Chapter 10

10-1 Solving Two-Step Equations

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpSolve.

1. x + 12 = 35

2. 8x = 120

3. = 7

4. –34 = y + 56

x = 23

x = 15

y = 63

Course 3


y = –90

y9

Problem of the Day

x is an odd integer. If you triple x and then subtract 7, you get a prime number. What is x? (Hint: Think about what the prime number must be in order for x to be an odd.)

x = 3

Course 3


Learn to solve two-step equations.

Course 3


Sometimes more than one inverse operation is needed to solve an equation. Before solving, ask yourself, “What is being done to the variable, and in what order?” Then work backward to undo the operations.

Course 3


The mechanic’s bill to repair Mr. Wong’s car was $650. The mechanic charges $45 an hour for labor, and the parts that were used cost $443. How many hours did the mechanic work on the car?

Additional Example 1: Problem Solving Application

Course 3


Additional Example 1 Continued

11 Understand the Problem

The answer is the number of hours the mechanic worked on the car.

List the important information:

Let h represent the hours the mechanic worked.

• The parts cost $443.• The labor cost $45 per hour.• The total bill was $650.

Total bill = Parts + Labor

650 = 443 + 45hCourse 3


Think: First the variable is multiplied by 45, and then 443 is added to the result. Work backward to solve the equation. Undo the operations in reverse order: First subtract 443 from both sides of the equation, and then divide both sides of the new equation by 45.

22 Make a Plan


Course 3


650 = 443 + 45h

Solve33

–443 –443 Subtract to undo the addition.

207 = 45h

4.6 = h

The mechanic worked for 4.6 hours on Mr. Wong’s car.


Course 3


Divide to undo multiplication.207 45h45 45=

If the mechanic worked 4.6 hours, the labor would be $45(4.6) = $207. The sum of the parts and the labor would be $443 + $207 = $650.

Look Back44


Course 3


The mechanic’s bill to repair your car was $850. The mechanic charges $35 an hour for labor, and the parts that were used cost $275. How many hours did the mechanic work on your car?

Try This: Example 1

Course 3


Try This: Example 1 Continued

11 Understand the Problem

The answer is the number of hours the mechanic worked on your car.

List the important information:

Let h represent the hours the mechanic worked.

• The parts cost $275.

• The labor cost $35 per hour.

• The total bill was $850.

Total bill = Parts + Labor

850 = 275 + 35hCourse 3


Think: First the variable is multiplied by 35, and then 275 is added to the result. Work backward to solve the equation. Undo the operations in reverse order: First subtract 275 from both sides of the equation, and then divide both sides of the new equation by 35.

22 Make a Plan


Course 3


850 = 275 + 35h

Solve33

–275 –275 Subtract to undo the addition.

575 = 35h

16.4 h

The mechanic worked for about 16.4 hours on your car.


Course 3


Divide to undo multiplication.575 35h35 35=

If the mechanic worked 16.4 hours, the labor would be $35(16.4) = $574. The sum of the parts and the labor would be $275 + $574 = $849.

Look Back44


Course 3


Additional Example 2A: Solving Two-Step Equations

A. + 7 = 22

Solve.

Think: First the variable is divided by 3, and then 7 is added. To isolate the variable, subtract 7, and then multiply by 3.

– 7 – 7 Subtract to undo addition.

n3

+ 7 = 22n3

Multiply to undo division.3 = 3 15n3n = 45

Course 3


n3

= 15

Additional Example 2A Continued

Substitute 45 into the original equation.

Check + 7 = 22n3

? + 7 = 2245

315 + 7 = 22

?

Course 3


Additional Example 2B: Solving Two-Step Equations

B. 2.7 = –1.3m + 6.6

Think: First the variable is multiplied by –1.3, and then 6.6 is added. To isolate the variable, subtract 6.6, and then divide by –1.3.

2.7 = –1.3m + 6.6–6.6 –6.6 Subtract to undo addition.–3.9 = –1.3m

Divide to undo multiplication.

3 = m

–1.3 –1.3–3.9 = –1.3m

Course 3


Additional Example 2C: Solving Two-Step Equations

C. = 9y – 4 3

Think: First 4 is subtracted from the variable, and then the result is divided by 3. To isolate the variable, multiply by 3, and then add 4.

= 9y – 43

y – 4 = 27 + 4 + 4 Add to undo subtraction.

y = 31

3 · 3 · Multiply to undo division.= 9y – 43

Course 3


Try This: Example 2A

A. + 5 = 29

Solve.

Think: First the variable is divided by 4, and then 5 is added. To isolate the variable, subtract 5, and then multiply by 4.

– 5 – 5 Subtract to undo addition.

n4

+ 5 = 29n4

Multiply to undo division.4 = 4 24n4

n = 96Course 3


Substitute 96 into the original equation.

