Integrated Math Concepts - Module 8mrallens.pbworks.com/f/book 8 - equations.pdf · Solving Two-Step Equations 15 Solving Complex Equations 17 Solving Multi-Step Equations 29 Translating

Integrated Math Concepts

Module 8

Equations

Second Edition

National PASS Center 2006


Solve Problems

Organize

Model

Compute

Communicate

Measure

Reason

Analyze

National PASS Center BOCES Geneseo Migrant Center 27 Lackawanna Avenue Mount Morris, NY 14510 (585) 658-7960 (585) 658-7969 (fax) www.migrant.net/pass

Authors: Justin Allen Diana Harke Editor: Sally Fox Desk Top Publishing: Sally Fox Developed for Project MATEMÁTICA ((Math Achievement Toward Excellence for Migrant Students And Professional Development for Teachers in Math Instruction Consortium Arrangement), a Migrant Education Program Consortium Incentive project, by the National PASS Center under the leadership of the National PASS Coordinating Committee with funding from Region 20 Education Service Center, San Antonio, Texas. Copyright © 2006 by the National PASS Center. All rights reserved. No part of this book may be reproduced in any form without written permission from the National PASS Center.


Module 8

Equations

Second Edition

National PASS Center 2006

BOCES Geneseo Migrant Center 27 Lackawanna Avenue Mount Morris NY 14510


Solve Problems

Organize

Model

Compute

Communicate

Measure

Reason

Analyze

Acknowledgements The materials included in this Integrated Math Concepts course were gathered, in part, from the National PASS Center’s Algebra I and Geometry courses which were written by Diana Harke. Ms. Harke currently is an instructor of mathematics at the State University of New York at Geneseo where she also supervises student teachers. She is a former junior and senior high school math teacher with experience in the United States and Canada. Ms. Harke’s courses produced thus far for the National PASS Center (NPC) have been very well received across the country, increasing the percentage of PASS mathematics courses being utilized throughout the migrant education network and beyond. It should be noted that two of the recent National Migrant PASS Students of the Year, Benancio Galvin of Marana, Arizona (2004) and Yesenia Medina of San Juan, Texas, and Wild Rose, Wisconsin (2006), have moved ahead toward their dreams of completing their high school graduation requirements thanks to their success with Ms. Harke’s Algebra I course. To meet the needs of migrant students requiring a more condensed resource to strengthen their math skills, the original curriculum materials were adapted, edited, modified, and expanded by Mr. Justin Allen. Mr. Allen is a certified secondary level math teacher and is currently pursuing a graduate degree in secondary education at the State University of New York at Geneseo. He taught middle school math and Algebra in Canandaigua, New York, for three years and, most recently, high school math in Livonia, New York. Mr. Allen assisted in the editing of the PASS Algebra II course which was released early in 2006. Acknowledgement is offered also to Ms. Sally Fox, Coordinator, National PASS Center, for her commitment to the development of quality curriculum. As with all materials produced by the NPC, her involvement with Integrated Math Concepts at all levels has played a key role in the addition of this offering to the growing number of courses available to migrant students and others seeking to master the necessary skills to become productive members of society.

Robert Lynch, Director

Module 8 – Order of Equations

Table of Contents

Page

Introduction i

Objectives 1

Translating 2

Solving One-Step Equations 10

Solving Two-Step Equations 15

Solving Complex Equations 17

Solving Multi-Step Equations 29

Translating and Graphing Inequalities 34

Solving Inequalities 36

Review 45

Practice Problems 47

Answers to “Try It” Problems 50

Answers to Practice Problems 60

Glossary of Terms 81

i

Integrated Math Concepts – Introduction

The PASS Concept PASS (Portable Assisted Study Sequence) is a study program created to help you earn

credit through semi-independent study with the help of a teacher/mentor. Your teacher/mentor

will meet with you on a regular basis to: answer your questions, review and discuss

assignments and progress, and administer tests. You can undertake courses at your own pace

and may begin a course in one location and complete it in another.

Strategy

Mathematics is not meant to be memorized; it is meant to be understood. This course

has been written with that goal in mind. Mathematics must not be read in the same way that a novel is read. In order to read a

mathematics text most effectively you must pay close attention to the structure of each

expression and to the order that operations are performed. You might think of mathematics as

you would a foreign language. Every symbol in a mathematical expression is meant to

communicate a message in that language; therefore, to understand the language you must

understand the symbolism. Always read with a pencil and scrap paper in hand. Make notes in the margins of your

book where you have questions and write “what if” variations to problems to discuss with your

teacher/mentor.

ii

Course Content Integrated Math Concepts is divided into ten modules. Each module teaches concepts

and strategies that are essential for establishing a firm foundation in each content area.

The following is a description of the ten modules in Integrated Math Concepts:

Module 1 Real Numbers

Learn to recognize and differentiate between natural numbers, whole numbers, integers,

rational numbers, irrational numbers, and real numbers.

Relate the number line to the collection of real numbers. Module 2 Sets

Recognize a well-defined set

Learn set notation and terminology

Study some subsets of real numbers – prime and composite numbers Module 3 Variables and Axioms

Learn

• why, when, and how to use a variable

• the definition of an axiom

• some specific axioms Module 4 Properties of Real Numbers

Learn the characteristics and uses of the following properties of real numbers:

• the commutative property

• the associative property

• the distributive property

• identity elements

• inverses

• the multiplication property of zero

• to understand why division by zero is not allowed

• to introduce the uniqueness and existence properties

iii

Module 5 Fractions

Become comfortable with fractions by

• understanding their make-up

• comparing their sizes

Prepare for operations with algebraic fractions

• by understanding the concepts behind the algorithms

• by determining if solutions are reasonable

Module 6 Decimals

Become comfortable with decimals and decimal operations

• by understanding the relative size of decimals

• by understanding why the algorithms or rules dealing with decimals work

• by testing answers for reasonableness

Module 7 Order of Operations

Understand why problems need to be performed in a certain order

Evaluate numerical expressions using order of operations

Evaluate variable expressions for specific values

Module 8 Equations

Translate algebraic expressions and equations, as well as consecutive integer questions

Solve:

• One-step equations

• Two-step equations

• Complex equations (combining like terms, use of the distributive property,

variables on both sides)

• Multi-step equations

Translate algebraic inequalities

Solve and graph solutions to one and two-step inequalities

iv

Module 9 Geometry

Describe points, lines, and planes

Sketch and label points, lines, and planes

Use problem solving to explore points, lines, and planes

Define line segments, rays, and angles

Recognize and examine types of angles

Explore problems using angle properties

Explore line relationships

Module 10 Properties of Polygons

Recognize and classify 2-dimensional shapes –

circles, triangles and quadrilaterals

Find 2-dimensional shapes in the environment

Explore the sum of the measures of the angles of triangles and quadrilaterals

Classify a polygon according to the number of its sides

Count diagonals in polygons

Find the measures of the interior and exterior angles in polygons

Course Organization Each module begins with a list of the objectives. This is a short list of what you will

learn. Definitions, theorems, and

mathematical properties appear as

strips of paper tacked to the page so that

they may be easily found. Examples are used to illustrate each new concept. These are

followed immediately by “Try It” problems to see if you

understand the concept. You are to write the answers to the “Try

It” problems right in your book and then check your answers with

the detailed solutions farther back in the module.

A set is a collection of objects.

v

Many lessons include the following types of inserts.

“Think Back” boxes – denoted with an arrow pointing backwards. These are

reminders of things that you have probably already learned.

“Problem solving tips” – denoted with a light bulb

“Calculator tips” – denoted with a small calculator

“Algorithms” – denoted with a fancy capital A. An algorithm is a rule (or step by

step process) used to solve a specific type of problem.

"Facts” – denoted by a small flashlight

At the end of each module you will be asked to highlight parts of the lesson as a way to

review the terminology and concepts that you just studied. You will also be asked to write

about something that you learned in your own words or list any questions to ask your

teacher/mentor about something that you did not understand. This last step is extremely

important. You should not continue on to the next activity or module until all your questions

have been answered and you are sure that you thoroughly understand the concept you just

finished. Finally, you will be asked to practice what you have learned. Athletes in every sport

must practice their skills to become better at their sport. The same is true of mathematicians.

In order to become a good mathematician, you must practice what you have learned so that it

becomes easier and easier to solve problems. You should keep a math journal or notebook

where you will do your practice problems. Detailed answers to the practice problems will be

found toward the end of the module just ahead of the glossary section.

vi

A glossary / index of the mathematical terms used in this course has been provided at

the end of each module. It contains definitions as a reference to help your understanding of

these specialized mathematical terms. Unlike other PASS courses, there is no separate Mentor Manual for this course as all of

the answers to practice problems are provided within each module. Should you require

additional support, do not hesitate to ask your mentor or teacher. That is why they are there.

Testing When you have completed all the exercises and practice problems in a module and you

and your teacher/mentor feel that you have a good grasp of the material, you will take a test

covering what you should have learned in that module.

Test taking tips 1) Make sure all of your questions have been answered and that you feel confident that

you understand the concepts on which you are to be tested.

2) Do not rush.

3) Be neat. Sometimes handwritten numbers or letters are misread.

4) Be organized. Do computations on a separate piece of paper or, if there is room on

your test sheet, in the space provided, so as to keep the flow of the problem clearly

in focus.

5) Check your answers to see

a) if you actually answered the question that was asked, and

b) that the answer is reasonable.

6) Be aware of the particular types of errors that you are prone to make. Arithmetic

mistakes are often repeated if you merely repeat the computations. Use your

calculator to prevent these types of errors and concentrate on

a) choosing the correct operations,

b) following the proper order of operations, and

c) applying valid mathematical techniques.

National PASS Center

Module 8 - Equations 1

Equations

Objectives

Translate algebraic expressions and equations, as well as

consecutive integer questions

Solve:

• One-step equations

• Two-step equations

• Complex equations (combining like terms, use of the distributive property,

variables on both sides)

• Multi-step equations

Translate algebraic inequalities

Solve and graph solutions to one and two-step inequalities


Solve Problems

Organize

Model

Compute

Communicate

Measure

Reason

Analyze

MATEMATICA August 2006

Integrated Math Concepts 2

Knowing these differences will allow you to take sentences and translate them into

mathematical expressions and equations.

Translating

In your Math Journal make charts like the ones provided here, or feel free to use these:

+ −

Add Subtract

Think Back

Remember the difference between “expression” and “equation”? An algebraic expression is a mathematical phrase that represents a quantity and it

does not contain an equal sign. An algebraic equation is a mathematical sentence which shows that two expressions

are equal and it does contain an equal sign. Also, you can only compute answers for an expression; in an equation you can solve

for the specific value of the variable.



× ÷ = Multiply Divide Equal

Now take a few minutes and write words or phrases associated with the mathematical symbols

and operations. As you can see, the most obvious words have been filled in for you. Think

back to all that you have learned up until this point and apply that knowledge here.

Okay, let’s see how you did. Keep in mind that your list doesn’t have to match the ones on the

next page. You may have come up with other words that aren’t even on the following lists.

They are examples of the words or phrases most often associated with these symbols. Observe

that the important ones, or ones used most often, have been put in capitol letters and bolded.



+ −

Add Subtract

Any variation on the word add:

Addition, added, add to, added on…

Any variation on the word subtract:

Subtraction, subtracted, subtracted from…

SUM DIFFERENCE

Plus Minus

*More than *Less than

Deposit Withdraw, withdrawal

Increased by Decreased by

Reduce, reduced

* - Indicates special words that switch the order (or arrangement) of the numbers and variables.

They also happen to be words associated with the inequality signs > and <, so be careful and

use context clues in the sentences to determine their meaning. We will examine this more

closely in the upcoming examples.

× ÷ = Multiply Divide Equal

Any variation on the word multiply:

Multiplication, multiplied,

multiplied by, multiplies…

Any variation on the word divide:

Divided, divided by, division… Equals to

TIMES Dividend Results in

PRODUCT OF Divisor The answer

QUOTIENT IS

Fraction – infers division



Example 1

Translate the following sentences; some will be expressions and other will be equations. DO

NOT solve the equations just yet. Keep in mind that if a

specific variable has not been stated for the unknown, you

will need to pick one.

