Week 2. Due for this week… Homework 2 (on MyMathLab – via the Materials Link) Monday night at 6pm. Read Chapter 3 and 8.1, 8.2 Do the MyMathLab

Week 2

Due for this week…

Homework 2 (on MyMathLab – via the Materials Link) Monday night at 6pm.

Read Chapter 3 and 8.1, 8.2 Do the MyMathLab Self-Check for week 2. Learning team planning for week 5.

Slide 2Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Introduction to Equations

Basic Concepts

The Addition Property of Equality

The Multiplication Property of Equality

2.1

The Linear Equation

It has one variable : x

ax + b = 0

Where a and b are real numbers and a is not equal to 0.

A photo album of linear equations

2x+3=0 (say cheese!) x+8=0 3x=7 2x+5=9-5x 3+5(x-1)=-7+x 3=8x+5 Etc.

How to legally play with linear equations.

Most of the magic we work with are really obvious tricks. Remember what happens if we multiply by things

that equal 1? e.g. 5/5 10/10 ?

Adding the SAME THING to both sides of an equation does not change the equation…

The Addition Property of Equality

So if you start with a = b You don’t change anything if you take a new

number or variable and add it to both sides!

a+c = b+c

The GOAL

We want x all by itself on one side and the rest of the junk on the other (so we can use our calculator or fingers).

The technical term for this is: solving for x



EXAMPLE

Solution

Using the addition property of equality

Solve each equation.

a. x + 21 = 9 b. n – 8 = 17

b. a. x + 21 = 9

x + 21 + (−21) = 9 + (−21)

x + 0 = −12

x = −12

n – 8 = 17

n – 8 + 8 = 17 + 8

n = 25

n + 0 = 25

The solution is −12. The solution is 25.

** Try exercises 17-26


EXAMPLE

Solution

Solving and checking a solution

Solve the equation −7 + x = 12 and then check the solution.

Check:

Isolate x by adding 7 to each side.

−7 + x = 12

−7 + 7 + x = 12 + 7

0 + x = 19

x = 19

12 = 12The solution is 19.

The answer checks.

−7 + x = 12

−7 + 19 = 12?




EXAMPLE

Solution

Using the multiplication property of equality

Solve each equation.

a. b. −11a = 33

b. a.

18

6x

18

6x

8 61

66 x

1 48x

48x

−11a = 33

11 1

3

1

11 3a

3a

The solution is 48.

The solution is −3.


Slide 14Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley-58

EXAMPLE

Solution

Solving and checking a solution

Solve the equation and then check the solution.

Check:

The solution isThe answer checks.

2 2

5 9x

2 2

5 9x

9 2 2 9

2 5 9 2x

9

15

x

9

5x

9.

5

2 2

5 9x

? 9

5

2 2

5 9

2 2

5 5



EXAMPLE

Solution

Application

A veterinary assistant holds a cat and steps on a scale. The scale reads 153 lbs. Alone the assistant weighs 146 lbs.

a. Write a formula to show the relationship of the weight of the cat, x, and the assistant.

b. Determine how much the cat weighs.

a. 146 + x = 153 146 + x = 153

146 + (−146) + x = 153 − 146

b.

x = 7

The cat weighs 7 lbs.** Try exercises 61-70


Linear Equations

Basic Concepts

Solving Linear Equations

Applying the Distributive Property

Clearing Fractions and Decimals

Equations with No Solutions or Infinitely Many Solutions

2.2


If an equation is linear, writing it in the form ax + b = 0 should not require any properties or processes other than the following:

• using the distributive property to clear parentheses,• combining like terms,• applying the addition property of equality.


EXAMPLE

Solution

Determine whether an equation is linear

Determine whether the equation is linear. If the equation is linear, give values for a and b.

a. 9x + 7 = 0 b. 5x3 + 9 = 0 c.

b. This equation is not linear because it cannot be written in the form ax + b = 0. The variable has an exponent other than 1.

a. The equation is linear because it is in the form ax + b = 0 with a = 9 and b = 7.

