.48. What was the price of the tennis racket without tax? Write an equation and solve.

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Assessment

5. After installing a new light fixture, Petra notices a sticker on the box that advises “Only use light bulbs of 40 watts or less.” Write an algebraic inequality to show what wattages of light bulbs can be used and graph the solution set on the number line below.

0 10 20 30 40 50 60 70 80 90 100x

6. Graph the solution set of –2x + 3 > 6 on the number line below.

–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3x

7. Solve the quadratic equation x2 + 9x + 14 = 0 by factoring.

8. Solve the quadratic equation –x2 – 5x + 6 = 0 by factoring.

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Assessment

9. Two angles are supplementary. If one has a measure of 15°, what is the measure of the other angle?

10. Two angles are complementary. One angle measures 40° more than the other angle. What are the measures of the angles?

11. A triangle has one angle that measures 30° and another angle that measures 12°. What is the measure of the third angle? Write an equation and solve.

12. Graph the linear equation y = x – 3. Identify the slope and the y-intercept.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

slope = _____________

y-intercept = _________________

continued


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Assessment

13. Use the points (0, 1) and (1, 4) to find the slope, y-intercept, and the equation of the line. Then graph the line.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

slope = _____________

y-intercept = _________________

14. A rectangular trampoline has a length of 12 feet and a width of 16 feet. What is the diagonal length of the trampoline? Write an equation and solve.

15. Jennifer walked 21 meters east and 28 meters south. How far is Jennifer from her starting point? Write an equation and solve.



Unit Post-Assessment 1 Answer Key 1. 65x 2. 2x + 5 3. 2x + x = 9; Hillary plays tennis for 6 hours each week. 4. 1.07x = 68.48; x = $64 5. x ≤ 40

0 10 20 30 40 50 60 70 80 90 100x

6.

–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3x

7. (x + 2)(x + 7); x = –2 and x = –7 8. (x – 6)(x + 1); x = 6 and x = –1 9. 165° 10. 25° and 65° 11. 180 = 30 + 12 + x; x = 138° 12. slope = 1; y-intercept: (0, –3)

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

13. slope = 3; y-intercept: (0, 1); equation: y = 3x + 1

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

14. x2 = 122 + 162; x = 20 ft 15. x2 = 212 + 282; x = 35 m


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Assessment

Solve the following problems. Show all your work.

1. On your way to work, you traveled at a speed of 45 miles per hour. Write an algebraic expression to represent the number of miles you traveled in x hours.

2. Your favorite restaurant is having a special—$3 per meal plus a flat fee of $7 for unlimited beverages for everyone at the table. Write an algebraic expression to represent the cost of $3 per x meals eaten plus a $7 flat fee for unlimited beverages.

3. Michael plays basketball 3 times as many hours as Bryan. Bryan plays basketball x hours a week. Together, they play a total of 12 hours. How many hours a week does Michael play basketball? Write an equation and solve.

4. Elena purchased a new snowboard and paid 6% in sales tax on her purchase. The total cost of her snowboard was $365.70. What was the price of the snowboard without tax? Write an equation and solve.

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Assessment

5. The label on a container of paint recommends painting when temperatures are 40°F or above, but lower than 60°F. Write an algebraic inequality to show the temperature range in which painting should occur. Then, graph the solution set on the number line below.

0 10 20 30 40 50 60 70 80 90 100x

6. Graph the solution set of –5 – 2x < –8 on the number line below.

–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3x

7. Solve the quadratic equation x2 + 13x + 36 = 0 by factoring.

8. Solve the quadratic equation x2 – 6x – 16 = 0 by factoring.

9. Two angles are supplementary. If one has a measure of 35°, what is the measure of the other angle?

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Assessment

10. Two angles are complementary. One angle measures 20° more than the other angle. What are the measures of the angles?

11. A triangle has one angle that measures 33° and another angle that measures 18°. What is the measure of the third angle? Write an equation and solve.