A. Five times a number x.

Solution

( )5 , 5 , 5 , 5x x x x× ⋅ .

All of these are correct but the most efficient one is the last one: 5x. Note that

( )5 , 5x x ⋅ are not appropriate orders.

B. The quotient of a number and six.

Solution

Let: x = a number

6,6xx ÷ .

These are both correct but the most efficient one is 6x

.

C. Five is less than twice a certain number; the result is eleven.

Solution

Let: n = a certain number Five is less than twice a certain number the result is eleven

2 5 11n − = .

Notice that the wording that comes after “less than” goes before the subtraction sign,

except for the words following the equal sign, and the words prior to the subtraction

sign go after it.

Think Back

A variable in algebra is a letter or symbol used to

represent a number or group of numbers.

The most commonly used ones are: x, y, z, a, n

A “Let” statement identifies the variable you are going to

use and its purpose. You should always write a “Let” statement when you have to

pick the variable.



Translating Consecutive Integer Problems

To obtain a set of consecutive integers, start with any integer and count by ones. Each number

in the set is 1 more than the previous number in the set.

Think Back

An integer is any whole number or its opposite, including zero.

Examples: 5, -3, -27, 17, and 0.

Consecutive integers are integers that follow one another in order.

1. Translate the following sentences into expressions and/ or

equations. DO NOT SOLVE.

a. Three times a number x is eighteen.

b. Seventeen less than a number.

c. Twenty-four is a number m divided by three.



Each of the following is a set of consecutive integers:

1. }{5,6,7,8,...

2. }{ 7, 6, 5, 4, 3,...− − − − −

3. }{100,101,102...

To obtain a set of consecutive even integers, start with any even integer and count by twos.

Each number in the set is 2 more than the previous number in the set. Each of the following is

a set of consecutive even integers:

1. }{2, 4,6,8,...

2. }{ 12, 10, 8, 6, 4,...− − − − −

3. }{100,102,104...

Consecutive Even Integers:

1. Let x = any even integer.

2. The next number after x would have to be 2 more than it, thus: x + 2.

3. The next number would have to be 2 more than the previous one, so

therefore: x + 2 + 2, or x + 4, and so.

4. Therefore, the rule is: }{ , 2, 4, 6,...x x x x+ + +

Algorithm

Consecutive even integers are even integers that follow one another in order.

Consecutive Integers:

1. Let x = any integer.





Algorithm



To obtain a set of consecutive odd integers, start with any odd integer and count by twos. Each

number in the set is 2 more than the previous number in the set. Each of the following is a set

of consecutive odd integers:

1. }{1,3,5,7,...

2. }{ 5, 3, 1,1,3,...− − −

3. }{75,77,79...

Consecutive Odd Integers:

1. Let x = any odd integer.





Algorithm

Consecutive odd integers are odd integers that follow one another in order.

Problem Solving Tip Yes, consecutive even and consecutive odd integers have the same rule.

Think about it: If the number is even, say 16, we know the next even number is 18. How do you get

from 16 to 18? You need to add 2 to it. If the number is odd, say 7, we know the next odd number is 9. How do you get

from 7 to 9? You can’t add 1 to it, that would make it 8. You can’t add 3 to it; that would make it 10. So the only logical solution is to add 2 to it.



Example 2

Translate the following sentences, DO NOT SOLVE.

A. The sum of three consecutive integers is sixty.

Solution Let: x = 1st consecutive integer

x + 1 = 2nd consecutive integer

x + 2 = 3rd consecutive integer

Sometimes it helps to write out in words what the equation will look like:

( ) ( ) ( )( ) ( ) ( )

st nd rd1 consecutive integer + 2 consecutive integer + 3 consecutive integer 6060

1 2 60x x x

=

+ + =

+ + + + =

Recall from module 7 that the first step is to replace all the variables, or phrases in this

case, with parentheses and then substitute.

B. Find three consecutive even integers such that the sum of the first and third is forty.

Solution

Let: x = 1st consecutive even integer

x + 2 = 2nd consecutive even integer

x + 4 = 3rd consecutive even integer


( ) ( )( ) ( )

st rd1 consecutive even integer + 3 consecutive even integer 4040

4 40x x

=

+ =

+ + =

2. Translate the following sentence, DO NOT SOLVE:

Find four consecutive odd integers whose sum is 112.



Problem Solving Tip The easiest way to remember the inverse operation

is to “undo” what is being done to it. For example the opposite of addition is subtraction.

Keep in mind though, that equations are “balanced” – equal on both sides. What you do to

one side, you must do to the other.

Solving one-step equations

When solving an equation, you are trying to find the value(s) for the variable(s) in the equation

that makes the equation true. To do this you must isolate (get alone) the variable on one side

of the equal sign. There should be a number remaining on the other side. Then you will know

what the value of the variable is.

To isolate a variable we will need to use inverse operations.

While viewing the next examples, see if you can derive an algorithm for solving one-step

equations.

Example 3

A. Solve: 7 16x + =

Solution

Since x is being added to 7, the opposite is to subtract, but we want to move the

numbers not the variable. So subtract 7 from the left side to get x alone, but what you

do to one side, you have to do to the other. So it’s easier to say, “Subtract 7 from both

sides.”

7 167 7

9

x

x

+ =− −

=

the 7s on the left cancel since 7 7 0+ − =

Think Back

Inverse operations can be found in Module 4.

They are opposite operations of what is

happening to the variable.



Therefore the value of x that makes this equation true is 9.

Always perform a check by substituting the value you found for the variable:

( )

( )?

7 16

9 7 1616 16

+ =

+ =

=

Then you are all set; circle the answer as your last step.

B. Solve: 8 3x − =

Solution

Since x is being subtracted by 8, the opposite is to add. So add 8 to both sides.

8 38 8

11

x

x

− =+ +

=

check:

( )

( )?

8 3

11 8 33 3

− =

− =

=

Therefore the value of x that makes this equation true is 11. If for some reason the two

numbers at the end are not the same, then you haven’t found the right answer. Go back

and try again.

For example:

If you had accidentally multiplied both sides by 8 thinking that would cancel the

8 on the left, you would have gotten x = 24.

Now perform the check:

Notice the two numbers at the end are not equal to each other. Therefore the

answer that you found is incorrect. You’ll need to go back and try again.

( )

( )?

8 3

24 8 316 3

− =

− =

≠



Example 4

A. Translate and solve: Ten times a number is 120.

Solution

Let: n = a number

10 120n =

n is being multiplied by 10, the opposite is to divide. So divide by 10 on both sides.

10 12010 10

12

n

n

=

= check:

( )

( )?

10 120

10 12 120120 120

=

=

=

Therefore the value of n that makes this equation true is 12. When you have written a

“Let” statement it is often properly expected of you to go back into the “Let” statement

and write in the correct answer, like this:

Let: n = a number = 12

B. Translate and solve: Two-thirds of x is thirty-two.

Solution

2 323

x =

x is being multiplied by 23

, the opposite is to divide. So divide by 23

on both sides.

Recall that you can’t divide by fractions, so let the calculator help you. Make sure that

you use the fraction key, or recall the division algorithm: “Copy, change, flip”.

2323

2 23 3

48

x

x

=

=

check:

( )

( )?

2 323

2 48 323

32 32

=

=

=


Think Back

Coefficients go before

the variables.



C. Translate and solve: If Mary’s age is divided by 3 it would be equal to 12.

Solution

Let: a = Mary’s age

123a

=

a is being divided by 3, the opposite is to multiply. So multiply by 3 on both sides.

( ) ( )3 12 33

36

a

a

=

= check:

( )

( ) ?

123

3612

312 12

=

=

=

Therefore the value of n that makes this equation true is 36.

Let: a = Mary’s age = 36

Did you come up with an algorithm for solving one-step equations? Here’s a pretty good one:

Solving One-Step Equations:

1. Translate if necessary.

2. Get the variable alone by inverse operation, “undo it”.

3. Check. Algorithm



3. Translate (if necessary) and solve:

a. 1 12 4

x + =

b. A number subtracted by 5.5 is 7.

c. 6 37 8

n = −

d. 2.5 17.5y =



Solving two-step equations

Notice in the last section it only took one step to undo the operation and solve for the value of

the variable. Take a guess at how many steps it’s going to take to solve these equations – you

got it: 2! Although now it’s a little more important to make sure that you follow some sort of

an order to isolate the variable.

Here is an algorithm that should be able to aid you in solving these types of problems.

Example 5

A. Solve: 3 8 26x + =

Solving Two-Step Equations:

1. Translate if necessary.

2. Get the variable alone by performing 2 inverse operations, SADMEP.

3. Check the answer. Algorithm

Problem Solving Tip The easiest way to think about the order of the 2 inverse operations is by performing

PEMDAS backwards. Some have coined the phrase “SADMEP”, its the order of operations backwards.

You want to undo addition/subtraction first and then undo multiplication/division next.



3 8 268 8

3 18

x

x

+ =− −

=

( )

( )?

?

3 8 26

3 6 8 26

18 8 2626 26

+ =

+ =

+ ==

3 183 3

6

x

x

=

=

When you practice

enough, you’ll be able to

do each step, one right

after the other like this.

Solution

The ultimate goal is to get x alone. Ask yourself: “If I knew what x was, what would I

do first?” Hopefully you said, multiply it by 3 and then add it to 8. So go backwards

undoing those operations, subtract 8 from both sides and then divide both sides by 3.

It should look like this: then: check:


B. Translate and solve: Nine less than half of Ron’s age is 11.

Solution

Let: r = Ron’s age

9 112r

− =

Remember SADMEP, undo subtraction of 9 first and then undo the division of 2.

( )

( ) ?

?

check: 9 112

409 11

2

20 9 1111 11

− =

− =

− ==

Therefore the value of r that makes this equation true is 40. Thus Ron’s age is 40.

( ) ( )

9 112

9 9 Step1:add 9 to both sides

2 20 2 Step 2:multiply both sides by 22

40

r

r

r

− =

+ + ←

= ←

=



Solving Complex Equations: Combining like terms

Notice in the last section it took two steps to undo the operations and solve for the value

of the variable. With a title like “complex equation,” you know they are going to be a little

more involved than that.


Combine like terms means to group together terms that are the same (numbers with numbers / variables with variables) and are on the same side of the equal sign.

Solving Complex Equations - combining like terms:

1. Combine like terms.

2. Get the variable alone by performing inverse operations, SADMEP

if possible.

3. Check the answer in the original equation.

Algorithm

4. Translate the following and solve:

a. Twenty-four equals the quotient of x and 3− added to 4.

b. 5 7 33y − =



Example 6

A. Solve: 4 3 56a a+ =

Solution

Notice that there are two sets of a’s on the left-hand side of the equation. You should

be asking yourself, “How many a’s altogether?” There are 7 a’s, not 2. Thus:

Therefore the value of a that makes this equation true is 8.

Like terms are 5x and -3x or -27y and 6y. Their coefficients might be different but they have the same variables-19xy and -19z are not like terms. Even though they have the same

coefficient, they don’t have the same variables.

Problem Solving Tip A good way to think about combining like terms is to make a phrase to go along with it.For example, in the last problem, 4 3 56a a+ = . 4 “apples” plus 3 “apples” is a total

of 7 “apples” and that’s equal to 56.

( ) ( )

( ) ( )?

?

check: 4 3 56

4 8 3 8 56

32 24 5656 56

+ =

+ =

+ ==

4 3 567 56 Divide both sides by 7.7 7

8

a aa

a

+ =

= ←

=



B. Solve: 2 6 40p p− =

Solution

Notice that there are two sets of p’s on the left-hand side of the equation. You should

be asking yourself, “How many p’s altogether?” “I have 2 ‘plums’ and I take away 6

‘plums’ then I have 4− ‘plums’ left.” Thus:

Therefore the value of p that makes this equation true is 10− .

5. Solve: 5 9 20c c− + = − .

( ) ( )

( ) ( )?

?