57 0

x

c. This equation is not linear because it cannot be written in the form ax + b = 0. The variable appears in the denominator of a fraction. ** Try exercises 13-26


EXAMPLE

Solution

Using a table to solve an equation

Complete the table for the given values of x. Then solve the equation 4x – 6 = −2.

From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1.

x −3 −2 −1 0 1 2 3

4x − 6 −18

x −3 −2 −1 0 1 2 3

4x − 6 −18 −14 −10 −6 −2 2 6


Slide 20

Rules for Happiness

First do whatever adding and subtracting you can do…

THEN do whatever multiplication and division you can do.

THEN you should be done!


EXAMPLE

Solution

Solving linear equations

Solve each linear equation.

a. 12x − 15 = 0 b. 3x + 19 = 5x + 5

b. a. 12x − 15 = 0

12x − 15 + 15 = 0 + 15

12x = 15 12 15

12 12

x

15

12x

5

4

3x + 19 = 5x + 5

3x − 3x + 19 = 5x − 3x + 5

19 = 2x + 5

19 − 5 = 2x + 5 − 5

14 = 2x 14

2 2

2x

7 x** Try exercises 39-42


EXAMPLE

Solution

Applying the distributive property

Solve the linear equation. Check your solution. x + 5 (x – 1) = 11

6x − 5 = 11

16

6x

6 16

6 6

x

x + 5 (x – 1) = 11 x + 5 x – 5 = 11

6x − 5 + 5 = 11 + 5 6x = 16

8

3


EXAMPLE continued

Checkx + 5 (x – 1) = 11

8 85 1 11

3 3

8 8 35 11

3 3 3

8 55 11

3 3

8 2511

3 3

3311

3

11 11

The answer checks, the solution is

8.

3



EXAMPLE

Solution

Clearing fractions from linear equations

Solve the linear equation. 1 1

52 6

x x

1 156 6

2 6x x

3 30x x

2 30x

15x

1 15

2 6x x

The solution is 15.



EXAMPLE

Solution

Clearing decimals from linear equations

Solve the linear equation. 5.3 0.8 7x

5.3 0.8 7x 5.3 0.10 18 7 0x

53 8 70x 53 8 753 0 53x

8 17x 8 17

8 8

x

17

8x

The solution is17

.8


Equations with No Solutions or Infinitely Many Solutions

An equation that is always true is called an identity and an equation that is always false is called a contradiction.

Slide 27


EXAMPLE

Solution

Determining numbers of solutions

Determine whether the equation has no solutions, one solution, or infinitely many solutions.

a. 10 – 8x = 2(5 – 4x)b. 7x = 9x + 2(12 – x) c. 6x = 4(x + 5)

a. 10 – 8x = 2(5 – 4x)

10 – 8x = 10 – 8x

– 8x = – 8x

0 = 0

Because the equation 0 = 0 is always true, it is an identity and there are infinitely many solutions.


EXAMPLE continued

b. 7x = 9x + 2(12 – x)

Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions.

c. 6x = 4(x + 5)

7x = 9x + 24 – 2x

7x = 7x + 24

0 = 24

6x = 4x + 20

2x = 20

x = 10

Thus there is one solution.


Slide 30


Introduction to Problem Solving

Steps for Solving a Problem

Percent Problems

Distance Problems

Other Types of Problems

2.3



EXAMPLE

Solution

Translating sentences into equations

Translate the sentence into an equation using the variable x. Then solve the resulting equation.

a. Six times a number plus 7 is equal to 25.b. The sum of one-third of a number and 9 is 18.c. Twenty is 8 less than twice a number.

a. 6x + 7 = 25

6x = 186 18

6 6

x

3x

b. 19 18

3x

27x

19

3x

c. 20 = 2x − 8

28 = 2x

14 = x

28 2

2 2

x



EXAMPLE

Solution

Solving a number problem

The sum of three consecutive integers is 126. Find the three numbers.