Check + 5 = 29n4

? + 5 = 2996

424 + 5 = 29

?

Try This: Example 2A Continued

Course 3


Try This: Example 2B

B. 4.8 = –2.3m + 0.2

Think: First the variable is multiplied by –2.3, and then 0.2 is added. To isolate the variable, subtract 0.2, and then divide by –2.3.

4.8 = –2.3m + 0.2–0.2 –0.2 Subtract to undo addition.4.6 = –2.3m

Divide to undo multiplication.

–2 = m

–2.3 –2.3 4.6 = –2.3m

Course 3


Try This: Example 2C

C. = 8y – 2 4

Think: First 2 is subtracted from the variable, and then the result is divided by 4. To isolate the variable, multiply by 4, and then add 2.

= 8y – 24

y – 2 = 32 + 2 + 2 Add to undo subtraction.

y = 34

4 · 4 · Multiply to undo division.= 8y – 24

Course 3


Solve.

1. – 3 = 10

2. 7y + 25 = –24

3. –8.3 = –3.5x + 13.4

4. = 3

5. The cost for a new cell phone plan is $39 per month plus a one-time start-up fee of $78. If you are charged $1014, how many months will the contract last?

Lesson Quiz

y = –7

x = –117

Insert Lesson Title Here

x = 6.2

y = 28

24 months

x–9

y + 5 11

Course 3


10-2 Solving Multistep Equations

Course 3

Warm UpWarm Up



Warm UpSolve.

1. 3x = 102

2. = 15

3. z – 100 = –1

4. 1.1 + 5w = 98.6

x = 34

y = 225

z = 99

Course 3


w = 19.5

y15

Problem of the Day

Ana has twice as much money as Ben, and Ben has three times as much as Clio. Together they have $160. How much does each person have?Ana, $96; Ben, $48; Clio, $16

Course 3


Learn to solve multistep equations.

Course 3


To solve a complicated equation, you may have to simplify the equation first by combining like terms.

Course 3


Solve.

8x + 6 + 3x – 2 = 37

Additional Example 1: Solving Equations That Contain Like Terms

11x + 4 = 37 Combine like terms.

– 4 – 4 Subtract to undo addition.11x = 33

x = 3

Divide to undo multiplication.3311

11x11

=

Course 3


Check


8x + 6 + 3x – 2 = 37

8(3) + 6 + 3(3) – 2 = 37?

24 + 6 + 9 – 2 = 37?

37 = 37?

Course 3


Substitute 3 for x.

Solve.

9x + 5 + 4x – 2 = 42

Try This: Example 1

13x + 3 = 42 Combine like terms.

– 3 – 3 Subtract to undo addition.13x = 39

x = 3

Course 3


Divide to undo multiplication.3913

13x13

=

Check


9x + 5 + 4x – 2 = 42

9(3) + 5 + 4(3) – 2 = 42?

27 + 5 + 12 – 2 = 42 ?

42 = 42?

Course 3


Substitute 3 for x.

If an equation contains fractions, it may help to multiply both sides of the equation by the least common denominator (LCD) to clear the fractions before you isolate the variable.

Course 3


Solve.

A. + = –

Additional Example 2: Solving Equations That Contain Fractions

34

74

5n4

Multiply both sides by 4 to clear fractions, and then solve.

( ) ( )74

–3 4

5n4

4 + = 4

( ) ( ) ( )5n4

74

–3 44 + 4 = 4

5n + 7 = –3

Course 3


Distributive Property.


5n + 7 = –3 – 7 –7 Subtract to undo addition.

5n = –10

5n5

–10 5

= Divide to undo multiplication.

n = –2

Course 3



The least common denominator (LCD) is the smallest number that each of the denominators will divide into.

Remember!

Course 3


Solve.

B. + – =

Additional Example 2B: Solving Equations That Contain Fractions

23

The LCD is 18.

x 2

7x9

17 9

18( ) + 18( ) – 18( ) = 18( )7x9

x2

17 9

23

14x + 9x – 34 = 12

23x – 34 = 12 Combine like terms.

( )x2

23

7x9

17 918 + – = 18

Course 3



Multiply both sides by the LCD.

Additional Example 2B Continued

23x = 46

= 23x23

4623 Divide to undo multiplication.

x = 2

+ 34 + 34 Add to undo subtraction.


Course 3



69

69=

?

Check

x 2

7x9

17 9

+ – = 23

23 Substitute 2 for x.7(2)

9 + – =(2) 2

17 9

?

23

149 + – =2

2 17 9

?

23

149 + – =

17 9

?1

The LCD is 9.69

149 + – =9

9 17 9

?

Course 3


Solve.

A. + = –


14

54

3n4

Multiply both sides by 4 to clear fractions, and then solve.

