Course 3 11-2 Solving Multi-Step Equations 11-2 Solving Multi-Step Equations Course 3 Warm Up Warm...

Course 3

11-2 Solving Multi-Step Equations11-2 Solving Multi-Step Equations

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Course 3

11-2 Solving Multi-Step Equations

Warm UpSolve.

1. 3x = 102

2. = 15

3. z – 100 = 21

4. 1.1 + 5w = 98.6

x = 34

y = 225

z = 121

w = 19.5

y15

Course 3


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 multi-step equations.

Course 3


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


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.

+ = –

Additional Example 2A: 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 ( ) ( ) ( ) ( ) ( )5n

474

–3 44 + 4 = 4

5n + 7 = –3

Distributive Property.

Course 3


Additional Example 2A Continued

5n + 7 = –3 – 7 –7 Subtract 7 from both sides.

5n = –10

5n5

–10 5

= Divide both sides by 5

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.

+ – =

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


Multiply both sides by 18.

Course 3


Additional Example 2B Continued

23x = 46

= 23x23

4623 Divide both sides by 23.

x = 2

+ 34 + 34 Add 34 to both sides.


Course 3


Additional Example 2B Continued

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.

+ = –

Check It Out: Example 2A

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


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.

+ – =

Check It Out: Example 2B

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( )


Multiply both sides by 9.

Course 3


8x = 16

= 8x8

16 8 Divide both sides by 8.

x = 2

+ 13 + 13 Add 13 to both sides.


Check It Out: 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

?

Check It Out: Example 2B Continued

Course 3


On Monday, David rides his bicycle m miles in 2 hours. On Tuesday, he rides three times as far in 5 hours. If his average speed for two days is 12 mi/h, how far did he ride on the second day? Round your answer to the nearest tenth of a mile.

Additional Example 3: Travel Application

David’s average speed is his combined speeds for the two days divided by 2.

average speed

=Day 1 speed Day 2 speed+

2

Course 3



Multiply both sides by the LCD 10.

+Multiply both sides by 2.

22 = 2(12)

3m5

m2

m2

10 = 10(24) 3m5

+

m 2Substitute for Day 1 speed

and for Day 2 speed.3m 5

m2

2= 12

3m5

+

1

1

Course 3



5m + 6m = 240

On the second day David rode 3 times m (3m) or approximately 65.5 miles.

Combine like terms. Divide both sides by 11.

m ≈ 21.82

Simplify.

240 11

11m11

=

Course 3


On Saturday, Penelope rode her scooter m miles in 3 hours. On Sunday, she rides twice as far in 7 hours. If her average speed for two days is 20 mi/h, how far did she ride on Sunday? Round your answer to the nearest tenth of a mile.

Check It Out: Example 3

Penelope’s average speed is her combined speeds for the two days divided by 2.

average speed

=Day 1 speed Day 2 speed+

2

Course 3



Multiply both sides by the LCD 21.

+Multiply both sides by 2.

22 = 2(20)

2m7

m3

m 3Substitute for Day 1 speed

and for Day 2 speed.2m 7

m3

21 = 21(40) 2m7

+

m3

2= 20

2m7

+

1

1

Course 3



7m + 6m = 840

On Sunday Penelope rode 2 times m, (2m), or approximately 129.2 miles.

Combine like terms. Divide both sides by 13.

m ≈ 64.62

Simplify.

840 13

13m13

=

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 11-2 Solving Multi-Step Equations 11-2 Solving Multi-Step Equations Course 3 Warm Up Warm...

Documents

Transcript of Course 3 11-2 Solving Multi-Step Equations 11-2 Solving Multi-Step Equations Course 3 Warm Up Warm...