?

check: 2 6 40

2 10 6 10 40

20 60 40

20 60 4040 40

− =

− − − =

− − − =

− + ==


10

p pp

p

− =−

= ← −− −

= −



Solving Complex Equations: Using the Distributive Property


Example 7

A. Solve: ( )2 4 3 46x + =

Solution

Notice that there is a 2 outside a set of parentheses and an operation taking place inside

the parentheses; in this case it is addition. This is an indicator that the distributive

property has to be used. Thus:

The Distributive Property of

Multiplication over Addition

a b c a b a c( )+ = ⋅ + ⋅

for any real numbers a b and c, ,

The Distributive Property of

Multiplication over Subtraction

a b c a b a c( )− = ⋅ − ⋅

for any real numbers a, b, and c

Think Back

Solving Complex Equations - Distributive property:

1. Use the distributive property.

2. Get the variable alone by performing inverse operations, SADMEP

if possible.


Algorithm



( )( )( )( )

( )

?

?

check: 3 2 18

3 8 2 18

3 6 1818 18

− =

− =

=

=

( )( ) ( )

( )( )( )( )( )

( )

?

?

?

2 4 3 46 Distribute the 2 to each of the terms inside.

2 4x 2 3 468 6 46 6 is being added, so subtract it.

6 6

8 40 "Undo" the 8 by division.8 8

5 check: 2 4 3 46

2 4 5 3 46

2 20 3 46

2 23 4646 46

x

x

x

x

+ = ←

+ =

+ = ←− −

= ←

= + =

+ =

+ =

=

=


B. Solve: ( )3 2 18x − =

Solution

Notice that there is a 3 outside a set of parentheses and an operation taking place inside

the parentheses; in this case it is subtraction. This is an indicator that the distributive

property has to be used. Thus:

( )( ) ( )

3 2 18 Distribute the 3 to each of the terms inside.

3 3 2 183 6 18 6 is being subtracted, so add it.

6 6


8

x

xx

x

x

− = ←

− =

− = ←+ +

= ←

=




( ) ( )

( ) ( )?

?

check: 7 3 8 24

7 8 3 8 8 24

56 24 8 2424 24

− − =

− − =

− − ==

When the distributive property and combining like terms are being used together in an

equation, there is a new set of rules to follow in solving these types of equations.

Example 8

A. Solve: 7 3 8 24x x− − =

Solution

First determine if you need to use the distributive property. Since there are no sets of

parentheses, it is not necessary to use it; move on to combine like terms. There appears

to be 2 x’s so, “If I have 7 x’s and I take away 3 x’s, how many are left?” There should

be 4 x’s left. Notice that you can not do anything with the minus 8, since there are no

other numbers on the left that don’t have x’s attached to them (24 is on the other side of

the equal sign so it does not count); move on to solving a 2-step equation. Thus:

7 3 8 24 Combine the 's.

4 8 24 Notice now it's a 2-step eqn, 8 is being subtracted, so add it.8 8


8

x x xx

x

x

− − = ←− = ←+ +

= ←

=


Solving Complex Equations

(When distribution and combine like terms are involved):

1. Use the distributive property whenever parentheses are involved.

2. Then combine like terms if possible.

3. Get the variable alone by performing inverse operations, SADMEP.


Algorithm



B. Solve: ( )5 3 4 28x x+ + =

Solution

First determine if you need to use the distributive property. Since there is a set of

parentheses it is necessary to do this first, and then move on to combine like terms.

Don’t be fooled by the 2 x’s. They can not be combined yet since one of them is inside

a set of parentheses. Distribute the 3 to both terms inside the parentheses. Then there

appears to be 2 x’s so, “If I have 5 x’s and I add 3 x’s, how many are there?” There

should be 8 x’s altogether. Notice that you can not do anything with the plus 12, since

there are no other numbers that don’t have x’s attached to them (28 is on the other side

of the equal sign, so it does not count). Thus:

( )5 3 4 28 Distribute first.5 3 12 28 Now combine like terms.

8 12 28 "Undo" the addition of 12 by subtracting.12 12


2

x xx x

x

x

x

+ + = ←

+ + = ←+ = ←− −

= ←

=

( ) ( )( )( ) ( )( )

( )

?

?

?

check: 5 3 4 28 Replace variables with parentheses.

5 2 3 2 4 28 Substitute the value you found for and solve.

10 3 6 28 As long as you obey PEMDAS,

10 18 28 you can do multiple steps at once. 28 28

x

+ + = ←

+ + = ←

+ = ←

+ ==




Solving Complex Equations: With Variables on Both Sides

The title indicates exactly that, solving equations where there will be variables on both

sides of the equal sign. You’ll follow a very similar pattern to what you’ve already learned,

keeping in mind that the ultimate task is to get all the variables on one side of the equal sign

and all the numbers on the other.

Solving Complex Equations - Variables on both sides:


2. Then combine like terms if possible.


4. Move the variable to one side of the equation and the numbers to the

other.


Algorithm

6. Solve: 3 7 5y y− + = .

7. Solve: ( )22 2 3 5 66x x+ + =



Example 9

A. Solve: 5 12 9 16n n+ = −

Solution


parentheses, it is not necessary to use it. Move on to combine like terms. There

appears to be 2 n’s but 5n is on one side of the equal sign and 9n is on the other,

therefore you can not combine like terms. So then you need to move a set of n’s to one

side, but which one do you move? In the past students have found it easier to move the

smaller of the two choices, so ask yourself, “Which one is smaller? 5n or 9n?”

Hopefully you said 5n. “How do you move a positive 5n?” By subtracting it,

remember what you do to one side you have to do to the other. Notice that it cancels

the 5n on the left and then there are 4 n’s left on the right. Then you are back to a 2-

step equation. Thus:

5 12 9 16 Move the smaller of the two 's.5 5

12 4 16 Notice now it's a 2-step equation.16 16 16 is being subtracted, so add it.


7

n n nn n

n

n

n

+ = − ←− −

= − ←+ + ←

= ←

=

Here is where the check becomes VERY important, because it’s the first time that you

have to check both sides of the equation to ensure that you get the same result.

( ) ( )

( ) ( )?

?

check: 5 12 9 16

5 7 12 9 7 16

35 12 63 1647 47

+ = −

+ = −

+ = −=

Therefore the value of n that makes this equation true is 7.



B. Solve: 4 3 2 8 3x x x x− + = + −

Solution


parentheses, it is not necessary to use it; move on to combine like terms. There appears

to be a whole lot of x’s. You should be thinking of combining like terms because there

appears to some x’s on the left and some x’s on the right. After that step you should be

trying to isolate the variable to one side of the equation and move the numbers to the

other. Thus:

Therefore the value of x that makes this equation true is 6− .

Problem Solving Tip For the check, as long as PEMDAS (order of operations) is being followed you can perform multiple steps at the same time. It also conveniently saves papers and cuts

down on the amount of calculations you have to perform.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )?

?

check: 4 3 2 8 3

4 6 3 2 6 8 6 3 6

24 3 12 48 3 639 39

− + = + −

− − + − = − + − −

− − + − =− + +− = −

4 3 2 8 3 No ( ), so combine like terms on the same side.6 3 7 3 4 2 6 ,8 76 3 7 3 Move the smaller of the 2 's.6 6

3 3 You moved the 's to the right, so move the numbers to the

x x x xx x x x x x x xx x xx x

x x

− + = + − ←− = + ← + = − =− = + ←

− −

− = + ← left.3 3

6 x

− −

− =



C. Solve: ( ) ( )4 3 2 6x x+ = −

Solution


parentheses, it is necessary to distribute. Distribute the 4 to the x + 3 on the left and

distribute the 2 to the x – 6 on the right. Determine if you need to combine like terms.

Since there are no more x’s on the same side of the equal sign, it is not necessary to do

this. Think about trying to isolate the variable to one side of the equation and move the

numbers to the other. Thus:

Therefore the value of x that makes this equation true is 12− .

( )( ) ( )( )( )( ) ( )( )

( ) ( )

?

?

check: 4 3 2 6

4 12 3 2 12 6

4 9 2 1836 36

+ = −

− + = − −

− = −

− = −

( ) ( )( ) ( ) ( ) ( )

4 3 2 6 Distribute.

4 4 3 2 2 64 12 2 12 No more 's on the same side, so can't combine.2 2 Move the smaller of the 2 's.

2 12 12 You moved the 's to the left so move the numbers to the right

x x

x xx x xx x x

x x

+ = − ←

+ = −

+ = − ←− − ←

+ = − ← .12 12


12

x

x

− −

−= ←

= −



8. Solve and check the following:

a. 1 16 182 4

n n+ = − +

b. 6 12 3 2 18x x x+ − = +

c. ( )6 3 4 8m m+ = +



Solving Multi-Step Equations

You have now worked with many different types of equations and have learned about different

steps that must be used to solve them: adding like terms, using inverse operations, using the

distributive property, and working with variables on both sides of the equation. Now you will

put all those skills together to solve equations that involve multiple steps. Here is an algorithm

to follow that should ensure continued success:

Example 10

A. Solve: ( ) ( )8 3 1 4 2x x x− − = +

Solution


parentheses, it is necessary to distribute. Distribute the 3− to the x – 1 on the left and

distribute the 4 to the x + 2 on the right. Determine if you need to combine like terms.

Since there are more x’s on the same side of the equal sign, it is necessary to do this on

the left. Then think about trying to isolate the variable to one side of the equation and

move the numbers to the other. Thus:

Solving Multi-Step Equations:

1. Translate when necessary.


3. Then combine like terms (if on the same side of the equal sign).

4. Get the variables on one side of the equal sign and numbers on the

other.

5. Make sure you undo addition/subtraction first.

6. Undo multiplication/division last.

7. Check you answer in the original equation.

Algorithm



( ) ( )

( ) ( ) ( ) ( )8 3 1 4 2 Use the distributive property.

8 3 3 1 4 4 28 3 3 4 8 Combine like terms on the left.

5 3 4 8 At this point you should be able to start combining steps.4 3 4 3 Move the 4 to the

x x x

x x xx x x

x xx x x

− − = + ←

− − − = +

− + = + ←+ = + ←

− − − − ← left and the 3 to the right.

5x =

( ) ( )( ) ( )( )( ) ( )( ) ( )( )

( ) ( )

?

?

?

check: 8 3 1 4 2

8 5 3 5 1 4 5 2

40 3 4 4 7

40 12 2828 28

− − = +

− − = +

− =

− ==


B. Translate and Solve:

The perimeter of a rectangle is 48cm. If the length is twice the width, find the dimensions of

the rectangle.

Solution

First draw a picture; it generally helps to organize thoughts. Also don’t forget “Let”

statements. Thus:

Let: x = width

2x = length

Length = 2x

Length = 2x

Width = x Width = x



( ) ( ) ( )( ) ( ) ( )

2 2 Recall perimeter formula.2 2

48 2 2 2 Substitute the values.48 4 2 Distribute and then combine like terms.48 6 "Undo" the 6.6 6

8cm width Solve for the dimensions.16cm 2 length

P l w

x xx xx

xx

= + ←

= +

= + ←

= + ←

= ←

= = ←= =

( ) ( )

( ) ( )?

?

check: 2 248 2 2

48 2 16 2 8

48 32 1648 48

P l w= +

= +

= +

= +=


C. Translate and Solve:

Recall consecutive integer questions:

Find three consecutive positive even integers such that four times the first decreased by the

second is twelve more than twice the third.

Solution Let: x = 1st consecutive even integer






( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

st nd rd4 1 2 2 3 124 2 12 replace the phrases with parentheses

4 2 2 4 12 substitute in the variable values

4 2 2 2 4 12 distribute where necessary4 2 2 8 12 simplify

3 2 2 20 combine

x x x

x x xx x x

x x

× − = × +

− = + ←

− + = + + ←

− + − = + + ←

− − = + + ←− = + ←

st

nd

rd

like terms on each side of the equal sign22 move 's to one side and number's to the other

1 consecutive even integer 222 2 consecutive even integer 244 3 consecutive even integer 26

x x

xxx

= ←

= =

+ = =

+ = =

( ) ( ) ( )

( ) ( ) ( )

st nd rd

?