Step 1: Assign a variable to an unknown quantity.

n + (n + 1) + (n + 2) = 126

n + 1: next integer

n + 2: largest integer

Step 2: Write an equation that relates these unknown quantities.

n: smallest of the three integers


EXAMPLE continued

Step 3: Solve the equation in Step 2.

n + (n + 1) + (n + 2) = 126

Step 4: Check your answer. The sum of these integers is 41 + 42 + 43 = 126.

The answer checks.

(n + n + n) + (1 + 2) = 126

3n + 3 = 1263n = 123n = 41

So the numbers are 41, 42, and 43.


Note: To write x% as a decimal number, move the decimal point in the number x two places to the left and then remove the % symbol.

Slide 36


EXAMPLE

Solution

Converting percent notation

Convert each percentage to fraction and decimal notation.

a. 47% b. 9.8% c. 0.9%

a. Fraction Notation:47

47% .100

Decimal Notation: 47% 0.47.

b. Fraction Notation:9.8 98 49 2 49

9.8% .100 1000 500 2 500

Decimal Notation: 9.8% 0.098.

c. Fraction Notation:0.9 9

0.9%100 1000

Decimal Notation: 0.9% 0.009. ** Try exercises 35-42


EXAMPLE

Solution

Converting to percent notation

Convert each real number to a percentage.

a. 0.761 b. c. 6.3

a. Move the decimal point two places to the right and then insert the % symbol to obtain 0.761 = 76.1%

2

5

b. 2 2

0.40, so 40%.5 5

c. Move the decimal point two places to the right and then insert the % symbol to obtain 6.3 = 630%. Note that percentages can be greater than 100%.



EXAMPLE

Solution

Calculating a percent increase

The price of an oil change for an automobile increased from $15 to $24. Calculate the percent increase.

old value

ol

-

d

new

valu

e

value100 15

1

24

5

- 100 60%



EXAMPLE

Solution

Solving a percent problem

A car salesman sells a total of 85 cars in the first and second quarter of the year. In the second quarter, he had an increase of 240% over the previous quarter. How many cars did the salesman sell in the first quarter?

Step 1: Assign a variable.

x: the amount sold in the first quarter.

Step 2: Write an equation.

x + 2.4x = 85


EXAMPLE continued


Step 4: Check your answer. An increase of 240% of 25 is 2.4 × 25 = 60.

Thus the amount of cars sold in the second quarter would be 25 + 60 = 85.

In the first quarter the salesman sold 25 cars.

x + 2.4x = 85

3.4x = 85

x = 25



EXAMPLE

Solution

Solving a distance problem

A truck driver travels for 4 hours and 30 minutes at a constant speed and travels 252 miles. Find the speed of the truck in miles per hour.

Step 1: Let r represent the truck’s rate, or speed, in miles.

Step 2: The rate is to be given in miles per hour, so change the 4 hours and 30 minutes to 4.5 or 9/2 hours.

9252

2r

d = rt


EXAMPLE continued

Step 3: Solve the equation.9

2522

r

2 2

9 9

9252

2r

56 r

The speed of the truck is 56 miles per hour.

Step 4:

d = rt9

56 252 miles2

The answer checks.** Try exercises 71-74


EXAMPLE

Solution

Mixing chemicals

A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used?

Concentration Solution Amount (milliliters)

Pure alcohol

0.28 100 28

0.40 x 0.4x

0.36 x + 100 0.36x + 36

Step 1: Assign a variable. x: milliliters of 40%

x + 100: milliliters of 36%

Step 2: Write an equation.

0.28(100) + 0.4x = 0.36(x + 100)


EXAMPLE continued


200 mL of 40% alcohol solution was added to the 100 mL of the 28% solution.

0.28(100) + 0.4x = 0.36(x + 100)

28(100) + 40x = 36(x + 100)2800 + 40x = 36x + 36002800 + 4x = 3600

4x = 800 x = 200


EXAMPLE continued

Step 4: Check your answer. If 200 mL of 40% solution are added to the 100 mL of 28% solution, there will be 300 mL of solution.