12. Graph the linear equation y = x + 4. Identify the slope and the y-intercept.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

Slope: _____________

y-intercept: _____________

continued


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Assessment

13. Use the points (0, 3) and (2, 7) to find the slope, y-intercept, and equation of the line. Then graph the line.

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

Slope: _____________

y-intercept: _____________

Equation: _____________

14. A rectangular picture frame has a length of 35 inches and a width of 12 inches. What is the diagonal length of the picture? Write an equation and solve.

15. Ming walked 8 meters west and 15 meters south. How far is Ming from her starting point? Write an equation and solve.




Unit Post-Assessment 2 Answer Key 1. 45x 2. 3x+7 3. 3x+x=12;Michaelplaysbasketballfor9hourseachweek. 4. 1.06x=365.70;x =$345 5. 40<x <60

0 10 20 30 40 50 60 70 80 90 100x

6. x>1.5

–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5 3x

7. (x+4)(x+9);x =–4andx =–9 8. (x+8)(x–2);x =–8andx =2 9. 145° 10. 35°and55° 11. 180=33+18+x;x=129° 12. slope=1;y-intercept:(0,4)

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

13. slope=2;y-intercept:(0,3);equation:y=2x+3

x

y

–10 –8 –6 –4 –2 0 2 4 6 8 10

10

8

6

4

2

–2

–4

–6

–8

–10

14. x2=352+122;x=37in. 15. x2=82+152;x=17m

Unit 3 • Equations, inEqualitiEs, and word problEmsStation Activities Set 1: Solving Equations


InstructionGoal: To provide opportunities for students to develop concepts and skills related to

solving equations

Student Activities Overview and Answer KeyStation 1

Students are given a set of cards with linear equations written on them. They are given another set of cards with values of the variable written on them. Students work together to match each equation to its solution. Once students have paired the cards, they reflect on the strategies they used.

Answers: The cards should be paired as follows: x + 8 = 5 and x = –3; –4x = 24 and x = –6; x⁄2 and – 2 and x = –4; x –7 = –4 and x = 3; 3 = x⁄2 and x = 6; –24 = –6x and x = 4.

Possible strategies: Choose an equation. Check each value of x in the equation to see if it is a solution. Alternatively, solve the equation and look for its solution among the values of x.

Station 2

Students work together to use algebra tiles to represent linear equations. Then they use the tiles to help them solve the equations. Students explain the strategies they used to manipulate the tiles and solve the equations.

Answers: 1. x = 6; 2. x = –5; 3. x = –2; 4. x = –5; 5. x = 4; 6. x = 3

Possible strategies: Remove the same number of yellow tiles or the same number of red tiles from both sides of the mat. Add the same number of red tiles or yellow tiles to both sides, and then remove zero pairs. Divide the tiles on each side of the mat into the same number of equal groups, and remove all but one of the groups on each side. The remaining group shows the solution.




Instruction

Station 3

In this activity, students use cups and counters to model linear equations. In the given pictures, each cup is holding an unknown number of counters. Students use this idea to write the equation that is modeled by each picture. Then they use actual cups and counters, as well as logical reasoning, to help them find the unknown number of counters in each cup. This is equivalent to solving the corresponding equation.

Answers: 1. x + 1 = 10, x = 9; 2. 2x = 12, x = 6; 3. 2x + 3 = 7, x = 2; 4. 10 = 3x + 1, x = 3

Station 4

Students are given a set of equations and a set of real-life situations. They work together to match each situation to an equation. Then they solve the equation. At the end of the activity, students explain the strategies they used to match the equations to the situations.

Answers: 1. 2x + 3 = 25, x = 11; 2. 3x – 25 =2, x = 9; 3. 2x –3 = 25, x = 14; 4. 3x + 2 = 25, x = 7 2⁄3

Possible strategies: Use the words or phrases that refer to arithmetic operations as clues to identifying the corresponding equations. For example, gathering equal groups of objects corresponds to multiplication.