( ) ( )54

–1 4

3n4

4 + = 4

( ) ( ) ( )3n4

54

–1 44 + 4 = 4

3n + 5 = –1

Course 3




3n + 5 = –1 – 5 –5 Subtract to undo addition.

3n = –6

3n3

–6 3

= Divide to undo multiplication.

n = –2

Course 3


Solve.

B. + – =


13

The LCD is 9.

x 3

5x9

13 9

9( ) + 9( )– 9( ) = 9( )5x9

x3

13 9

13

5x + 3x – 13 = 3


( )x3

13

5x9

13 9 9 + – = 9( )

Course 3



Multiply both sides by the LCD.

8x = 16

= 8x8

16 8 Divide to undo multiplication.

x = 2

+ 13 + 13 Add to undo subtraction.


Try This: Example 2B Continued

Course 3


39

39=

?

Check

x 3

5x9

13 9

+ – = 13

13 Substitute 2 for x.5(2)

9 + – =(2) 3

13 9

?

13

109 + – =2

3 13 9

?

The LCD is 9.39

109 + – =6

9 13 9

?


Course 3


When Mr. and Mrs. Harris left for the mall, Mrs. Harris had twice as much money as Mr. Harris had. While shopping, Mrs. Harris spent $54 and Mr. Harris spent $26. When they arrived home, they had a total of $46. How much did Mr. Harris have when he left home?

Additional Example 3: Money Application

Let h represent the amount of money that Mr. Harris had when he left home. So Mrs. Harris had 2h when she left home.

h + 2h – 26 – 54 = 46Mr. Harris $+ Mrs. Harris $ – Mr. Harris spent – Mrs. Harris spent = amount left

Course 3



3h – 80 = 46 Combine like terms.

+ 80 +80 Add 80 to both sides.

3h = 126

3h3

1263= Divide both sides by 3.

h = 42

Mr. Harris had $42 when he left home.

Course 3


When Mr. and Mrs. Wesner left for the store, Mrs. Wesner had three times as much money as Mr. Wesner had. While shopping, Mr. Wesner spent $50 and Mrs. Wesner spent $25. When they arrived home, they had a total of $25. How much did Mr. Wesner have when he left home?

Try This: Example 3

Let h represent the amount of money that Mr. Wesner had when he left home. So Mrs. Wesner had 3h when she left home.

h + 3h – 50 – 25 = 25Mr. Wesner $ + Mrs. Wesner $ – Mr. Wesner spent – Mrs. Wesner spent = amount left

Course 3



4h – 75 = 25 Combine like terms.

+ 75 +75 Add 75 to both sides.

4h = 100

4h4

1004= Divide both sides by 4.

h = 25

Mr. Wesner had $25 when he left home.

Course 3


Solve.

1. 6x + 3x – x + 9 = 33

2. –9 = 5x + 21 + 3x

3. + =

5. Linda is paid double her normal hourly rate for each hour she works over 40 hours in a week. Last week she worked 52 hours and earned $544. What is her hourly rate?

Lesson Quiz

x = –3.75

x = 3


x = 2858

x8

33 8

6x 7

4. – =2x21

2521

$8.50

x = 1 916

Course 3


10-3 Solving Equations with Variables on Both Sides

Course 3

Warm UpWarm Up



Warm UpSolve.

1. 2x + 9x – 3x + 8 = 16

2. –4 = 6x + 22 – 4x

3. + = 5

4. – = 3

x = 1

x = -13

x = 34

Course 3


27

x7 7

1

9x16

2x4

18

x = 50

Problem of the Day

An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle?22.5 in.

Course 3


Learn to solve equations with variables on both sides of the equal sign.

Course 3


Some problems produce equations that have variables on both sides of the equal sign.

Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.

Course 3


Solve.

A. 4x + 6 = x

Additional Example 1A: Solving Equations with Variables on Both Sides

4x + 6 = x– 4x – 4x

6 = –3x

Subtract 4x from both sides.

Divide both sides by –3.

–2 = x

6–3

–3x–3=

Course 3


Solve.

B. 9b – 6 = 5b + 18

Additional Example 1B: Solving Equations with Variables on Both Sides

9b – 6 = 5b + 18– 5b – 5b

4b – 6 = 18

4b 4

24 4 =

Subtract 5b from both sides.

Divide both sides by 4.

b = 6

+ 6 + 6

4b = 24Add 6 to both sides.

Course 3


Solve.

C. 9w + 3 = 5w + 7 + 4w

Additional Example 1C: Solving Equations with Variables on Both Sides

9w + 3 = 5w + 7 + 4w

3 ≠ 7

9w + 3 = 9w + 7 Combine like terms.

– 9w – 9w Subtract 9w from both sides.