200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol.

The concentration is or 36%. 108

0.36,300



Formulas

Basic Concepts

Formulas from Geometry

Solving for a Variable

Other Formulas

2.4


EXAMPLE

Solution

Calculating mileage of a trip

A tourist starts a trip with a full tank of gas and an odometer that reads 59,478 miles. At the end of the trip, it takes 8.6 gallons of gas to fill the tank, and the odometer reads 59,715 miles. Find the gas mileage for the car.

DM

G

The distance traveled is 59,715 – 59, 478 = 237 miles and the number of gallons used is G = 8.6. Thus,

6

237

8. 27.6 miles per gallon.



EXAMPLE

Solution

Calculating area of a region

A residential lot is shown.Find the area of this lot.

RA LWThe area of the rectangle:

205 ft

372 ft

116 ft

372 205RA 76,260 square feetRA

The area of the triangle: 12TA bh

12 116 372TA 21,576 square feetTA

Total area = 76,260 + 21,576 = 97,836 square feet.** Try exercises 19-29


EXAMPLE

Solution

Finding angles in a triangle

In a triangle, the smaller angles are equal in measure and are one-third of the largest angle. Find the measure of each angle.

Let x represent the measure of each of the two smaller angles. Then the measure of the largest angle is 3x, and the sum of the measures of the three angles is given by

3 180x x x

The measure of the largest angle is 3x, thus 36 ∙ 3 = 108°.The measure of the three angles are 36°, 36°, and 108°.

5 180x

5 180

5 5

x

36x



EXAMPLE

Solution

Finding the volume and surface area of a box

Find the volume and the surface area of the box shown.

V = LHW

The volume of the box is

2 2 2S LW WH LH

12 cm5 cm

6 cm

V = 12 ∙ 6 ∙ 5 V = 360 cm3

The surface area of the box is

2(12)(5) 2(5)(6) 2(12)(6)S 120 60 144S 324 square centimetersS ** Try exercises 41-46


EXAMPLE

Solution

Calculating the volume of a soup can

A cylindrical soup can has a radius of 2 ½ inches and a height of inches. Find the volume of the can.

2V r h

∙h

r

245

8

5

2V

585

1125

32V

110.45 cubic inchesV

** Try exercises 51


EXAMPLE

Solution

Solving for a variable

Solve each equation for the indicated variable.

a. b. 3 for 5

y zx z

for np nm nq p

a. 3 5

y zx

15 x y z

15 x y z

15 z x y

b. for np nm nq p np nq nm

( )np n q m ( )n q m

pn

p q m


Other Formulas

To calculate a student’s GPA, the number of credits earned with a grade of A, B, C, D, and F must be known. If a, b, c, d, and f represent these credit counts respectively, then

4 3 2.

a b c dGPA

a b c d f

Slide 54


EXAMPLE

Solution

Calculating a student’s GPA

A student has earned 18 credits of A, 22 credits of B, 8 credits of C and 4 credits of D. Calculate the student’s GPA to the nearest hundredth.

Let a = 18, b = 22, c = 8, d = 4 and f = 0

4 18 3 22 2 8 4

18 22 8 4 0GPA

158

52 3.04

The student’s GPA is 3.04.



EXAMPLE

Solution

Converting temperature

The formula is used to convert degrees Fahrenheit to degrees Celsius.

Use this formula to convert 23°F to an equivalent Celsius temperature.

= −5°C

59 32C F

532

9C F

25

329

3C

59

9C



Linear Inequalities

Solutions and Number Line Graphs

The Addition Property of Inequalities

The Multiplication Property of Inequalities

Applications

2.5


Solutions and Number Line Graphs

A linear inequality results whenever the equals sign in a linear equation is replaced with any one of the symbols <, ≤, >, or ≥.

x > 5, 3x + 4 < 0, 1 – y ≥ 9

A solution to an inequality is a value of the variable that makes the statement true. The set of all solutions is called the solution set.