Materials List/SetupStation 1 set of 6 index cards with the following equations written on them:

x + 8 = 5 –4x = 24 x⁄2 = –2 x –7 = –4 3 = x⁄2 –24 = –6x set of 6 index cards with the following values of x written on them: x = –6, x = –4, x = –3, x = 3, x = 4, x = 6

Station 2 algebra tiles; equation mat

Station 3 3 paper cups; 12 counters or other small objects, such as pennies or beans

Station 4 none



Instruction

Discussion GuideTo support students in reflecting on the activities and to gather some formative information about student learning, use the following prompts to facilitate a class discussion to “debrief” the station activities.

Prompts/Questions

1. What are some different tools, objects, or drawings that you can use to help you solve equations?

2. How do you solve an equation using inverse operations?

3. How do you know which operation to use first when you solve a two-step equation?

4. How can you check your solution to an equation?

Think, Pair, Share

Have students jot down their own responses to questions, then discuss with a partner (who was not in their station group), and then discuss as a whole class.

Suggested Appropriate Responses

1. You can use algebra tiles, cups and counters, drawings of balance scales, etc.

2. Isolate the variable by applying inverse operations to both sides of the equation.

3. You usually add or subtract on both sides of the equation before you multiply or divide on both sides of the equation. (You reverse the order of operations to “undo” the operations on the variable.)

4. Substitute the value for the variable in the equation and simplify. If the solution is correct, the two sides of the equation should be equal.

Possible Misunderstandings/Mistakes

• Using an incorrect operation to solve an equation (e.g., solving x + 3 = 12 by adding 3 to both sides)

• Attempting to solve an equation such as x + 4 = 9 by subtracting x from both sides

• Applying an operation that does not isolate the variable (e.g., solving 9 = 3x by dividing both sides by 9)


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Station 1At this station, you will find a set of cards with the following equations written on them:

You will also find a set of cards with the following values of x written on them:

Work with other students to match each equation with its solution.

Work together to check that each equation is paired with its correct solution. Write the pairs below.

_____________________________

_____________________________

_____________________________

_____________________________

_____________________________

Explain the strategies you used to match up the cards. __________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

x = –6 x = –4 x = –3 x = 3 x = 4 x = 6

x + 8 = 5 –4x = 24 = –2 x –7 = –4 3 = –24 = –6x x2

x2


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Station 2You can use algebra tiles to help you solve equations.

Each square yellow tile shows +1. Each square red tile shows –1. Each rectangular yellow tile shows x. Use the equation mat to show the two sides of an equation.

Work together using algebra tiles to show each equation.

Then work together using the tiles to solve the equation. Write the answers below.

1. x + 4 = 10 ________________

2. x + 7 = 2 ________________

3. x – 5 = –7 ________________

4. 2x = –10 ________________

5. 2x + 1 = 9 ________________

6. 8 = 4x –4 ________________

Explain at least two strategies you used to solve the equations using algebra tiles.

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________


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Station 3In each picture, the cup is holding an unknown number of counters, x. If there is more than one cup, every cup is holding the same number of counters.

Each picture shows an equation. This picture shows x + 5 = 7. To make the two sides equal, there must be 2 counters in the cup. This means x = 2.

Work with other students to write an equation for each picture. Then find the number of counters in each cup. You can use the cups and counters at the station to help you.

1. Equation: __________________

Solution: __________________

2. Equation: __________________

Solution: __________________

3. Equation: __________________

Solution: __________________

4. Equation: __________________

Solution: __________________

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=


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Station 4At this station, you will match equations to real-life situations and then solve the equations.

Work with other students to match each situation to one of the following equations. When everyone agrees on the correct equation, write it on the line. Then work together to solve it.

2x – 3 = 25 2x + 3 = 25 3x + 2 = 25 3x – 25 = 2

1. Rosa bought some notebooks that cost $2 each. She also bought a compass that cost $3. She spent a total of $25. How many notebooks did she buy?

Equation: ________________________

Solution: ________________________

2. Ms. Chen brought 3 packages of pencils for her class. Each package contained the same number of pencils. The 25 students in her class each took one pencil. There were 2 pencils left over. How many pencils were in each package?

Equation: ________________________

Solution: ________________________

3. Tyler bought two copies of a DVD to give as gifts. He had a coupon for $3 off his total purchase. The final cost of the DVDs was $25. How much did each DVD cost?