No solution. There is no number that can be substituted for the variable w to make the equation true.

Course 3


Solve.

A. 5x + 8 = x


5x + 8 = x– 5x – 5x

8 = –4x

Subtract 4x from both sides.


–2 = x

8–4

–4x–4=

Course 3


Solve.

B. 3b – 2 = 2b + 123b – 2 = 2b + 12

– 2b – 2b

b – 2 = 12

Subtract 2b from both sides.

+ 2 + 2

b = 14Add 2 to both sides.


Course 3


Solve.

C. 3w + 1 = 10w + 8 – 7w3w + 1 = 10w + 8 – 7w

1 ≠ 8

3w + 1 = 3w + 8 Combine like terms.

– 3w – 3w Subtract 3w from both sides.

No solution. There is no number that can be substituted for the variable w to make the equation true.


Course 3


To solve multistep equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

Course 3


Solve.

A. 10z – 15 – 4z = 8 – 2z - 15

Additional Example 2A: Solving Multistep Equations with Variables on Both Sides

10z – 15 – 4z = 8 – 2z – 15

+ 15 +15

6z – 15 = –2z – 7 Combine like terms.+ 2z + 2z Add 2z to both sides.

8z – 15 = – 7

8z = 8

z = 1

Add 15 to both sides.

Divide both sides by 8.8z 88 8=

Course 3


B.

Additional Example 2B: Solving Multistep Equations with Variables on Both Sides

Multiply by the LCD.

4y + 12y – 15 = 20y – 14

16y – 15 = 20y – 14 Combine like terms.

y5

34

3y5

710

+ – = y –

y5

34

3y5

710

+ – = y –

20( ) = 20( )y5

34

3y5

710

+ – y –

20( ) + 20( ) – 20( )= 20(y) – 20( )y5

3y5

34

710

Course 3




–15 = 4y – 14

–1 = 4y

+ 14 + 14

–1 4

4y4 = Divide both sides by 4.

-14 = y

16y – 15 = 20y – 14

– 16y – 16y Subtract 16y from both sides.

Course 3


Solve.

A. 12z – 12 – 4z = 6 – 2z + 32


12z – 12 – 4z = 6 – 2z + 32

+ 12 +12

8z – 12 = –2z + 38 Combine like terms.+ 2z + 2z Add 2z to both sides.

10z – 12 = + 38

10z = 50

z = 5


Divide both sides by 10.10z 5010 10=

Course 3


B.

Multiply by the LCD.

6y + 20y + 18 = 24y – 18

26y + 18 = 24y – 18 Combine like terms.

y4

34

5y6

68

+ + = y –

y4

34

5y6

68

+ + = y –

24( ) = 24( )y4

34

5y6

68

+ + y –

24( ) + 24( )+ 24( )= 24(y) – 24( )y4

5y6

34

68


Course 3


Subtract 18 from both sides.

2y + 18 = – 18

2y = –36

– 18 – 18

–36 2

2y2 = Divide both sides by 2.

y = –18

26y + 18 = 24y – 18

– 24y – 24y Subtract 24y from both sides.


Course 3


Additional Example 3: Consumer Application

Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

Course 3



First solve for the price of one doughnut.

1.25 + 2d = 0.50 + 5dLet d represent the price of one doughnut.

– 2d – 2d

1.25 = 0.50 + 3dSubtract 2d from both sides.

– 0.50 – 0.50Subtract 0.50 from both sides.

0.75 = 3d

0.753

3d3= Divide both sides by 3.

0.25 = d The price of one doughnut is $0.25.

Course 3



Now find the amount of money Jamie spends each morning.

1.25 + 2d Choose one of the original expressions.

Jamie spends $1.75 each morning.

1.25 + 2(0.25) = 1.75

0.25n0.25

1.75 0.25 =

Let n represent the number of doughnuts.

Find the number of doughnuts Jamie buys on Tuesday.

0.25n = 1.75

n = 7; Jamie bought 7 doughnuts on Tuesday.

Divide both sides by 0.25.

Course 3


Try This: Example 3

Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?

Course 3



First solve for distance around the track.

2x + 4 = 4x + 2Let x represent the distance around the track.

– 2x – 2x

4 = 2x + 2Subtract 2x from both sides.

– 2 – 2 Subtract 2 from both sides.

2 = 2x

22

2x2= Divide both sides by 2.

1 = x The track is 1 mile around.

Course 3



Now find the total distance Helene walks each day.

2x + 4 Choose one of the original expressions.

Helene walks 6 miles each day.2(1) + 4 = 6

Let n represent the number of 1-mile laps.

Find the number of laps Helene walks on Saturdays.

1n = 6

Helene walks 6 laps on Saturdays.

n = 6

Course 3


Lesson Quiz

Solve.