?

check: 4 1 2 2 3 124 2 12

4 22 24 2 26 12

88 24 52 1264 64

× − = × +

− = +

− = +

− = +=





a. ( )7 6 5 7r r− − =

b. The larger of two numbers is 8 more than three times the smaller.

Their sum is 28. Find the numbers.

c. Find three consecutive even integers such that the sum of the smallest and twice the

second is twenty more than the third.



Translating and Graphing Inequalities

You have worked with many different types of equations; now it’s time to take a look at

inequalities.

Here is a chart of words or phrases associated with the inequality symbols:

< ≤ ≥ >

Less than Less than or equal to

Greater than or equal to More than

At most At least

Maximum value Minimum value

Example 11: Translate and graph the solution set for the following sentences:

A. A number q is less than 2.

Solution

2q <

5 4 3 2 1 0 1 2 3 4 5− − − − −

Since 2 is not less than 2 it cannot be part of the solution, so the dot is not shaded. This

is called an open dot. Also notice that it is shaded on the number line to the left of 2

because all of those numbers are less than 2. Keep in mind that there are infinitely

many solutions that are less than 2 which is why the arrow has been drawn to indicate

that.

Open dot means the number is not part of the solution set, thus it is not shaded.

Inequality is a mathematical sentence that compares two unequal expressions.



B. A number n is greater than or equal to –4.

Solution

4n ≥ −

5 4 3 2 1 0 1 2 3 4 5− − − − −

Since –4 is equal to –4 it can be included in the solution which is why the dot is shaded.

This is called a closed dot.

C. A number t is at most –1.

Solution

1n ≤ −

5 4 3 2 1 0 1 2 3 4 5− − − − −

Since –1 is equal to –1 it can be included in the solution which is why there is a closed

dot.

Closed dot means the number is part of the solution set, thus it is shaded.

Problem Solving Tip

Keep in mind that when graphing on a number line the most important features on it are the arrows, showing negative and positive infinity; zero, which is the starting

point of the number line and then the number you are talking about.



D. A number m is greater than 5.

Solution

5m >

5 4 3 2 1 0 1 2 3 4 5− − − − −

Since 5 is not greater than 5 it cannot be part of the solution, so the dot is not shaded.

Solving Inequalities

You can solve inequalities the same way that you solve equations, by using inverse operations.

The only difference is that you are not finding one solution for the variable; you are finding

infinitely many solutions, which is why you graph the solution on the number line.

Example 12: Solve and graph the solution set for the following problems.

A. 7 8x + >

Solution 7 8 7 is being added so subtract.

7 7

1 The solution is any number greather than 1.

x

x

+ > ←− −

> ←

Problem Solving Tip Notice that anytime you can say “equal to” the dot is closed.

Otherwise, the dot is open. Therefore, the open dots are < and > and ≠ .

The closed dots are and≤ ≥ and =.



Perform the operations necessary to get the variable alone. Then with inequalities you

should perform several checks to ensure that you are graphing (shading) in the correct

direction.

( )( ) ( ) ( )

Check: 7 8 Check any number greater than 1.

2 7 8 5 7 8 0 7 8 etc...9 8 12 8 7 8 Therefore you don't shade to the left.

+ > ←

+ > + > + >

> > > ←/ Graph your solution.

5 4 3 2 1 0 1 2 3 4 5− − − − −

Since 1 is not greater than 1 it cannot be part of the solution, so the dot is not shaded.

B. 2 3x − ≤

Solution

2 3 2 is being subtracted, so add.2 2

5 The solution is any number less than or equal to 5.

x

x

− ≤ ←+ +

≤ ←

Perform the operations necessary to get the variable alone. Then, with inequalities you


direction.

( )( ) ( ) ( )

Check: 2 3 Check any number less than or equal to 5.

1 2 3 6 2 3 9 2 3 etc...1 3 8 8 7 3 Therefore you don't shade to the right.

− ≤ ←

− ≤ − − ≤ − ≤

− ≤ − ≤ ≤ ←/ Graph your solution.

5 4 3 2 1 0 1 2 3 4 5− − − − −

Since 5 is equal to 5 it can be included in the solution, which is why there is a closed

dot.



Problem Solving Tip Notice that in the first example the

closed end of the inequality sign points at the x in both statements.

Notice in the second example that the open end of the inequality sign points

at the n in both statements.

Doing this will ensure that you always maintain the inequality value.

The examples you have done so far have had the variable on the left. Inequalities can also

have the variables on the right. However, many students get confused when they try to read

and interpret inequalities that are written with the variable on the right. To make things easier,

inequalities can be rewritten so the variable is on the left. To do this, switch the location of the

numbers and variables and reverse the inequality sign.

Example 13: Rewrite the variable on the left and state the inequality verbally.

A. 4 x>

Solution

4x < verbal: “the set of all real numbers less than 4”

B. 3 n− ≤

Solution

3n ≥ − verbal: “the set of all real numbers greater than or equal to –3”



10. Solve and graph the solution set. Rewrite the inequality, if

necessary, to ensure the variable is on the left:

a. 7 16y + ≥

0

b. 12 6x− > −

0

c. 4 is less than twice a number.

0



So far you have not run into any problems with regard to solving inequalities and graphing

them. The inverse operation has worked thus far, however when you divide or multiply by a

negative number it will require a little more work to get to the appropriate answer.

Example 14: Solve and graph the solution set for the following problems.

A. 2 6x− >

Solution

2 6 "Undo" the 2 with division.2 2

3 The solution appears to be any number greather than 3.

x

x

−> ← −

− −> − ← −

Perform the operations necessary to get the variable alone. Then, with inequalities you


direction.

( )( ) ( ) ( )

Check: 2 6 Check any number greather than 3.

2 0 6 2 5 6 2 9 6 etc...0 6 10 6 18 6 This is the only one that works.

− > ← −

− > − > − − >

> − > > ←/ /

The problem is that when you divide or multiply by a negative number, you are

changing the sign of the numbers. You make positive numbers negative and negative

numbers positive … the opposite of what they are. So, to make an equivalent

inequality, you must change the direction of the inequality sign.

Therefore, the answer should be: 3x < −

Graph your solution.

5 4 3 2 1 0 1 2 3 4 5− − − − −



B. 1 52

n− ≤

Solution

15 12 "Undo" the multiplication of .1 1 2

2 210 Remember to change the direction of the inequality sign.10 The solution is any number greater than or equal to 10.

n

nn

−≤ ← −

− −

≤ − ←≥ − ← −

( )

( ) ( ) ( )

1Check: 5 Check any number greather than or equal to 10.21 1 14 5 6 5 12 5 etc...2 2 2

2 5 3 5 6 5 Therefore, you don't shade to the left.

− ≤ ← −

− − ≤ − ≤ − − ≤

≤ − ≤ ≤ ←/ Graph your solution.

10 8 6 4 2 0 2 4 6 8 10− − − − −

Problem Solving Tip When you divide or multiply by a negative number, you must reverse

the direction of the inequality sign.



C. 2 1 4x + ≤

Solution

2 1 4 "Undo" the addition first.1 1

2 3 "Undo" multiplication.2 2

1 11 The solution is any number less than or equal to 1 .2 2

x

x

x

+ ≤ ←− −

≤ ←

≤ ←

( )

( ) ( ) ( )

1Check: 2 1 4 Check any number less than or equal to 1 .2

2 0 1 4 2 10 1 4 2 8 1 4 etc...0 1 4 20 1 4 16 1 4

1 4 19 4 17 4 Therefore, you don't shade to the right.

+ ≤ ←

+ ≤ − + ≤ + ≤

+ ≤ − + ≤ + ≤≤ − ≤ ≤ ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −

D. 3 5 11n− + <

Solution

3 5 11 "Undo" the addition first.5 5


2 Remember to change the direction of the inequality sign since you divided by a .

2 The solution is any number greater tha

n

n

n

n

− + < ←− −

−< ←

− −< − ←

−> − ← n 2.−



( )( ) ( ) ( )

Check: 3 5 11 Check any number greater than 2.

3 0 5 11 3 4 5 11 3 5 5 11 etc...0 5 11 12 5 11 15 5 11

5 11 7 11 20 11 Therefore, you don't shadeto the left.

− + < ← −

− + < − + < − − + <

+ < − + < + << − < < ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −

In case you haven’t developed your own algorithm for solving inequalities, here it is:

Solving One and Two-step Inequalities:

1. Translate when necessary.


3. Undo addition/ subtraction first.

4. Undo multiplication/ division last.

5. Remember that you must reverse the direction of the inequality sign

if you multiply or divide by a negative number.

6. Check and graph the solution set.

Algorithm



11. Solve and graph the solution set. Rewrite the inequality, if

necessary, to ensure the variable is on the left:

a. 4 3 9x − >

0

b. 25 22d≤ +

0

c. Negative five times a certain number is less than fifty-five. Find the numbers.

0



Review 1. Highlight the following words and their definitions:

a. consecutive integers

b. consecutive even integers

c. consecutive odd integers

d. combine like terms

e. inequality

f. open dot

g. closed dot

2. Go back and highlight the algorithms for the following:

a. consecutive integers

b. consecutive even integers

c. consecutive odd integers

d. solving one-step equations

e. solving two-step equations

f. solving complex equations – combining like terms

g. solving complex equations – distributive property

h. solving complex equations – when combine like terms and distributive are

being used together

i. solving complex equations – variables on both sides

j. solving multi-step equations

k. solving one and two-step inequalities



3. Go back and highlight the following boxes:

a. The “Think Back” box on the difference between expression and equation.

b. The “Think Back” box with the definition for variable.

c. The “Fact” box about “Let” statements.

d. The “Think Back” box with the distributive property.

e. Any “Problem Solving Tip” boxes you feel will help you be successful in this

module.

4. Write one question that you would like to ask your mentor or one new thing you learned in

this module.



Practice Problems Equations

Directions: Write your answers in your math journal.

Label these exercises Equations: Set A, Set B, Set C, Set D, Set E, and Set F.

Set A

1. Translate, do not solve:

a. The product of negative four and a number x.

b. Twice your weight.

c. Eight more than 3 times a number.


a. Sixty-three is equal to the product of seven and number.

b. A number z decreased by three is thirty-nine.

c. Two-thirds of a number x is equal to negative eighteen.

d. x decreased by negative nine equals three.

e. Thirteen less than a number y is forty-two.


a. Find two consecutive integers such that the sum is negative one hundred and five.

b. Two sides of a triangle are equal in length. The length of the third side exceeds the

others by three centimeters. The perimeter of the triangle is ninety-three

centimeters.

c. The smaller of two numbers is two less than the larger. If five times the smaller is

decreased by four times the larger the result is ten.



Set B

1. Solve “Practice Problems” from Set A numbers 2 a – e.

2. Solve: 358

y− =

3. Solve: 54w

= −

4. Solve: 1 32 14 4

a + = −

5. Solve: 23 109x+ =

Set C

1. Solve: 3 1 8x − =

2. Solve: 5 13x

− + = −

3. Solve: 1 3 82

h − =

4. Solve: 37 5 2c= − +

5. Solve: 6 14 2x= −

Set D

1. Solve “Practice Problems” from Set A numbers 3 a – c.

2. Solve “Example” numbers 2 A and B on page 9.

3. Solve: ( )4 2 8 56g g+ + = .

4. Solve: 3 14 4 180x x x− + + + = .

5. Solve: ( )12 3 2 10x− = − .



Set E

1. Solve: 9 7 6 52y y+ = − + .

2. Solve: ( )3 1 19 5a a+ = − .

3. Solve: ( )3 4 1 2 23x x+ − = .

4. Solve: ( ) ( )3 2 4 10x x+ = − .

5. Solve: ( )3 6 20 25x + + − = .

6. Solve: The length of a rectangle exceeds its width by five meters. The perimeter is sixty-

six meters. Find the dimensions.

7. Find two consecutive integers such that four times the larger is twenty-three more than

three times the smaller.

8. The perimeter of a square is three times larger than the perimeter of an equilateral triangle.

If the length of one side of the square is ten meters more than the length of the side of the

triangle, find the dimensions of the triangle and square.