Equation: ________________________

Solution: ________________________

4. A bowl can hold 25 fluid ounces of liquid. Omar empties a full teacup of water into the bowl 3 times. Then he adds another 2 fluid ounces of water to fill the bowl. How many fluid ounces of liquid does the teacup hold?

Equation: ________________________

Solution: ________________________

Explain the strategies you used to match the equations to the situations. ____________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

Unit 3 • Equations, inEqualitiEs, and word problEmsStation Activities Set 2: Solving Inequalities



InstructionGoal: To provide opportunities for students to develop concepts and skills related to

solving inequalities


Students are given a series of inequalities and a number cube. For each inequality, they roll the number cube and then work together to decide if the number shown on the cube is a solution of the inequality. Students explain the strategies they used to decide whether each value was a solution.

Answers: 1–3. Answers will depend upon numbers rolled. 4. Yes; 5. No

Station 2

In this activity, students work together to use number lines to help them solve inequalities. To do so, they test various values of the variable in the given inequalities, and check to see whether each value is a solution. They keep track of the values that are solutions by marking them on a number line. After testing enough values to see a pattern, students shade the values that represent all solutions of the inequality. Then they write the solution algebraically.

Answers:

1. x < 2

–5 –4 –3 –2 –1 0 1 2 3 4 5

2. x ≥ –2

–5 –4 –3 –2 –1 0 1 2 3 4 5

3. x > 1

–5 –4 –3 –2 –1 0 1 2 3 4 5

4. x ≤ 0

–5 –4 –3 –2 –1 0 1 2 3 4 5



Instruction

5. x < 2

–5 –4 –3 –2 –1 0 1 2 3 4 5

Station 3

In this activity, students work together to match a set of given inequalities with a set of given solutions. Once students have paired each inequality with its correct solution, they discuss the strategies they used to solve the problem.

Answers: The cards should be paired as follows: 5x + 2 < 12 and x < 2; 4x + 3 < –5 and x < –2; –3x < 6 and x > –2; –x⁄2 > 2 and x < –4; 3x + 1 > –11 and x > –4; x⁄4 + 1 > 2 and x > 4.

Station 4

Students are given a set of inequalities and a set of real-world situations. They work together to match each situation to an inequality. Then they solve the inequality. At the end of the activity, students explain the strategies they used to match the inequalities to the situations.

Answers: 1. 10x + 5 ≤ 105, x ≤ 10; 2. 5x + 10 < 105, x < 19; 3. 10x – 5 ≥ 105, x ≥ 11; 4. 5(x – 10) < 105, x < 31.

Station 5

Students work together to match inequalities with their solution graphs. Then they discuss the strategies they used.

Answers:

x + 5 < 7

–5 –4 –3 –2 –1 0 1 2 3 4 5

2x – 5 > –3

–5 –4 –3 –2 –1 0 1 2 3 4 5

4 – 3x ≤ 7

–5 –4 –3 –2 –1 0 1 2 3 4 5




Instruction

–2x + 7 ≥ 3

–5 –4 –3 –2 –1 0 1 2 3 4 5

4x – 5 ≥ 3

–5 –4 –3 –2 –1 0 1 2 3 4 5

–11 < 6x – 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

7 – 7x ≤ 0

–5 –4 –3 –2 –1 0 1 2 3 4 5

5x + 12 ≤ 7

–5 –4 –3 –2 –1 0 1 2 3 4 5



Instruction

Materials List/SetupStation 1 number cube (numbers 1-6)

Station 2 none

Station 3 set of index cards with the following inequalities written on them: 5x + 2 < 12 –x⁄2 > 2

4x + 3 < –5 3x + 1 > –11

–3x < 6 x⁄4 + 1 > 2

set of index cards with the following solutions written on them: x < –4, x > –4, x < –2, x > –2, x < 2, x > 4

The two sets of cards should be placed in two piles, face-up, on a table or desk at the station.