1. 4x + 16 = 2x

2. 8x – 3 = 15 + 5x

3. 2(3x + 11) = 6x + 4

4. x = x – 9

5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each?

x = 6

x = –8


no solution

x = 3614

12

An orange has 45 calories. An apple has 75 calories.

Course 3


10-4 Solving Multistep Inequalities

Course 3

Warm UpWarm Up



Warm UpSolve.

1. 6x + 36 = 2x

2. 4x – 13 = 15 + 5x

3. 5(x – 3) = 2x + 3

4. + x =

x = –9

x = –28

x = 6

Course 3


78

316

1116

x = –

Problem of the Day

Find an integer x that makes the following two inequalities true:4 < x2 < 16 and x < 2.5

x = –3

Course 3


Learn to solve two-step inequalities and graph the solutions of an inequality on a number line.

Course 3


Solving a multistep inequality uses the same inverse operations as solving a multistep equation. Multiplying or dividing the inequality by a negative number reverses the inequality symbol.

Course 3


Solve and graph.

Additional Example 1A: Solving Multistep Inequalities

A. 4x + 1 > 13

4x + 1 > 13 – 1 – 1 Subtract 1 from both sides.

4x > 124x4

> 124


x > 3 1 2 3 4 5 6 7

Course 3


Additional Example 1B: Solving Multistep Inequalities

B. –7 < 3x + 8

–7 < 3x + 8


–15 < 3x

– 15 3

< 3x 3 Divide both sides by 3.

–5 < x -7 -6 -5 -4 -3 -2 -1

Course 3


Additional Example 1C: Solving Multistep Inequalities

C. -9x + 7 25

–9x + 7 25


–9x 18

–9x–9

18–9

Divide each side by –9; change to .

x –2-6 -5 -4 -3 -2 -1 0

Course 3


Solve and graph.


A. 5x + 2 > 12

5x + 2 > 12 – 2 – 2 Subtract 2 from both sides.

5x > 105x5

> 105


x > 2 1 2 3 4 5 6 7

Course 3


B. –5 < 2x + 9

–5 < 2x + 9


–14 < 2x

– 14 2

< 2x 2 Divide both sides by 2.

–7 < x -7 -6 -5 -4 -3 -2 -1


Course 3


C. -4x + 2 18

–4x + 2 18


–4x 16

–4x–4

16–4

Divide each side by –4; change to .

x –4-6 -5 -4 -3 -2 -1 0


Course 3


Solve and graph.

Additional Example 2A: Solving Multistep Inequalities

A. 10x + 21 – 4x < –15

10x + 21 – 4x < –15


6x6

< –36 6 Divide both sides by 6.

x < –6-8 -7 -6 -5 -4 -3 -2

6x + 21 < –15 Combine like terms.

6x < –36

Course 3


Additional Example 2B: Solving Multistep Inequalities

Multiply by LCD, 20.

8x + 15 18


8x 3

B. + 2x5

34

910

+ 2x5

34

910

20( + ) 20( )2x5

34

910

20( ) + 20( ) 20( )2x5

34

910

Course 3



x 38

8x8

38 Divide both sides by 8.

8x 3

Course 3


0 1

38

Additional Example 2C: Solving Multistep Inequalities

C. 8x + 8 > 11x – 1

8x + 8 > 11x – 1– 8x – 8x Subtract 8x from both sides.

8 > 3x – 1

93

> 3x 3

Add 1 to each side.

3 > x

-1 0 1 2 3 4 5

+1 +1

9 > 3x

Course 3



Solve and graph.


A. 15x + 30 – 5x < –10

15x + 30 – 5x < –10


10x10

< –40 10 Divide both sides by 10.

x < –4-8 -7 -6 -5 -4 -3 -2

10x + 30 < –10 Combine like terms.

10x < –40

Course 3



Multiply by LCD, 20.

12x + 5 10


12x 5

B. + 3x5

14

510

+ 3x5

14

510

20( + ) 20( )3x5

14

510

20( ) + 20( ) 20 ( )3x5

14

510

Course 3



x 512

12x12

512 Divide both sides by 12.

12x 5

0 5 12

Course 3



C. 4x + 3 > 8x – 1

4x + 3 > 8x – 1– 4x – 4x Subtract 4x from both sides.

3 > 4x – 1

44

> 4x 4

Add 1 to each side.

1 > x

-1 0 1 2 3 4 5

+1 +1

4 > 4x

Course 3



Additional Example 3: Business Application

A school’s Spanish club is selling bumper stickers. They bought 100 bumper stickers for $55, and they have to give the company 15 cents for every sticker sold. If they plan to sell each bumper sticker for $1.25, how many do they have to sell to make a profit?Let R represent the revenue and C represent the cost. In order for the Spanish club to make a profit, the revenue must be greater than the cost.

