Course 3 11-2 Solving Multi-Step Equations To solve a multi-step equation, you may have to simplify the equation first by combining like terms

Course 3

11-2 Solving Multi-Step Equations

To solve a multi-step 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 4 from both sides.

11x = 33

x = 3

Divide both sides by 11.3311

11x11

=

Course 3


Check

Additional Example 1 Continued

8x + 6 + 3x – 2 = 37

8(3) + 6 + 3(3) – 2 = 37?

24 + 6 + 9 – 2 = 37?

37 = 37?

Substitute 3 for x.

Course 3


Solve.

9x + 5 + 4x – 2 = 42

Check It Out: Example 1

13x + 3 = 42 Combine like terms.

– 3 – 3 Subtract 3 from both sides.13x = 39

x = 3

Divide both sides by 13.3913

13x13

=

Course 3


Check

Check It Out: Example 1 Continued

9x + 5 + 4x – 2 = 42

9(3) + 5 + 4(3) – 2 = 42?

27 + 5 + 12 – 2 = 42 ?

42 = 42?

Substitute 3 for x.

Course 3


Solve.

+ = –

Check It Out: Example 2A

1 4

5 4

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

Distributive Property.

Course 3


Check It Out: Example 2A Continued

3n + 5 = –1 – 5 –5 Subtract 5 from both sides.

3n = –6

3n3

–6 3

= Divide both sides by 3.

n = –2

Course 3


Solve.

1.6x + 3x – x + 9 = 33

2. + =

3. 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

x = 2858

x8

33 8

$8.50

Course 3


Answers 1 - 71. x= -2.2 2. w = 2.75

3. x = 11

4. b = -7 5. m = 1

6. x = 25

7. m = 15

Course 3


Solve.

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

11-3 Solving Equations with Variables on Both Sides

Course 3


Course 3


Check your solution by substituting the value back into the original equation. For example, 4(2) + 6 = 2 or 2 = 2.

Helpful Hint

Course 3


Solve.

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


Course 3


Solve.

5x + 8 = x


5x + 8 = x– 5x – 5x

8 = –4x

Subtract 5x from both sides.

Divide both sides by –4.

–2 = x

8–4

–4x–4=

Course 3


Course 3


Solve.

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.

Check It Out: Example 1B

Course 3


Course 3


Solve.

10z – 15 – 4z = 8 – 2z - 15

Additional Example 2A: Solving Multi-Step 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


Course 3


Solve.

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

Add 12 to both sides.

Divide both sides by 10.10z 5010 10=

Course 3


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.

Check It Out: Example 2B Continued

Course 3


Course 3


Additional Example 3: Business Application

Daisy’s Flowers sell a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florist’s bouquets cost the same price.

Course 3


Course 3


Additional Example 3 Continued

39.95 + 2.95r = 26.00 + 4.50rLet r represent the price of one rose.

– 2.95r – 2.95r

39.95 = 26.00 + 1.55r

Subtract 2.95r from both sides.

– 26.00 – 26.00 Subtract 26.00 from both sides.

13.95 = 1.55r 13.951.55

1.55r 1.55= Divide both sides by 1.55.

9 = r

The two services would cost the same when using 9 roses.Course 3


Course 3


Lesson Quiz

Solve.

1. 4x + 16 = 2x

2. 8x – 3 = 15 + 5x

3. x = x – 9

x = 6

x = –8

Insert Lesson Title Here

x = 3614

12

Course 3


Course 3


Answers 1 - 9

1. x = 9 2. k = 10.2

3. d = 2 4. a = -4

5. x = 3 6. d = ¾

7. x = 7 8. y = -2/5

9. x = 4

Course 3


Warm UpSolve.

1. 6x + 36 = 2x

2. 4x – 13 = 15 + 5x

3. 5(x – 3) = 2x + 3

x = –9

x = –28

x = 6

Course 3

11-5 Solving Two-Step Inequalities

Course 3


Solve and graph.

Additional Example 1A: Solving Two-Step Inequalities

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


Course 3


Course 3


If both sides of an inequality are multiplied or divided by a negative number, the inequality symbol must be reversed.

Remember!

Course 3


Additional Example 1B: Solving Two-Step Inequalities

–9x + 7 25

–9x + 7 25

– 7 – 7 Subtract 7 from both sides.

–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.

Course 3


Solve and graph.


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


Course 3


–4x + 2 18

–4x + 2 18

– 2 – 2 Subtract 2 from both sides.

–4x 16

–4x–4

16–4

Divide each side by –4; change to .

x –4-6 -5 -4 -3 -2 -1 0

Check It Out: Example 1B

Course 3


Course 3


Check your Understanding

Solve and graph.

1. 4x – 6 > 10

2. 7x + 9 < 3x – 15

3. w – 3w < 32

x < –6

x > 4

Insert Lesson Title Here

w > –16

1 2 3 4 5 6 7

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

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

Course 3


Course 3


1. 2.

3. 4.

5. 6.

7. 9.

Documents

Course 3 11-2 Solving Multi-Step Equations To solve a multi-step equation, you may have to simplify the equation first by combining like terms