Station 4 none

Station 5 8 note cards with the following inequalities on them: x + 5 > 7 2x – 5 > –3 4 – 3x ≤ 7 –2x + 7 ≥ 3 4x – 5 ≥ 3 –11 < 6x – 5 7 – 7x ≤ 0 5x + 12 ≤ 7

8 note cards with the graphs in the Station 5 answers on the previous two pages




Instruction


Prompts/Questions

1. What is the difference between < and ≤?

2. How do you check to see if a value of the variable is a solution of an inequality?

3. How is the solution of an inequality different from the solution of an equation?

4. How do you solve an inequality using algebra?

5. What strategies can you use to prevent confusion when the variable of an inequality is on the right side of the inequality symbol? (Example: 2 < x)

Think, Pair, Share



1. The symbol < means “less than.” The symbol ≤ means “less than or equal to.”

2. Substitute the value for the variable in the inequality. Check to see if the resulting inequality is true. If it is, the value is a solution.

3. In general, the solution of an inequality is itself an inequality (a range of values). The solution of an equation is usually a single value (or several discrete values).

4. Use inverse operations, as when solving an equation, to isolate the variable on one side of the inequality. If you multiply or divide by a negative number, reverse the direction of the inequality.

5. Rearrange the inequality so that the variable is on the left. (Using the example above, x > 2.) Students could also read the inequality from right to left. (Using the example above, “x is greater than 2.”)


• Using an incorrect operation to solve an inequality (e.g., solving x + 2 < 5 by adding 2 to both sides)

• Incorrectly translating verbal expressions to inequalities (e.g., representing the phrase “no more than” by < rather than ≤)

• Forgetting to reverse the direction of the inequality when multiplying or dividing by a negative number


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Station 1You will find a number cube at this station.

For each inequality, roll the number cube and write the number in the box. Then work together to decide if this value of the variable is a solution of the inequality. Write “yes” or “no” on the line provided.

1. 2x + 1 < 7 Solution? ________

2. 3x – 4 ≥ 5 Solution? ________

3. –3x < –12 Solution? ________

4. x2

+ 1 < 5 Solution? ________

5. 1 – x > 0 Solution? ________

6. Explain the strategies you used to decide whether each value was a solution of the inequality.

______________________________________________________________________

______________________________________________________________________

______________________________________________________________________


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Station 2You can use number lines to help you solve inequalities.

For each inequality, work together to test different values of the variable to see if they are solutions of the inequality. If a value is a solution, draw a solid dot at that value on the number line. Test at least five different values for each inequality.

When you think you know what the solution set of an inequality looks like, shade the correct part of the number line to show all the solutions.

Finally, write the solution in the space provided.

1. 2x – 3 < 1

Solution: ________

2. 3x + 1 ≥ –5

Solution: ________

3. –2x < –2

Solution: ________

4. –3x ≥ 0

Solution: ________

5. x2

+ 3 < 4

Solution: ________

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5


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Station 3At this station, you will work with other students to match inequalities to their solutions.

You will find a set of cards with the following inequalities written on them:

You will also find a set of cards with these solutions written on them:

Work together to match each inequality with its solution. When everyone agrees on the answers, write the matching pairs below.

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

Explain the strategies you used to match up the cards.

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

x2

x4

5x + 2 < 12 4x + 3 < –5 –3x < 6 – > 2 3x + 1 > –11 + 1 > 2

x < –4 x > –4 x < –2 x > –2 x < 2 x > 4


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Station 4At this station, you will match inequalities to real-world situations and then solve the inequalities.

Work with other students to match each situation to one of the following inequalities. When everyone agrees on the correct inequality, write it on the line provided. Then work together to solve it.

5x + 10 < 105 10x + 5 ≤ 105 5(x – 10) < 105 10x – 5 ≥ 105

1. Mai rents DVDs by mail. There is a one-time membership fee of $5 and the service costs $10 per month. Mai wants to spend no more than $105. For how many months can she rent DVDs with this service?

Inequality: ____________________

Solution: ________

2. Tyrone bought 5 trays of food for a party. The price of each tray of food was the same. He also spent $10 on paper plates, napkins, and utensils. Altogether, he spent less than $105. What was the price of each tray of food?