R > C

Course 3



The revenue from selling x bumper stickers at $1.25 each is 1.25x. The cost of selling x bumper stickers is the fixed cost plus the unit cost times the number of bumper stickers sold, or 55 + 0.15x. Substitute the expressions for R and C.

1.25x > 55 + 0.15x Let x represent the number of bumper stickers sold. Fixed cost is $55. Unit cost is 15 cents.

Course 3



– 0.15x – 0.15x Subtract 0.15x from both sides.

1.10x > 55

x > 50

The Spanish club must sell more than 50 bumper stickers to make a profit.


1.25x > 55 + 0.15x

1.10x1.10

551.10>

Course 3


Try This: Example 3

R > C

A school’s Spanish club is selling bumper stickers. They bought 200 bumper stickers for $45, and they have to give the company 25 cents for every sticker sold. If they plan to sell each bumper sticker for $2.50, how many do they have to sell to make a profit?Let R represent the revenue and C represent the cost. In order for the Spanish club to make a profit, the revenue must be greater than the cost.

Course 3



The revenue from selling x bumper stickers at $2.50 each is 2.5x. The cost of selling x bumper stickers is the fixed cost plus the unit cost times the number of bumper stickers sold, or 45 + 0.25x. Substitute the expressions for R and C.

2.5x > 45 + 0.25x Let x represent the number of bumper stickers sold. Fixed cost is $45. Unit cost is 25 cents.

Course 3


– 0.25x – 0.25x Subtract 0.25x from both sides.

2.25x > 45

x > 20

The Spanish club must sell more than 20 bumper stickers to make a profit.


2.5x > 45 + 0.25x

2.25x2.25

452.25>


Course 3


Lesson Quiz: Part 1

Solve and graph.

1. 4x – 6 > 10

2. 7x + 9 < 3x – 15

3. w – 3w < 32

4. w +

x < –6

x > 4


w > –1623

14

12

w 38

1 2 3 4 5 6 7

-10 -9 -8 -7 -6 -5 -4

-18 -17 -16 -15 -14 -13 -12

0 38

Course 3


Lesson Quiz: Part 2

5. Antonio has budgeted an average of $45 a month for entertainment. For the first five months of the year he has spent $48, $39, $60, $48, and $33. How much can Antonio spend in the sixth month without exceeding his average budget?

no more than $42

Course 3


10-5 Solving for a Variable

Course 3

Warm UpWarm Up



Warm UpSolve.

1. 8x – 9 = 23

2. 9x + 12 = 4x + 37

3. 6x – 8 = 7x + 3

4. x + 3 =

x = 4

x = 5

x = –11

Course 3


12

18

23 4x = – , or –5 3

4

Problem of the Day

The formula A = 4r2 gives the surface area of a geometric figure. Solve the formula for r.

Can you identify what the geometric figure is?sphere

Course 3


Learn to solve an equation for a variable.

Course 3


If an equation contains more than one variable, you can sometimes isolate one of the variables by using inverse operations. You can add and subtract any variable quantity on both sides of an equation.

Course 3


Solve for the indicated variable.

Additional Example 1A: Solving for a Variable by Addition or Subtraction

A. Solve a – b + 1 = c for a.

a – b + 1 = c

+ b – 1 + b – 1 Add b and subtract 1 from both sides.

a = c + b – 1 Isolate a.

Course 3



Additional Example 1B: Solving for a Variable by Addition or Subtraction

Course 3


B. Solve a – b + 1 = c for b.

a – b + 1 = c– a – 1 – a – 1 Subtract a and 1 from both

sides.

–b = c – a – 1 Isolate b.

Multiply both sides by –1.–1 (–b) = –1 (c – a – 1)

b = –c + a + 1 Isolate b.



A. Solve y – b + 3 = c for y.

y – b + 3 = c

+ b – 3 + b – 3 Add b and subtract 3 from both sides.

y = c + b – 3 Isolate y.

Course 3




Course 3


B. Solve p – w + 4 = f for w.

p – w + 4 = f– p – 4 – p – 4 Subtract p and 4 from both

sides.

–w = f – p – 4 Isolate w.

Multiply both sides by –1.–1 (–w) = –1 (f – p – 4)

w = –f + p + 4 Isolate w.

Course 3


To isolate a variable, you can multiply or divide both sides of an equation by a variable if it can never be equal to 0. You can also take the square root of both sides of an equation that cannot have negative values.

Solve for the indicated variable. Assume all values are positive.

Additional Example 2A: Solving for a Variable by Division or Square Roots

A. Solve A = s2 for s.

A = s2

Course 3


√A = s Isolate s.

√A = √s2 Take the square root of both sides.

Additional Example 2B: Solving for a Variable by Division or Square Roots

B. Solve V = IR for R.

V = IR

= Divide both sides by I.IRI

VI

IV= R Isolate R.