Set F: Solve and graph the solutions (make your own number lines).

1. 11 9e − ≥

2. 5 35t− >

3. 53u

≤

4. 7 10a + <

5. The product of three and a number n is less than or equal to 15.

6. The quotient of a number divided by negative six is greater than one.



1. Translate the following:

a. Three times a number x is eighteen.

3x = 18

b. Seventeen less than a number.

Let: n = a number

n – 17

c. Twenty-four is a number m divided by 3

243m

=

7. Carol weighs three times as much as Sue. Both weights are whole numbers. The sum of

their weights is at most one hundred and sixty pounds. Find the greatest possible weight in

pounds for each girl.

8. 11 12 45x + ≤

9. 2 4 63

k− + > −

10. In the set of positive integers, what is the solution set of the inequality: 2 3 5x − < ?

a. {0,1,2,3} b. {1,2,3} c. {0,1,2,3,4} d. {1,2,3,4}



2. Find four consecutive odd integers whose sum is 112.

Let: x = 1st consecutive odd integer

x + 2 = 2nd consecutive odd integer

x + 4 = 3rd consecutive odd integer

x + 6 = 4th consecutive odd integer

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

st nd rd th1 odd# + 2 odd# + 3 odd# + 4 odd# 112112

2 4 6 112x x x x

=

+ + + =

+ + + + + + =


a.

b. A number subtracted by 5.5 is 7.

Let: n = a number

c.

( )?

1 1check: 2 4

1 1 14 2 4

1 14 4

+ =

⎛ ⎞ + =⎜ ⎟⎝ ⎠

=

1 12 41 12 2

14

x

x

+ =

− −

= −

( )

( )?

check: 5.5 7

12.5 5.5 77 7

− =

− =

=

5.5 75.5 5.5

12.5

n

n

− =+ +

=

( )?

6 3check: 7 8

6 7 37 16 8

3 38 8

= −

⎛ ⎞− = −⎜ ⎟⎝ ⎠

− = −

6 37 86 67 7

716

n

n

−=

= −



d.

4. Translate the following and solve:

a. Twenty-four equals the quotient of x and 3− added to 4.

24 43x

= +−

then

b.

5. Solve:

( )

( )?

check: 2.5 17.5

2.5 7 17.517.5 17.5

=

=

=

2.5 17.52.5 2.5

7

y

y

=

=

( ) ( )

24 43

4 4

3 20 33

60

x

x

x

= +−

− −

− = −−

− =

( )

( )?

?

check: 24 4360

24 43

24 20 424 24

= +−−

= +−

= +=

5 7 337 7

5 405 5

8

y

y

y

− =+ +

=

=

( )

( )?

?

check: 5 7 33

5 8 7 33

40 7 3333 33

− =

− =

− ==


5

c cc

c

− + = −−

= ←

= −

( ) ( )

( ) ( )?

?

check: 5 9 20

5 5 9 5 20

25 45 2020 20

− + = −

− − + − =−

+ − =−− = −



6. Solve:

7. Solve:

3 7 5 combine like terms, 3 44 7 5 undo subtraction with addition

+7 +7

4 12 undo multiplication with division4 4

3

y y y y yy

y

y

− + = ← + =− = ←

= ←

= ( ) ( )

( ) ( )

( )

( )

?

?

?

check: 3 7 5

3 3 7 3 5

9 7 3 5

2 3 55 5

− + =

− + =

− + =

+ =

=

( )( ) ( )

22 2 3 5 66 distribute the 2 to each term in the inside

22 2 3 2 5 66 perform multiplication for the distributive property22 6 10 66 combine like terms

28 10 66 undo addition with subtraction10 10

2

x x

x xx x

x

+ + = ←

+ + = ←

+ + = ←+ = ←− −

8 56 undo multiplication with division28 28

2

x

x

= ←

= ( ) ( )( )( ) ( )( )

( )

( )

?

?

?

?

check: 22 2 3 5 66

22 2 2 3 2 5 66

44 2 6 5 66

44 2 11 66

44 22 6666 66

+ + =

+ + =

+ + =

+ =

+ ==



8. Solve and check the following:

a.

b.

←Move the smaller of the two n’s,

that is definitely the negative one.

←Since it is negative, the opposite is positive.

←The opposite of multiplication is division;

don’t forget that you can’t divide by fractions.

←Use the calculator to assist you with any

troublesome calculation.

( ) ( )

( ) ( )?

?

1 1check: 6 182 4

1 116 6 16 182 4

8 6 4 1814 14

+ = − +

+ =− +

+ =− +=

1 16 182 41 14 43

1843 34 4

16

n n

n n

n

n

+ = − +

+ +

=

=

6 12 3 2 18 Combine like terms on the left.3 12 2 18 Move the smaller of the two 's.2 2

12 18 's went to the left so move #'s to the right.12 12

6

x x xx x xx x

x x

x

+ − = + ←+ = + ←

− −

+ = ←− −

= ( ) ( ) ( )

( ) ( ) ( )?

?

check: 6 12 3 2 18

6 6 12 3 6 2 6 18

36 12 18 12 1830 30

+ − = +

+ − = +

+ − = +=



c.


a.

b. The larger of two numbers is 8 more than three times the smaller.

Their sum is 28. Find the numbers.

Let: x = smaller #

3x + 8 = larger #

(continued on next page)

( )( ) ( )

6 3 4 8 Distribute the 6 on the left.

6 6 3 4 86 18 4 8 Move the smaller of the two 's.

4 4

2 18 8 's were moved to the left so move #'s to the right.18 18

2 102 2

5

m m

m mm m m

m m

m m

m

m

+ = + ←

+ = +

+ = + ←− −

+ = ←− −

−=

= −

( )( ) ( )( )( ) ( )

( )

check: 6 3 4 8

6 5 3 4 5 8

6 2 20 812 12

+ = +

− + = − +

− = − +

− = −

( ) ( )( )( ) ( )( )

( )

( )

?

?

?

check: 7 6 5 7

7 2 6 2 5 7

14 12 5 7

14 7 77 7

− − =

− − =

− − =

− =

=

( )( ) ( )

7 6 5 7 Distribute the negative.

7 6 5 7 Combine like terms on the left.5 7 Move the 5 to the right by subtraction.

2

r r

r rr

r

− − = ←

− − − = ←

+ = ←=



( ) ( )( ) ( )

small# larger# 28 Write it in words first.28 Replace phrases with parentheses.

3 8 28 Substitute the variable values in.4 8 28 Combine like terms, no distributive.

8 8 "Undo" addition with sub

x xx

+ = ←

+ = ←

+ + = ←

+ = ←− − ← traction.

4 20 "Undo" multiplication with division.4 4

5

x

x

= ←

=

x = smaller # = 5

3x + 8 = larger # = 3(5) + 8 = 23

( ) ( )

( ) ( )?

check: small# larger# 2828

5 23 2828 28

+ =

+ =

+ =

=

c. Find three consecutive even integers such that the sum of the smallest and twice the

second is twenty more than the third.

Let: x = 1st consecutive even integer






( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

st nd rd1 2 2 3 202 20 replace the phrases with parentheses

2 2 4 20 substitute in the variable values

2 2 2 4 20 distribute where necessary2 4 24 simplify3 4 24 combine like terms

x x x

x x xx x x

x x

+ × = +

+ = + ←

+ + = + + ←

+ + = + + ←

+ + = + ←+ = + ←

st

nd

rd

on each side of the equal sign2 20 move 's to one side and #'s to the other

10

1 consecutive even integer 102 2 consecutive even integer 124 3 consecutive even integer 14

x xx

xxx

= ←=

= =

+ = =

+ = =

( ) ( ) ( )

( ) ( ) ( )

st nd rd

?

?

check: 1 2 2 3 202 20

10 2 12 14 20

10 24 14 2034 34

+ × = +

+ = +

+ = +

+ = +=

10.

a.

7 16 "Undo" the addition with subtraction.7 7

9 The soultion is any number greater than or equal to 9.

y

y

+ ≥ ←− −

≥ ←

( )( ) ( ) ( )

Check: 7 16 Check any number greater than or equal to 9.

9 7 16 20 7 16 0 7 16 etc...16 16 27 16 7 16 Therefore you don't shade

to the left.

+ ≥ ←

+ ≥ + ≥ + ≥

≥ ≥ ≥ ←/


10 8 6 4 2 0 2 4 6 8 10− − − − −



b.

12 6 "Undo" the subtraction with addition.6 6

6 Switch the inequality around so that the variable is on the left.6 The solution is any number less than 6.

x

xx

− > − ←+ +

− > ←< − ← −

( )( ) ( ) ( )

Check: 12 6 Check any number less than 6.

12 10 6 12 7 6 12 0 6 etc...12 16 12 13 12 6 Thus, no shade to the right.

− > − ← −

− > − − − > − − − > −

− ≥ − − ≥ − − > − ←/


10 8 6 4 2 0 2 4 6 8 10− − − − −

c. 4 is less than twice a number.

Let: a number4 24 2 "Undo" the 2 with division.2 22 Switch the inequality around so that the variable is on the left.

2 The solution is any number greater than 2.

xx

x

xx

=<

< ←

< ←> ←

( )( ) ( ) ( )


4 2 5 4 2 10 4 2 0 etc...4 10 4 20 4 0 Thus, no shade to the left.

< ←

< < <

< < < ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −



11. Solve and graph the solution set. Rewrite the inequality, if necessary, to ensure the

variable is on the left:

a.

4 3 9 "Undo" the subtraction first.3 3


3 The solution is any number greater than 3.

x

x

x

− > ←+ +

> ←

> ←

( )( ) ( ) ( )

Check: 4 3 9 Check any number greater than 3.

4 4 3 9 4 10 3 9 4 0 3 9 etc...16 3 9 40 3 9 0 3 9

13 9 37 9 3 9 Therefore you don't shade to the left.

− > ←

− > − > − >

− > − > − >> > − > ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −

b.

25 22 "Undo" the addition first.22 22

3 Remember to rewrite the inequality so variable is on the left.3 The solution is any number greater than or equal to 3.

d

dd

≤ + ←− −

≤ ←≥ ←

( )( ) ( ) ( )


25 4 22 25 8 22 25 0 22 etc...25 26 25 30 25 22 Therefore you don't shade

to the left.

≤ + ←

≤ + ≤ + ≤ +

≤ ≤ ≤ ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −



c. Negative five times a certain number is less than fifty-five. Find the numbers.

Let: a number5 555 55 "Undo" the 5 with division.5 5

11 ince you divided by a , change the direction of the inequality.11 The solution is any number greater than 11.

xx

x

x Sx

=− <−

< ← −− −

< − ← −> − ← −

( )( ) ( ) ( )



− < ← −

− < − − < − − <

< < < ←/


14 12 9 6 3 0 3 6 9 12 14− − − − −

Answers to Practice Problems

Set A


a. The product of negative four and a number x.

4x−

b. Twice your weight.

Let: my weight2

ww

=

c. Eight more than 3 times a number.

Let: a certain number3 8nn=+




a. Sixty-three is equal to the product of seven and number.

Let: a certain number63 7n

n==

b. A number z decreased by three is thirty-nine.

3 39z − =

c. Two-thirds of a number x is equal to negative eighteen.

2 183

x = −

d. x decreased by negative nine equals three.

( )9 3x − − =

e. Thirteen less than a number y is forty-two.

13 42y − =


a. Find two consecutive integers such that the sum is negative one hundred and five.

Let: x = 1st consecutive integer



( ) ( )( ) ( )

st nd1 consecutive integer + 2 consecutive integer 105105

1 105x x

= −

+ = −

+ + = −



b. Two sides of a triangle are equal in length. The length of the third side exceeds the

others by three centimeters. The perimeter of the triangle is ninety-three

centimeters.

Let: x = 1st side

x = 2nd side

x + 3 = 3rd side


( ) ( ) ( )

st nd rd1 side + 2 side + 3 side 933 93

3 3 93x x x

or x

=

+ + + =

+ =

c. The smaller of two numbers is two less than the larger. If five times the smaller is

decreased by four times the larger the result is ten.