Inequality: ____________________

Solution: ________

continued


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3. Mr. Garcia ordered 10 copies of a novel for students in his English class. He had a coupon for $5 off the total price of the order. The total cost of the order, before tax, came to no less than $105. What was the price of each novel?

Inequality: ____________________

Solution: ________

4. Rachel bought 5 pairs of jeans. Each pair of jeans was the same price. She had a coupon for $10 off the price of each pair of jeans. The total cost of the jeans, before tax, came to less than $105. What was the price of each pair of jeans before the coupon was applied?

Inequality: ____________________

Solution: ________

Explain the strategies you used to match the inequalities to the situations.

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________


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Station 5At this station, you will match inequalities to their solution graphs. Write the correct inequality on the line next to each solution. At the end, discuss strategies that you used.

Inequalities:

Solutions:

___________________

___________________

___________________

___________________

___________________

___________________

___________________

___________________

–11 < 6x – 5

–2x + 7 ≥ 3

x + 5 < 7

5x + 12 ≤ 7

2x – 5 > 3

4x – 5 ≥ 3

4 – 3x ≤ 7

7 – 7x ≤ 0

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 –4 –3 –2 –1 0 1 2 3 4 5

continued


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What strategies did you use to match the inequalities to their solution graphs?

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

Unit 3 • Equations, inEqualitiEs, and word problEmsStation Activities Set 3: Solving Quadratics by Factoring



InstructionGoal: To provide opportunities for students to solve quadratic equations by factoring


Students work in groups to solve the quadratic equations by factoring. Students may use algebra tiles as needed.

Answers

1. x x= = −29

10;

2. x = –5; x = –24

3. x = 7; x = –6

Station 2

Students work with a partner to solve the quadratic equations by factoring.

Answers

1. x = –8; x = 1/9

2. x = 1; x = –1/12

3. x = –2; x = 6

Station 3

Students work alone or in pairs to solve the quadratic equations by factoring. Students should be increasingly comfortable finding real roots independently.

Answers

1. x x= − =854

;

2. x = –3; x = 3

3. x = 4; x = 2

4. x = 2



Instruction

Station 4

Students work alone or in pairs to solve quadratic equations by factoring. Students should be increasingly comfortable finding real roots independently.

Answers

1. x x= = −175

;

2. x = 9

3. x = 2; x = 12

4. x x= − =27

52

;

Materials List/SetupStation 1 algebra tiles

Station 2 none

Station 3 none

Station 4 none




Instruction


Prompts/Questions

1. What are quadratic equations?

2. When solving quadratic equations, what do the values of x represent?

3. How can you determine if your values of x are correct?

Think, Pair, Share



1. Quadratic equations are equations that can be written as ax2 + bx + c, where a ≠ 0.

2. The values for x represent the solutions to the equation.

3. Substitute the first value of x into the original equation and simplify. Check if both sides of the equation are equal. Then substitute the second value of x into the original equation and simplify. Check if both sides of the equation are equal. If both sides are not equal, the value(s) for x are incorrect.


• Incorrectly factoring quadratic expressions

• Incorrectly factoring constants and coefficients

• Not understanding factoring

• Not understanding polynomial factoring

• Making simple arithmetical errors in factoring or in applying the quadratic formula


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Station 1Work in groups to solve the quadratic equations by factoring. Use algebra tiles as needed. Show all your work.

1. 3 20 8813

2x x= −( )

2. 14

29 302x x+( )= −

3. 12

212x x−( )=


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Station 2Work alone or in pairs to solve the following quadratic equations by factoring. Show all your work.

1. − + =x x2 89

719

2. 4 11 113

2x x= +( )

3. x

x2

43− =


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Station 3Work alone or in pairs to solve the following quadratic equations by factoring. Show all your work.

1. 83

10 182x x−( ) = −

2. x2 – 9 = 0

3. 2x2 = 2(6x – 8)

4. 7x2 = 28(x – 1)


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Station 4Work alone or in pairs to solve the quadratic equations by factoring. Show all your work.