Course 3



Additional Example 2C: Solving for a Variable by Division or Square Roots

C. Solve the formula for the area of a trapezoid for h. Assume all values are positive.

A = h(b1 + b2)12 Write the formula.

2 • A = 2 • h(b1 + b2)12 Multiply both sides by 2.

2A = h(b1 + b2)

Course 3


2A(b1 + b2)

= h Isolate h.

2A(b1 + b2)

h(b1 + b2) (b1 + b2)

= Divide both sides by (b1 + b2).



A. Solve w = g2 + 4 for g.

w = g2 + 4

Course 3


√ w – 4 = g Isolate g.

√w – 4 = √g2

w – 4 = g2


Take the square root of both sides.

B. Solve A = lw for w.

A = lw

= Divide both sides by l.lwl

Al

lA= w Isolate w.

Course 3




C. Solve s = 180(n – 2) for n.

s = 180(n – 2)

Course 3




s 180(n – 2)180 180

= Divide both sides by 180.

s180

= (n – 2)


s180

+ 2 = n

+ 2 + 2

Course 3


To find solutions (x, y), choose values for x substitute to find y.

Remember!

Additional Example 3: Solving for y and Graphing

Solve for y and graph 3x + 2y = 8.

3x + 2y = 8

–3x –3x

2y = –3x + 8

–3x + 82

2y2

=

y = + 4–3x2

x y

–2 7

0 4

2 1

4 –2

Course 3



Course 3


3x + 2y = 8

Try This: Example 3

Solve for y and graph 4x + 3y = 12.

4x + 3y = 12

–4x – 4x

3y = –4x + 12

–4x + 12 3

3y3

=

y = + 4–4x3

x y

–3 8

0 4

3 0

6 –4

Course 3



x

y

Course 3


–4 –2 2 4 6

10

86

2

–2

–4

44x + 3y = 12

–6

Lesson Quiz: Part 1


1. P = R – C for C.

2. P = 2l+ 2w for l.

3. V = Ah for h.

4. R = for S.

C = R - P


C – Rt = S

13C – S

t

= h3VA

= lP – 2w2

Course 3


Lesson Quiz: Part 2

5. Solve for y and graph 2x + 7y = 14.


y = – + 2 2x 7

Course 3


10-6 Systems of Equations

Course 3

Warm UpWarm Up



Warm UpSolve for the indicated variable.

1. P = R – C for R

2. V = Ah for A

3. R = for C

R = P + C

Rt + S = C

Course 3


13C – S

t

= h3VA

Problem of the Day

At an audio store, stereos have 2 speakers and home-theater systems have 5 speakers. There are 30 sound systems with a total of 99 speakers. How many systems are stereo systems and how many are home-theater systems?

17 stereo systems, 13 home-theater systems

Course 3


Learn to solve systems of equations.

Course 3


Vocabulary

system of equationssolution of a system of equations


Course 3


A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions of all of the equations. If the system has two variables, the solutions can be written as ordered pairs.

Course 3


Determine if the ordered pair is a solution of the system of equations below.

5x + y = 7x – 3y = 11

Additional Example 1A: Identifying Solutions of a System of Equations

A. (1, 2)

5x + y = 7

5(1) + 2 = 7?

7 = 7

x – 3y = 11

1 – 3(2) = 11 ? Substitute for

x and y.–5 11

The ordered pair (1, 2) is not a solution of the system of equations.

Course 3


Additional Example 1B: Identifying Solutions of a System of Equations

B. (2, –3)

5(2) + –3 = 7 ?

7 = 7

2 – 3(–3) = 11? Substitute for

x and y.11 = 11

The ordered pair (2, –3) is a solution of the system of equations.


5x + y = 7x – 3y = 11

5x + y = 7 x – 3y = 11

Course 3


Additional Example 1C: Identifying Solutions of a System of Equations

C. (20, 3)

5(20) + (3) = 7 ?

103 7

20 – 3(3) = 11? Substitute for

x and y.11 = 11



5x + y = 7x – 3y = 11

5x + y = 7 x – 3y = 11

Course 3


Determine if each ordered pair is a solution of the system of equations below.

4x + y = 8x – 4y = 12


A. (1, 2)

4x + y = 8

4(1) + 2 = 8?

6 8

x – 4y = 12

1 – 4(2) = 12 ? Substitute for

x and y.–7 12


Course 3




4x + y = 8x – 4y = 12

B. (2, –3)

4(2) + –3 = 8 ?

5 8

2 – 4(–3) = 12? Substitute for

x and y.14 12

The ordered pair (2, –3) is not a solution of the system of equations.

4x + y = 8 x – 4y = 12

Course 3



C. (1, 4)



4x + y = 8x – 4y = 12

4(1) + 4 = 8 ?