Let: x = larger number

x – 2 = smaller number


( ) ( )5 smaller 4 larger 10

5 2 4 10x x× − × =

− − =

Set B

1. Solve “Practice Problems” from Set A numbers 2 a – e.

a.

b.

( )

( )?

Let: a certain number63 77 79 check: 63 7

63 7 9 63 63

n

n

n

=

=

= =

=

=

( )

( )?

3 393 3

42 check: 3 39

42 3 3939 39

z

z

− =+ +

= − =

− =

=



c.

d.

e.

2.

3.

( )

( )?

3 2 3182 3 2

227 check: 183

2 27 183

18 18

x

x

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − = −

− =−

− = −

( )

( ) ( )

( ) ( )?

9 3 rewrite it so you don't have double signs.9 3

9 9

6 check: 9 3

6 9 33 3

xx

x

− − = ←

+ =− −

= − − − =

− − − =

=

( )

( )?

13 4213 13

55 check: 13 42

55 13 4242 42

y

y

− =+ +

= − =

− =

=

( )?

35 8 Divide both sides by 5.5 5

3 3check: 540 8

3 3540 8

3 38 8

y

y

−= ← −

− −

= − − =

⎛ ⎞− − =⎜ ⎟⎝ ⎠

=( ) ( )

( )

( ) ?

4 5 44

20 check: 54

205

45 5

w

w

= −

= − = −

−=−

− = −



4.

5.

Set C

1.

( )

( )?

1 32 14 4

1 12 24 4

1 34 check: 2 14 41 34 2 14 4

3 31 14 4

a

a

+ = −

− −

= − + = −

− + =−

− = −

( )

( )?

23 10923 23

86 check: 23 109

23 86 109109 109

x

x

+ =− −

= + =

+ =

=

( )

( )?

?

3 1 8 "Undo" subtraction.1 1


3 check: 3 1 8

3 3 1 8

9 1 88 8

x

x

x

− = ←+ +

= ←

= − =

− =

− ==



2.

3.

4.

( ) ( )

( )

( ) ?

?

5 1 "Undo" the negative 5.3

5 5

3 4 3 "Undo" division.3

12 check: 5 13

125 1

3

5 4 11 1

x

x

x

− + = − ←

+ +

= ←

= − + = −

− + =−

− + =−− = −

( )

( )?

?

1 3 8 "Undo" subtraction.2

3 3

1112 "Undo" multiplication.1 1

2 2122 check: 3 82

1 22 3 82

11 3 88 8

h

h

h

− = ←

+ +

= ←

= − =

− =

− ==

( )

( )?

?

37 5 2 "Undo" addition.2 2

35 5 "Undo" multiplication.5 57 check: 37 5 2

37 5 7 2

37 35 237 37

c

c

c

= − + ←− −

−= ←

− −− = = − +

=− − +

= +=



5.

Set D

1. Solve “Practice Problems” from Set A numbers 3 a – c.

a. Let: x = 1st consecutive integer



( ) ( )( ) ( )

st nd1 consecutive integer + 2 consecutive integer 105105

1 1052 1 105

1 1

2 1062 2

53

x xx

x

x

= −

+ = −

+ + = −

+ = −− −

−=

= −

x = 1st consecutive integer = 53−

x + 1 = 2nd consecutive integer = 52−

( ) ( )

( ) ( )

st nd

?

check:1 consecutive integer + 2 consecutive integer 105

105

53 52 105105 105

= −

+ = −

− + − =−

− = −

( )

( )?

?

6 14 2 "Undo" positive 14.14 14

8 2 "Undo" multiplication.2 24 check: 6 14 2

6 14 2 4

6 14 86 6

x

x

x

= − ←− −

− −= ←

− −= = −

= −

= −=



b. Let: x = 1st side

x = 2nd side

x + 3 = 3rd side


( ) ( ) ( )

st nd rd1 side + 2 side + 3 side 933 93

3 3 933 3

3 903 3

30

x x xx

x

x

=

+ + + =

+ =− −

=

=

x = 1st side = 30cm

x = 2nd side = 30cm

x + 3 = 3rd side = 33cm

By the way, this is considered an isosceles triangle.

( ) ( ) ( )

( ) ( ) ( )

st nd rd

?

check:1 side + 2 side + 3 side 93

93

30 30 33 9393 93

=

+ + =

+ + =

=



c. Let: x = larger number

x – 2 = smaller number


( ) ( )( ) ( ) ( )

5 smaller 4 larger 105 2 4 10

5 5 2 4 105 10 4 10

10 1010 10

20

x x

x xx x

x

x

× − × =

− − =

− − =

− − =− =+ +

=

x = larger number = 20

x – 2 = smaller number = 18

( ) ( )

( ) ( )

?

?

check:5 smaller 4 larger 10

5 4 10

5 18 4 20 1090 80 10

10 10

× − × =

− =

− =

− ==



2. Solve “Example” number 2 A and B on page 9.

A. Let: x = 1st consecutive integer


x + 2 = 3rd consecutive integer


( ) ( ) ( )( ) ( ) ( )

st nd rd1 consecutive integer + 2 consecutive integer + 3 consecutive integer 6060

1 2 603 3 60

3 3

3 573 3

19

x x xx

x

x

=

+ + =

+ + + + =

+ =− −

=

=

x = 1st consecutive integer = 19

x + 1 = 2nd consecutive integer = 20

x + 2 = 3rd consecutive integer = 21

( ) ( ) ( )

( ) ( ) ( )

st nd rd

?

check:1 consecutive integer + 2 consecutive integer + 3 consecutive integer 60

60

19 20 21 6060 60

=

+ + =

+ + =

=



B. Let: x = 1st consecutive even integer




( ) ( )( ) ( )

st rd1 consecutive even integer + 3 consecutive even integer 4040

4 402 4 40

4 4

2 362 2

18

x xx

x

x

=

+ =

+ + =

+ =− −

=

=

x = 1st consecutive even integer = 18

x + 2 = 2nd consecutive even integer = 20

x + 4 = 3rd consecutive even integer = 22

( ) ( )

( ) ( )

st rd

?

check:1 consecutive even integer + 3 consecutive even integer 40

40

18 22 4040 40

=

+ =

+ =

=

3.

( )( ) ( )

4 2 8 56

4 4 2 8 564 8 8 56

12 8 568 8

12 4812 12

4

g g

g gg g

g

g

g

+ + =

+ + =

+ + =+ =− −

=

=

( )( ) ( )

( )( ) ( )

( )

?

?

?

check: 4 2 8 56

4 4 2 8 4 56

4 6 32 56

24 32 5656 56

+ + =

+ + =

+ =

+ ==



4.

5.

Set E

1.

2.

( )( ) ( )

12 3 2 10

12 3 2 3 1012 6 3030 30

18 66 63

x

xx

x

x

− = −

− = −

− = −+ +

=

=

( )( )( )( )

( )

( )

?

?

?

check: 12 3 2 10

12 3 2 3 10

12 3 6 10

12 3 412 12

− = −

− = −

− = −

− = −

− = −

9 7 6 526 6

15 7 527 7

15 4515 15

3

y yy y

y

y

y

+ = − ++ +

+ =− −

=

=

( ) ( )

( ) ( )?

?

check: 9 7 6 52

9 3 7 6 3 52

27 7 18 5234 34

+ = − +

+ =− +

+ =− +=

( )3 1 19 53 3 19 55 5

8 3 193 3

8 168 8

2

a aa aa a

a

a

a

+ = −

+ = −+ +

+ =− −

=

=

( )( ) ( )

( )( ) ( )

( )

?

?

check: 3 1 19 5

3 2 1 19 5 2

3 3 19 109 9

+ = −

+ = −

= −

=

3 14 4 1805 10 180

10 10

5 1905 5

38

x x xx

x

x

− + + + =− =+ +

=

=

( ) ( ) ( )

( ) ( ) ( )?

?

check: 3 14 4 180

3 38 14 38 38 4 180

114 66 180180 180

− + + + =

− + + + =

+ ==



3.

4.

5.

( )3 4 1 2 2312 3 2 23

10 3 233 3

10 202

x xx x

x

xx

+ − =

+ − =+ =− −

==

( )( ) ( )

( )( ) ( )

( )

( )

?

?

?

?

check: 3 4 1 2 23

3 4 2 1 2 2 23

3 8 1 4 23

3 9 4 23

27 4 2323 23

+ − =

+ − =

+ − =

− =

− ==

( ) ( )3 2 4 103 6 4 403 3

6 4040 40

46

x xx xx x

x

x

+ = −

+ = −− −

= −+ +

=

( )( ) ( )( )( )( ) ( )( )

( ) ( )

?

?

check: 3 2 4 10

3 46 2 4 46 10

3 48 4 36144 144

+ = −

+ = −

=

=

( )3 6 20 253 18 20 25

3 2 252 2

3 273 3

9

xx

x

x

x

+ + − =

+ − =− =+ +

=

=

( )( )( )( )

( )

?

?

?

check: 3 6 20 25

3 9 6 20 25

3 15 20 25

45 20 2525 25

+ + − =

+ − =

− =

− ==



6. Solve: The length of a rectangle exceeds its width by five meters. The perimeter is sixty-

six meters. Find the dimensions.

First draw a picture; it generally helps to organize thoughts. Also don’t forget “Let”

statements. Thus:

Let: x = width

x + 5 = length

( ) ( ) ( )( ) ( ) ( )

2 2 Recall perimeter formula.2 2

66 2 5 2 Substitute the values.66 2 10 2 Distribute and then combine like terms.66 4 10 "Undo" the 10.10 10

56 4 "Undo" the 4.4 414

14m width Solve for the

P l w

x xx xx

x

xx

= + ←

= +

= + + ←

= + + ←= + ←

− −

= ←

== = ← dimensions.

19m 5 lengthx= + =

( ) ( )

( ) ( )?

?

check: 2 266 2 2

66 2 14 2 19

66 28 3866 66

P l w= +

= +

= +

= +=

Length = x + 5

Length = x + 5

Width = x Width = x



7. Find two consecutive integers such that four times the larger is twenty-three more than

three times the smaller.

Let: x = 1st consecutive integer



( ) ( )( ) ( )

nd st4 2 3 1 204 3 20

4 1 3 204 4 3 203 3

4 204 4

16

x xx xx x

x

x

× = × +

= +

+ = +

+ = +− −

+ =− −

=

x = 1st consecutive integer = 16

x + 1 = 2nd consecutive integer = 17

( ) ( )

( ) ( )

nd st

?

?

check: 4 2 3 1 204 3 20

4 17 3 16 20

68 48 2068 68

× = × +

= +

= +

= +=



8. The perimeter of a square is three times larger than the perimeter of an equilateral triangle.

If the length of one side of the square is ten meters more than the length of the side of the

triangle, find the dimensions of the triangle and square.

Let: x + 10 = side of square Let: x = side of triangle

P = 4s P = 3s

( ) ( )( )3

4 10 3 3

4 40 94 4

40 55 58

sq triangleP P

x x

x xx x

x

x

= ×

+ =

+ =− −

=

=

x = side of triangle = 8m

x + 10 = side of square = 18m

( ) ( )( )( ) ( )( )

( )

?

?

check: 3

4 3 3

4 18 3 3 8

72 3 2472 72

sq triangleP P= ×

=

=

=

=



Set F: Solve and graph the solutions (make your own number lines).

1.

( )( ) ( ) ( )


21 11 9 30 11 9 0 11 9 etc...10 9 19 9 11 9 Therefore you don't shade to the left.

− ≥ ←

− ≥ − ≥ − ≥

≥ ≥ − ≥ ←/


25 20 15 10 5 0 5 10 15 20 25− − − − −

2.

( )( ) ( ) ( )



− > ← −

− − > − − > − >

> > > ←/


14 12 9 6 3 0 3 6 9 12 14− − − − −

11 9 "Undo" the subtraction with addition.11 11

20 The soultion is any number greater than or equal to 20.

e

e

− ≥ ←+ +

≥ ←

5 355 35 "Undo" the 5 with division.5 5

7 Since you divided by a , change the direction of the inequality.7 The solution is any number less than 7.

t

t

tt

− >−

> ← −− −

> − ← −< − ← −



3.