1. 5x2 = 7 – 2x

2. x2 = –81 + 18x

3. 132 8 42x x x−( ) = −

4. 31x = 14x2 – 10

Unit 3 • Equations, inEqualitiEs, and word problEmsStation Activities Set 4: Geometry


InstructionGoal: To provide opportunities for students to develop concepts and skills related to applying

properties of right triangles, specifically the Pythagorean theorem


Students draw their own right triangles and measure the sides. Then they fill out a table to gather information and use that information to draw conclusions about the Pythagorean theorem.

Answers: Answers will vary; a 2 + b 2 = c 2

Station 2

Students draw their own right triangles and measure the sides. Then they guess what the third side will be and measure to verify their guess. Students explain their strategy for guessing the third side.

Answers: Answers will vary; the Pythagorean theorem

Station 3

Students are introduced to Pythagorean triples. They try to find as many as possible using the numbers 1–26. Then they explain their strategy for completing this task.

Answers: 3, 4, 5; 6, 8, 10; 5, 12, 13; 9, 12, 15; 8, 15, 17; 12, 16, 20; 15, 20, 25; 7, 24, 25; 10, 24, 26; answers will vary

Number 1 2 3 4 5 6 7 8 9 10 11 12 13

Square 1 4 9 16 25 36 49 64 81 100 121 144 169

Number 14 15 16 17 18 19 20 21 22 23 24 25 26

Square 196 225 256 289 324 361 400 441 484 529 576 625 676




Instruction

Station 4

Students will complete a crossword puzzle involving the vocabulary used in the Pythagorean theorem. Students can work together.

Answers:

Materials List/SetupStation 1 calculator, ruler, and protractor for each group member

Station 2 calculator, ruler, and protractor for each group member

Station 3 calculator for each group member

Station 4 none

1

2

3

5 6

7

4

9

10

11

8

13

14

15

12

G

G

G

O

O

O

O

O

O

Q

Q

I

I I

I

II

RR

R

R

R

R

R

M

R

R

R

B

R

C

C

C

D D

D

DD

L L

L

L

E

E

E E

E E

N

N

N

N

A

A

A A

AA

A

A

A

A

A

AY

Y

S

H

H

H

S

S

S

S

S

S

E

E

E

T

T

TT

T

OO

O

OO

I

I

N

N

NT

T

T

T

T

T

U

U

U

P

G

G

AY HTP

P

P

P

P



Instruction


Prompts/Questions

1. When is the Pythagorean theorem useful in real life?

2. If you have a triangle with a hypotenuse of 24 inches and a leg of 9 inches, how long is the other leg?

3. One Pythagorean triple is 3, 4, 5. Another is 6, 8, 10. Is 36, 48, 60 a Pythagorean triple? How do you know?

Think, Pair, Share



1. Many possibilities—when trying to find the distance between two points in a city if you know how many blocks apart they are

2. about 22.25 inches

3. Yes, because 36, 48, and 60 are 12 times the original triple.


• Inaccurately measuring sides of a triangle

• Having trouble coming up with Pythagorean triples—guessing rather than having a strategy

• Inaccurately measuring the angles of a triangle


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Station 1At this station, you will find enough protractors, rulers, and calculators for each group member. Each member should complete the activity and discuss his or her observations and strategies with the group.

Draw a right triangle in the space below.

Label the legs a and b, and the hypotenuse c.

Fill in the table below.

What do you notice? __________________________________________________________

__________________________________________________________________________

a 2 + b 2 = ________

Group member

Length a

Length b

Length c

a 2 b 2 c 2 a 2 + b 2 a 2 + c 2 b 2 + c 2


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Station 2At this station, you will find enough protractors, rulers, and calculators for each group member. Each member should complete the activity and discuss his or her observations and strategies with the group.

Draw a right triangle in the space below.

Label the legs a and b, and the hypotenuse c.

Measure only sides a and b. Compile your group information in the table below. Fill in only the first three columns.

Work with your group members to guess the length of the hypotenuse. Record this in the table.

Measure the hypotenuse and record this data in the table.