8 = 8

1 – 4(4) = 12? Substitute for

x and y.–15 12

4x + y = 8 x – 4y = 12

Course 3


Course 3


When solving systems of equations, remember to find values for all of the variables.

Helpful Hint

Additional Example 2: Solving Systems of Equations

Solve the system of equations. y = x – 4y = 2x – 9

Solve the equation to find x.

x – 4 = 2x – 9– x – x Subtract x from both sides.

–4 = x – 9

5 = x

+ 9 + 9 Add 9 to both sides.

y = x – 4 y = 2x – 9

y = y

x – 4 = 2x – 9

Course 3



To find y, substitute 5 for x in one of the original equations.

y = x – 4 = 5 – 4 = 1

The solution is (5, 1).

Check: Substitute 5 for x and 1 for y in each equation.

y = x – 4 y = 2x – 9

1 = 5 – 4? 1 = 2(5) – 9

?

1 = 1 1 = 1

Course 3


Try This: Example 2

Solve the system of equations. y = x – 5y = 2x – 8

Solve the equation to find x.

x – 5 = 2x – 8– x – x Subtract x from both sides.

–5 = x – 8

3 = x

+ 8 + 8 Add 8 to both sides.

y = x – 5 y = 2x – 8

y = y

x – 5 = 2x – 8

Course 3



To find y, substitute 3 for x in one of the original equations.

y = x – 5 = 3 – 5 = –2

The solution is (3, –2).

Check: Substitute 3 for x and –2 for y in each equation.

y = x – 5 y = 2x – 8

–2 = 3 – 5 ? –2 = 2(3) – 8

?

–2 = –2 –2 = –2

Course 3


To solve a general system of two equations with two variables, you can solve both equations for x or both for y.

Course 3


Additional Example 3A: Solving Systems of Equations

Solve the system of equations.

A. x + 2y = 8 x – 3y = 13x + 2y = 8 x – 3y = 13

–2y –2y + 3y + 3y

Solve both equations for x.

x = 8 – 2y x = 13 + 3y

8 – 2y = 13 + 3y+ 2y + 2y

8 = 13 + 5y

Add 2y to both sides.

Course 3


Additional Example 3A Continued

8 = 13 + 5y

–13 –13

–5 = 5y


–55

5y 5 = Divide both sides

by 5.–1 = y

x = 8 – 2y = 8 – 2(–1) Substitute –1 for y. = 8 + 2 = 10The solution is (10, –1).

Course 3


Course 3


You can choose either variable to solve for. It is usually easiest to solve for a variable that has a coefficient of 1.

Helpful Hint

Additional Example 3B: Solving Systems of Equations


B. 3x – 3y = -3 2x + y = -53x – 3y = –3 2x + y = –5

–3x –3x –2x –2x

Solve both equations for y.

–3y = –3 – 3x y = –5 – 2x

–3–3

3x–3

–3y–3 = –

y = 1 + x

1 + x = –5 – 2x

Course 3



+ 2x + 2x Add 2x to both sides.

1 + 3x = –5–1 –1

3x = –6

1 + x = –5 – 2x


–6 3

3x3 =


x = –2y = 1 + x

= 1 + –2 = –1 Substitute –2 for x.

The solution is (–2, –1).

Course 3




A. x + y = 5 3x + y = –1x + y = 5 3x + y = –1

–x –x – 3x – 3x


y = 5 – x y = –1 – 3x

5 – x = –1 – 3x+ x + x

5 = –1 – 2x

Add x to both sides.

Course 3



5 = –1 – 2x

+ 1 + 1

6 = –2x



–3 = x

y = 5 – x = 5 – (–3) Substitute –3 for x. = 5 + 3 = 8The solution is (–3, 8).

Course 3




B. x + y = –2 –3x + y = 2x + y = –2 –3x + y = 2

– x – x + 3x + 3x


y = –2 – x y = 2 + 3x

–2 – x = 2 + 3x

Course 3


+ x + x Add x to both sides.

–2 = 2 + 4x–2 –2

–4 = 4x

–2 – x = 2 + 3x


Divide both sides by 4.–1 = x

y = 2 + 3x= 2 + 3(–1) = –1 Substitute –1

for x.The solution is (–1, –1).

Course 3



Lesson Quiz

1. Determine if the ordered pair (2, 4) is a solution of the system. y = 2x; y = –4x + 12

Solve each system of equations.

2. y = 2x + 1; y = 4x

3. 6x – y = –15; 2x + 3y = 5

4. Two numbers have a sum of 23 and a difference of 7. Find the two numbers.

yes


(–2,3)

15 and 8

( , 2)12

Course 3


Chapter 10

Technology

Transcript of Chapter 10