( )

( ) ( ) ( )

Check: 5 Check any number less than or equal to 15.312 0 21

5 5 5 etc...3 3 3

4 5 0 5 7 5 Therefore you don't shade to the right.

≤ ←

≤ ≤ ≤

≤ ≤ ≤ ←/


25 20 15 10 5 0 5 10 15 20 25− − − − −

4.

( )( ) ( ) ( )



+ < ←

+ < − + < + <

< < < ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −

( ) ( )

53

3 5 3 "Undo" the division of 3.3


u

u

u

≤

≤ ←

≤ ←

7 107 7 "Undo" the addition of 7.

3 The solution is any number less than 3.

a

a

+ <− − ←

< ←



( ) ( )

Let: a number

16

6 1 6 "Undo" the division of 6.6

6 Since you multiplied by a , change the direction of the inequality.6 The solution is any number less than 6 .

nn

n

nn

=

>−

− > − ← −−

> − ← −< − ← −

5. The product of three and a number n is less than or equal to 15.

( )( ) ( ) ( )

Check: 3 15 Check any number less than or equal to 5.

3 0 15 3 5 15 3 10 15 etc...0 5 15 5 30 15 Therefore you don't shade to the right.

≤ ←

≤ − ≤ ≤

≤ − ≤ ≤ ←/


25 20 15 10 5 0 5 10 15 20 25− − − − −

6. The quotient of a number divided by negative six is greater than one.

( )

( ) ( ) ( )

Check: 1 Check any number less than 6.612 24 0

1 1 1 etc...6 6 6

2 1 4 1 0 1 Thus, no shade to the right.

> ← −−− −

> > >− − −

> > > ←/


14 12 9 6 3 0 3 6 9 12 14− − − − −

Let: a number3 15

3 15 "Undo" the 3 with division.3 3


nn

x

x

=≤

≤ ←

≤ ←



7. Carol weighs three times as much as Sue. Both weights are whole numbers. The sum of

their weights is at most one hundred and sixty pounds. Find the greatest possible weight in

pounds for each girl.

Let: w = Sue’s weight

3w = Carol’s weight

3 1604 160 "Undo" the multiplication of 4.

4 1604 4


w ww

w

w

+ ≤≤ ←

≤

≤ ←

Since you are looking for the greatest possible weight, the answer must be 40.

Therefore:

w = Sue’s weight = 40lbs 3w = Carol’s weight = 120lbs

There is no need to graph this solution since we are looking for a specific answer. 8.

( )( ) ( ) ( )

Check: 11 12 45 Check any number less than or equal to 3.

11 2 12 45 11 5 12 45 11 6 12 45 etc...22 12 45 55 12 45 66 12 45

34 45 43 45 78 45 Thus, no shade to the right.

+ ≤ ←

+ ≤ − + ≤ + ≤

+ ≤ − + ≤ + ≤≤ − ≤ ≤ ←/


5 4 3 2 1 0 1 2 3 4 5− − − − −

11 12 45 "Undo" the addition of 12.12 12

11 33 "Undo" the multiplication of 11.11 11


x

x

x

+ ≤ ←− −

≤ ←

≤ ←



9.

( )

( ) ( ) ( )

2Check: 4 6 Check any number less than 15.32 2 212 4 6 0 4 6 21 4 6 etc...3 3 3

8 4 6 0 4 6 14 4 64 6 4 6 10 6 Therefore you don't shade to the right.

− + > − ←

− + > − − + > − − + > −

− + > − + > − − + > −− > − > − − > − ←/


25 20 15 10 5 0 5 10 15 20 25− − − − −

10. In the set of positive integers, what is the solution set of the inequality: 2 3 5x − < ?

a. {0,1,2,3} b. {1,2,3} c. {0,1,2,3,4} d. {1,2,3,4}

2 3 5 "Undo" the subtraction of 3.3 3

2 8 "Undo" the multiplication of 2.2 2

4 The solution is any number less than 4.

x

x

x

− < ←+ +

< ←

< ←

Since 4 is not part of the solution that eliminates choices (c) and (d), and since 0 (zero)

is not a positive answer, the correct choice is b.

End of Equations

2 4 6 "Undo" the addition of 4.3

4 4

3 2 3 210 "Undo" the mulitplication of .2 3 2 3

15 Since you multiplied by a , change the direction of the inequality.15 The solution is any num

k

k

kk

− + > − ←

− −

⎛ ⎞ ⎛ ⎞− − > − ← −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠> ← −< ← ber less than 15.


Module 8 - Equations

81

Glossary of Terms

Acute angle – an angle whose measure is between 0 o and 90 o . (Modules: 9, 10)

Acute triangle – a triangle with three acute angles. (Module 10)

Addition Operation – term + term = sum. (Modules: 5 – 10)

Additive Inverse (or opposite of a number, x) – the unique number -x, which when added to x

yields zero. ( ) 0x x+ − = . (Modules: 4, 8)

Adjacent angles – two angles with the same vertex and a common side between them. Angles

1 and 2 are adjacent angles. (Modules: 9, 10)

2

1

Algebraic Expression – a mathematical combination of constants and variables connected by

arithmetic operations such as addition, subtraction, multiplication, and division.

(Module 8)

Algorithm – a rule (or step by step process) used to help solve a specific type of problem.

(Modules: 5 – 10)

Alternate exterior angles – when a line intersects two parallel lines, eight angles are formed;

two angles that are outside (exterior) the parallel lines and on opposite sides (alternate)

of the intersecting line are called alternate exterior angles. (Module 9)



Alternate interior angles – when a line intersects two parallel lines, eight angles are formed;

two angles that are between (interior) the parallel lines and on opposite sides (alternate)

of the intersecting line are called alternate interior angles. (Module 9)

Altitude – the perpendicular from a vertex to the opposite side (extended if necessary) of a

geometric figure. (Module 10)

altit

ude

altit

ude

Angle – the union of two rays with a common endpoint; angles are measured in a counter-

clockwise direction; the angle’s rays are labeled as initial and terminal sides with the

terminal side counter-clockwise from the initial side. (Modules: 9, 10)

initial side

terminal side

Apothem – the apothem of a regular polygon is the radius of an inscribed circle. (Module 10)

apothem

Arc – any part of a circle that can be drawn without lifting the pencil. (Module 10)



83

Area – the measurement in square units of a bounded region. (Module 3)

Associative Property of Addition – this property of real numbers may be written using

variables in the following way: ( ) ( )a b c a b c+ + = + + . Terms to be combined may be

grouped in any manner. (Module 4)

Associative Property of Multiplication – this property of real numbers may be written in the

following way: ( ) ( )a b c a b c⋅ ⋅ = ⋅ ⋅ . Terms to be multiplied may be grouped in any

manner. (Module 4)

Axiom – a statement that is accepted as true, without proof. (Module 3)

Base – the numbers being used as a factor in an exponential expression. In the exponential

expression 2 5 , 2 is the base. (Module 7)

Base angles of an isosceles triangle – the angles opposite the equal sides of an isosceles

triangle are the base angles, which are also equal. (Module 10)

Base of an isosceles triangle – the congruent sides of an isosceles triangle are called the legs,

while the third side of the isosceles triangle is called the base. (Module 10)

Binary operation – an operation such as addition, subtraction, multiplication, or division that

changes two values into a single value. (Modules: 5 – 10)

Bisector – a line that divides a figure into two equal parts. (Module 10)

Centi – a prefix for a unit of measurement that denotes one one-hundredth 1100( ) of the unit.



Central angle – an angle whose vertex is the center of a circle and whose sides are radii of the

circle. (Module 10)

Chord – a line segment with endpoints on a circle. (Module 10)

Circle – the set of all points in a plane at a given distance (the radius) from a given point (the

center). (Module 10)

radius

center

Circumference – the distance around the edge of a circle. (Modules: 9, 10)

Closed dot – means the number is part of the solution set, thus it is shaded. (Module 8)

Coefficient – the numerical part of a term. (Module 8)

Combine like terms – means to group together terms that are the same (numbers with numbers

/ variables with variables) and are on the same side of the equal sign. (Module 8)

Complementary angles – two angles whose sum is 90 o . (Modules: 9, 10)

Common factor – identical part of each term in an algebraic expression; in the expression ab +

ac, the variable a is the common factor. (Module 8)

Commutative Property of Addition – terms to be combined may be arranged in any order; this

property of real numbers may be written using variables in the following way:

a b b a+ = + . (Module 4)



85

Commutative Property of Multiplication – terms to be multiplied may be arranged in any

order; this property of real numbers may be written using variables in the following

way: a b b a⋅ = ⋅ . (Module 4)

Comparison Axiom – if the first of three quantities is greater than the second and the second is

greater than the third, then the first is greater than the third; if a > b and b > c,

then a > c. (Module 3)

Composite number – a natural number greater than one that has at least one positive factor

other than 1 and itself. (Module 2)

Consecutive even integers – even integers that follow one another such as 2, 4, 6, etc.

(Module 8)

Consecutive integers – integers that follow each other on the number line such as 7, 8, 9, etc.

(Module 8)

Consecutive odd integers – odd integers that follow one another such as 5, 7, 9, etc.

(Module 8)

Constant – any symbol that has a fixed value such as 2 or π. (Modules: 3, 7, 8)

Coplanar – coplanar points are points in the same plane. (Module 9)

Corresponding angles – if a line intersects two parallel lines, eight angles are formed; two

non-adjacent angles that are on the same side of the intersecting line but one between

the parallel lines and one outside the parallel lines are called corresponding angles.

(Module 9)



Counting numbers (or natural numbers) – the set of numbers {1, 2, 3, 4, 5, …}. (Module 1)

Decagon – a ten-sided polygon. (Module 10)

Denominator – the bottom part of a fraction. (Modules: 5, 6, 7, 8)

Diagonal – a line segment with endpoints on two non-consecutive vertices of a polygon.

(Module 10)

Diameter – a line segment that passes through the center of a circle and whose endpoints are

points on the circle. (Module 10)

Difference – the answer to a subtraction problem. (Modules: 5, 6)

Distributive Property of Multiplication over Addition – a property of real numbers used to

write equivalent expressions in the following way: ( )a b c a b a c+ = ⋅ + ⋅ .

(Modules: 4, 8)

Distributive Property of Multiplication over Subtraction – a property of real numbers used to

write equivalent expressions in the following way: ( )a b c a b a c− = ⋅ − ⋅ .

(Modules: 4, 8)

Dividend – the number being divided in a quotient; in c

b a or a cb= , a is the dividend.

(Modules: 5, 6, 7, 8)

Division operation – Quotient

Dividend Quotient or Divisor DividendDivisor

= . (Modules: 5, 6, 7, 8)



87

Elements (of a set) – the objects that belong to a set. (Module 2)

Empty set – a set that has no elements in it. (Module 2)

Equal Quantities Axiom – quantities which are equal to the same quantity or to equal

quantities, are equal to each other. (Module 3)

Equation – a mathematical statement that two quantities are equal to one another. (Module 8)

Equiangular polygon – a polygon with all angles equal. (Module 10)

Equiangular triangle – a triangle with all angles equal. (Module 10)

Equilateral polygon – a polygon with all sides equal. (Module 10)

Equilateral triangle – a triangle with all sides equal. (Module 10)

Existence Property – a property that guarantees a solution to a problem. (Module 4)

Existential quantifier – ∀ is the existential quantifier; it is read “for all,” “for every,” or “for

each.” (Modules: 1, 2)

Exponent – tells how many times a number called the base is used as a factor; in 32 8= , three

(3) is the exponent. (Module 7)



Exterior angle – is an angle formed by one side of a polygon and an adjacent side extended.

(Modules: 9, 10)

A

B C

ED Factor – one of the numbers multiplied together in a product; if a b c⋅ = , then a and b are

factors of c. (Modules: 5, 6)

Fundamental Theorem of Arithmetic – every composite number may be written uniquely

(disregarding order) as a product of primes. (Module 2)

Geometry – the branch of mathematics that investigates relations, properties, and

measurements of solids, surfaces, lines, and angles. (Modules: 9, 10)

Gram (g) – a basic unit of mass in the metric system; 1 gram≈ .035 ounces.