How close were you? __________________________________________________________

__________________________________________________________________________

What was your strategy for guessing? ______________________________________________

__________________________________________________________________________

__________________________________________________________________________

Group member Length of a Length of b Guess for c Length of c


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Station 3Pythagorean triples are a set of three whole numbers that can be used in the Pythagorean theorem and therefore, could be the sides of a right triangle. An example of this is 12, 35, and 37. 122 + 352 = 372. Both sides come out to 1,369. Now work with your group to find some Pythagorean triples.

Fill in the table below.

Use this information to help you find some Pythagorean triples. Try to name three. (There are nine total using these numbers.)

__________________________________________________________________________

__________________________________________________________________________

What was your strategy for coming up with the Pythagorean triples?

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

Number 1 2 3 4 5 6 7 8 9 10 11 12 13

Square

Number 14 15 16 17 18 19 20 21 22 23 24 25 26

Square


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Station 4Complete the crossword puzzle below.

1

2

4

7 8

5 6

11

10

12

14

13

15

9

3

Across 2. One of the operations used when

solving for the side that is across from the 90˚ angle

4. What you do to each leg of the triangle

7. The name of the man who discovered the theorem

11. Number under the symbol used in solving the theorem

12. The side across from the right angle 13. Order matters 14. In a right triangle, the hypotenuse

squared is equal to one leg squared plus the second leg squared

15. Name of a polygon with three sides

Down 1. Always positive 3. One of the sides of the triangle that is

not across from the 90˚ angle 5. Across from 6. Symbol used in solving the theorem 8. One of the operations used when

solving for a side that is not across from the 90˚ angle

9. Name of the angle the triangle must have

10. You use this at the end of solving the theorem

Word BankadditionhypotenuselegoperationsoppositePythagorasPythagorean theoremradicalradicandrightsolutionsquaresquare rootsubtractiontriangle



Unit Glossaryaddition property of equality if the same number is added to both sides of an equation, the

two sides remain equal

algebraic expression an expression that has one or more variables

algebraic inequality an inequality that has one or more variables and contains at least one of the following symbols: <, >, ≤, ≥, or ≠

binomial an algebraic expression with two unlike terms which is the sum of two monomials

coefficient the number multiplied by a variable in an algebraic expression

complementary angles two angles whose sum is 90°

constant a fixed value that does not change, such as a number

equation a mathematical sentence that uses an equal sign (=) to show that two quantities are equal

expression a mathematical statement that includes numbers, operations, and/or variables to represent a number or quantity

hypotenuse the side opposite of the right angle of any right triangle

inequality a mathematical sentence that shows the relationship between quantities that are not equivalent

inverse operation pairs of opposite operations that undo each other; addition and subtraction are inverse operations; and multiplication and division are inverse operations

isolate steps taken to get the variable alone on one side of an equation

leg of a triangle either of the two shorter sides of any right triangle

linear equation an equation that can be written in the form ax + b = c. The solution of a linear equation is a straight line.

monomial an expression that contains only one term, such as 4x or 6bc

perfect square an integer that is a square of an integer

Pythagorean theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs of any right triangle; for a right triangle with legs a and b, and hypotenuse c, a2 + b2 = c2

quadratic equation an equation of degree 2, with two solutions at most

right triangle a triangle with exactly one right (90°) angle

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Unit 3 • Equations, inEqualitiEs, and word problEmsUnit Glossary

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slope of a line the ratio of vertical change to horizontal change of a line

solution the value or values that make an equation true

solution set the value or values that make a sentence or statement true

subtraction property of equality if the same number is subtracted from both sides of an equation, the two sides remain equal

supplementary angles two angles whose sum is 180°

variable a letter used to represent a value that can change or vary

x-intercept the point where the graph of a line intercepts, or crosses the x-axis

y-intercept the point where the graph of a line intercepts, or crosses, the y-axis

zero factor property If ab = 0, then a = 0, b = 0, or both a = 0 and b = 0.

Documents

ACR EA Unit 3 · 2017. 8. 24. · Version 2.0 ElEmEntary algEbra UnIt 3. ... Unit 3 Post-Assessment 1..... 195 Unit 3 Post-Assessment 2 ... and a stack of student activity sheets