EXAMPLE Graphing inequalities on a number line

Use a number line to graph the solution set to each inequality.

a.

b.

c.

d.

1x

1x

5x

2x


Each number line graphed on the previous slide represents an interval of real numbers that corresponds to the solution set to an inequality.Brackets and parentheses can be used to represent the interval. For example:


Interval Notation

1x (1, )

1x [1, )


EXAMPLE Writing solution sets in interval notation

Write the solution set to each inequality in interval notation.

a. b.6x 2y

(6, ) ( , 2]



EXAMPLE Checking possible solutions

Determine whether the given value of x is a solution to the inequality. 4 2 8, 7x x

?

?

4 2 <8

4( 87) 2

x

?

28 2 8

?

26 8 e Fals


The Addition Property of Inequalities



EXAMPLE

Solution

Applying the addition property of inequalities

Solve each inequality. Then graph the solution set.a. x – 2 > 3 b. 4 + 2x ≤ 6 + x

a. x – 2 > 3

x – 2 + 2 > 3 + 2

x > 5

b. 4 + 2x ≤ 6 + x

4 + 2x – x ≤ 6 + x – x

4 + x ≤ 6

4 – 4 + x ≤ 6 – 4

x ≤ 2



The Multiplication Property of Inequalities


EXAMPLE

Solution

Applying the multiplication property of inequalities

Solve each inequality. Then graph the solution set.a. 4x > 12 b.

a. 4x > 12 b.

12

4x

4 2

4 4

1x

3x

12

4x

4 (1

( 2)4

4) x

8 x



EXAMPLE

Solution

Applying both properties of inequalities

Solve each inequality. Write the solution set in set-builder notation.a. 4x – 8 > 12 b.

a. 4x – 8 > 12 b.

4 3 4 5x x

4 8 8 12 8x 4 20x

4 3 4 5x x 4 3 3 4 5 3x x x x

4 5x

5x

{ | 5}x x

4 5 5 5x 9 x

{ | 9}x x



Applications

To solve applications involving inequalities, we often have to translate words to mathematical statements.


EXAMPLE Translating words to inequalities

Translate each phrase to an inequality. Let the variable be x.a. A number that is more than 25.

b. A height that is at least 42 inches.

x > 25

x ≥ 42



EXAMPLE Calculating revenue, cost, and profit

For a snack food company, the cost to produce one case of snacks is $135 plus a one-time fixed cost of $175,000 for research and development. The revenue received from selling one case of snacks is $250.

a. Write a formula that gives the cost C of producing x cases of snacks.

b. Write a formula that gives the revenue R from selling x cases of snacks.

C = 135x + 175,000

R = 250x


EXAMPLE Calculating revenue, cost, and profit

For a snack food company, the cost to produce one case of snacks is $135 plus a one-time fixed cost of $175,000 for research and development. The revenue received from selling one case of snacks is $250.

c. Profit equals revenue minus cost. Write a formula that calculates the profit P from selling x cases of snacks.

d. How many cases need to be sold to yield a positive profit?

P = R – C= 250x – (135x + 175,000)

= 115x – 175,000

115x – 175,000 > 0 115x > 175,000 x > 1521.74

Must sell at least 1522 cases. ** Try exercises 119-120

End of week 2

You again have the answers to those problems not assigned

Practice is SOOO important in this course. Work as much as you can with MyMathLab, the

materials in the text, and on my Webpage. Do everything you can scrape time up for, first the

hardest topics then the easiest. You are building a skill like typing, skiing, playing a

game, solving puzzles. NEXT TIME: Linear Equations w/2 variables and

Graphing + slope and y-intercepts

Documents

Week 2. Due for this week… Homework 2 (on MyMathLab – via the Materials Link) Monday night at 6pm. Read Chapter 3 and 8.1, 8.2 Do the MyMathLab