Heptagon – a seven-sided polygon. (Module 10)

Hexagon – a six-sided polygon. (Module 10)

Hypotenuse – the side opposite the right angle in a right triangle. (Module 10)

Identity – an equation that is true for all values of the variable; every real number is a root of

an identity. (Module 4)

Identity Element for Addition – zero is the additive identity element because 0 may be added

to any number and the number keeps its identity; 0 0a a a+ = + = for any real number

a. (Module 4)



89

Identity Element for Multiplication – one (1) is the multiplicative identity element because

any number may be multiplied by 1 and the number keeps its identity; 1 1a a a⋅ = ⋅ =

for any real number a. (Module 4)

Improper fraction – a fraction in which the numerator (top #) is larger than the denominator

(bottom #). Improper fractions are greater than 1 and can be turned into mixed

numbers. (Module 5)

Inequality – a mathematical sentence that compares two unequal expressions.

(Modules: 2, 3, 8)

Inscribed angle – an angle whose vertex lies on a circle and whose sides are chords of the

circle. (Module 10)

Integers – the natural numbers, zero, and the additive inverses of the natural numbers;

{…-3, -2, -1, 0, 1, 2, 3…}. (Modules: 1 – 10)

Interior angle – an angle that lies inside a polygon and is formed by two adjacent sides of the

polygon. (Module 10)

Intersect – to cross; two lines in the same plane intersect if and only if they have exactly one

point in common. (Module 9)

Irrational number – a real number that cannot be written as the quotient of two integers; an

irrational number, written as a decimal, does not terminate and does not repeat.

(Module 1)



Isosceles trapezoid – a trapezoid whose non-parallel sides (or legs) are congruent.

(Module 10)

leg leg

Isosceles triangle – a triangle with two sides equal. (Module 10)

Kilo – a prefix for measurement that denotes one thousand (1000) units.

Kite – a quadrilateral with two pairs of adjacent sides congruent and no opposite sides

congruent. (Module 10)

Least Common Multiple (LCM) – the least common multiple of two or more positive values is

the smallest positive value that is a multiple of each. (Modules: 5, 6)

Legs of an isosceles triangle – the congruent sides of an isosceles triangle are called its legs.

(Module 10)

Like terms – terms which have identical variable factors. (Module 8)

Line – one of the undefined terms; consists of a set of points extending without end in opposite

directions. (Modules: 9, 10)



91

Line segment – a subset of a line that contains two points of the line and all points between

those two points. (Modules: 9, 10)

Liter (L) – a basic unit of volume in the metric system; 1 liter ≈ 1.06 liquid quarts.

Lowest common denominator (lcd)(of two or more fractions) – the least common multiple of

the denominators of the fractions. (Modules: 5, 6)

Major arc – an arc of a circle that is greater than a semicircle. (Module 10)

Meter (m) – a basic unit of length in the metric system; 1 meter ≈ 39.37 inches.

Milli – a prefix for a unit of measurement that denotes one one-thousandth 11000( ) of the unit.

Minor arc – an arc of a circle that is less than a semicircle. (Module 10)

Minuend – the number from which something is subtracted; in 5 3 2− = , five (5) is the

minuend. (Modules: 5 – 8)

Multiplicative inverse (or reciprocal of a real number x) – the unique number, 1x

, which,

when multiplied by x, yields 1. 1 1xx⋅ = if 0x ≠ . (Modules: 4, 8)

Multiplication operation – factor x factor = product. (Modules: 5 – 8)

Multiplicative property of zero – for any real number a , 0 0 0a a⋅ = ⋅ = . (Modules: 4 – 8)

Natural numbers (or counting numbers) – the set of numbers {1, 2, 3, 4, 5, …}. (Module 1)



Negative integers – the opposite of the natural numbers. (Modules: 1 – 8)

Nonagon – a nine-sided polygon. (Module 10)

Numerator – the top part of a fraction. (Module 5)

Obtuse angle – an angle that measures between 90 o and 180 o . (Modules: 9, 10)

Obtuse triangle – a triangle with one obtuse angle. (Module 10)

Octagon – an eight-sided polygon. (Module 10)

Open dot – means the number is not part of the solution set, thus it is not shaded. (Module 8)

Parallel lines – lines in the same plane that do not intersect; the two lines are everywhere

equidistant. (Modules: 9, 10)

Parallelogram – a quadrilateral whose opposite sides are parallel. (Module 10)

Pentagon – a five-sided polygon. (Module 10)

Percent – Percent means per 100 or divided by 100. The symbol for percent is %.

(Module 6)

Perfect square – a number whose square root is a natural number. (Module 1)

Perimeter – the sum of the lengths of the sides of a figure or the distance around the figure.

(Modules: 8, 10)



93

Perpendicular lines – two lines that form a right angle. (Modules: 9, 10)

Plane – one of the undefined terms; a set of points that form a flat surface extending without

end in all directions. (Modules: 9, 10)

Plane geometry – the branch of mathematics that deals with figures that lie in a plane or flat

surface. (Module 10)

Point – one of the undefined terms; a location with no width, length, or depth.

(Modules: 9, 10)

Polygon – a closed figure bounded by line segments. (Module 9)

Positive integers – the collection of numbers known as natural numbers. (Modules: 1 – 10)

Prime numbers – the natural numbers greater than one (1) that have exactly two factors, one

(1) and themselves. (Module 2)

Product – the result when two or more numbers are multiplied. (Modules: 3 – 10)

Quadrilateral – a polygon with four sides. (Module 10)

Quotient – the number resulting from the division of one number by another. (Modules: 1, 5)

Radical – the symbol that tells you a root is to be taken; denoted by . (Module 1)

Radicand – the number inside the radical sign whose root is being found; in 7x , 7x is the radicand. (Module 1)



Radius (radii) – a line segment with endpoints on the center of the circle and a point on the

circle. (Module 10)

Ratio – proportional relation between two quantities or objects in terms of a common unit.

(Module 5)

Rational numbers – the collection of numbers that can be expressed as the quotient of two

integers; when written as a decimal it will terminate or repeat. (Modules: 1, 5)

Ray – a subset of a line that consists of a point and all points on the line to one side of the

point. (Modules: 9, 10)

Real numbers – the combined collection of the rational numbers and the irrational numbers.

(Module 1)

Reciprocal (or multiplicative inverse of a real number x) – the unique number which, when

multiplied by x, yields 1; 1 1xx⋅ = if 0x ≠ . (Module 4)

Rectangle – a parallelogram with one right angle. (Modules: 3, 8, 10)

Reflex angle – an angle greater than a straight angle and less than two straight angles.

(Module 9)

Regular polygon – a polygon whose sides and angles are all equal. (Module 10)

Relatively prime – a pair of numbers with no common factor other than 1. (Module 5)



95

Repeating decimal – a decimal with an infinite number of digits to the right of the decimal

point created by a repeating set pattern of digits. (Modules: 1, 6)

Rhombus (rhombi) – a parallelogram having two adjacent sides equal. (Module 10)

Right angle – an angle whose sides are perpendicular; having a measure of 90 degrees.

(Modules: 9, 10)

Right triangle – a triangle with one right angle. (Module 10)

Scalene triangle – a triangle with no two sides of equal measure. (Module 10)

Secant – a straight line intersecting a circle in exactly two points. (Module 10)

Sector of a circle – the figure bounded by two radii and an included arc of the circle.

(Module 10)

Sector

Semicircle – an arc equal to half of a circle is called a semicircle. (Module 10)

Set – a collection of objects. (Module 2)

Sides of a polygon – the line segments forming a polygon are called the sides of the polygon.

(Module 10)



Similar figures – figures with the same shape but not necessarily the same size. (Module 10)

Similar polygons – polygons whose corresponding angles are congruent and whose

corresponding sides are proportional; the symbol ~ is used to indicate that figures are

similar. (Module 10)

Solution – a value that makes the two sides of an equation equal. (Modules: 5 – 10)

Solution set – the set of all roots of the equation. (Module 8)

Square – a rectangle having two adjacent sides equal. (Modules: 8, 10)

Square root – one of the two equal factors of a number. (Module 1)

Straight angle – an angle measuring 180 o . (Modules: 9, 10)

Subset – B is a subset of A, written B ⊆ A, if and only if every element of B is an element of A.

(Module 2)

Substitution Axiom – a quantity may be substituted for its equal in any expression. (Modules: 3, 4, 7 – 10)

Subtraction operation – Minuend

Subtrahend

Difference

− or Minuend – Subtrahend = Difference.

(Modules: 5 – 10)

Subtrahend – the number being subtracted in a subtraction problem; in 5 – 2 = 3, 2 is the

subtrahend. (Modules: 5, 6)



97

Sum – the result when two numbers are added. (Modules: 5 – 10)

Supplementary angles – two angles whose sum is 180 o . (Modules: 9, 10)

Term – a single number, a single variable, or a product of a number and one or more variables.

(Modules: 1 – 10)

Terminating decimal – a decimal with a finite (or countable) number of digits to the right of

the decimal point. (Module 6)

Transversal – a straight line that intersects two or more straight lines. (Module 9)

transversa l

Trapezoid – a quadrilateral with exactly one pair of parallel sides. (Module 10)

Triangle – a polygon with three sides. (Modules: 8, 10)

Trichotomy Property – for all real numbers, a and b, exactly one of the following is true;

a b= , a b< , or a b> . (Module 3)

Uniqueness Property – a property that guarantees that when two people work the same

problem they should get the same result. (Module 4)

Universal quantifier – ∃ is the universal quantifier. It is read, there exists or for some.

(Modules: 1, 2)

Variable – a letter or symbol used to represent a number or a group of numbers.

(Modules: 3, 7, 8)



Vertex – the turning point of a parabola; the common endpoint of the two intersecting rays of

an angle. (Module 10)

Vertex angle of an isosceles triangle – the angle formed by the equal sides of the triangle.

(Module 10)

Vertex of a polygon – a point where two sides of a polygon meet. (Module 10)

Vertical angles – two non-adjacent angles formed by two straight intersecting lines.

(Module 9)

Whole numbers – the collection of natural numbers including zero; {0, 1, 2, 3…}. (Modules: 1 – 10)

FORMULAS AND DISCOVERIES The Triangle Inequality:

The sum of two sides of a triangle must be greater than the third side. In ∆ABC

AB BC ACAB AC BCAC BC AB

+ >+ >+ >



99

Name Sketch Perimeter Area/ Surface Area Volume

Triangle

P a b c= + + 12A bh=

Does not have

volume

Square

P = 4s A = 2s Does not

have volume

Rectangle

P = 2l +2w A = lw Does not have

volume

Circle

C 2 rπ= 2A rπ= Does not

have volume

P 2 2a b= + A bh= Does not

have volume

1 1 2 2P s b s b= + + +

( )11 22A b b h= + Does not

have volume

P = r + s + t + u + v +

A = 12 ap

where p is the perimeter

Does not have

volume

The distance around a base

S. A. = area of bases ( 1 2B B+ ) + area of all lateral faces

V Bh= or

12V aph=

s

ss

sD C

BA

A B

CD

l

l

w w

A

B

C

a

b

ch

r

A B

CD

h

1b

2b

1s 2s

1B

2Blateral face

Parallelogram

Trapezoid

Regular Polygon

Prism

a

r s

t

u v

A

B C

D

h

b a



The distance around the base

S.A. = area of the base + area of all the lateral faces.

13V Bh=

or 16V aph=

C 2 rπ= S.A. = 22 2r rhπ π+

2V r hπ=

C 2 rπ= S.A. = 2 2r rlπ π+

213V r hπ=

Sphere

C 2 rπ= S.A. = 24 rπ 343V rπ=

End of Glossary

h

base

lateralface

h

r

r

h l

r

Pyramid

Cylinder

Cone

Documents

Integrated Math Concepts - Module 8mrallens.pbworks.com/f/book 8 - equations.pdf · Solving Two-Step Equations 15 Solving Complex Equations 17 Solving Multi-Step Equations 29 Translating