1. Solving Multi-Step Equations
11-2
Solving Multi-Step Equations
Course 3
Warm Up
Problem of the Day
Lesson Presentation

2. Warm Up
Solve.
1. 3x = 102
= 15
3. z – 100 = 21
w = 98.6
x = 34 y 15
y = 225
z = 121
w = 19.5

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

4. Learn to solve multi-step equations.

5. To solve a multi-step equation, you may have to simplify the equation first by combining like terms.

6. Additional Example 1: Solving Equations That Contain Like Terms
Solve.
8x x – 2 = 37
11x + 4 = 37 Combine like terms.
– 4 – 4 Subtract 4 from both sides.
11x = 33 33 11
11x =
Divide both sides by 11.
x = 3

7. Additional Example 1 Continued
Check
8x x – 2 = 37
8(3) (3) – 2 = 37 ?
Substitute 3 for x.
– 2 = 37 ?
37 = 37 ? 

8. Check It Out: Example 1
Solve.
9x x – 2 = 42
13x + 3 = 42 Combine like terms.
– 3 – 3 Subtract 3 from both sides.
13x = 39 39 13
13x =
Divide both sides by 13.
x = 3

9. Check It Out: Example 1 Continued
9x x – 2 = 42
9(3) (3) – 2 = 42 ?
Substitute 3 for x.
– 2 = 42 ?
42 = 42 ? 

10. 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.

11. Additional Example 2A: Solving Equations That Contain Fractions
Solve.
+ = – 5n 4 7 4 3 4
Multiply both sides by 4 to clear fractions, and then solve. 7 4 –3 5n
= 4
( ) ( )
( ) ( ) ( ) 5n 4 7 –3
= 4
Distributive Property.
5n + 7 = –3

12. Additional Example 2A Continued
– 7 –7 Subtract 7 from both sides.
5n = –10 5n 5
–10 =
Divide both sides by 5
n = –2

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

14. Additional Example 2B: Solving Equations That Contain Fractions
Solve.
+ – = x 2 7x 9 17 2 3
The LCD is 18.
( ) ( ) x 2 3 7x 9 17
– = 18
Multiply both sides by 18.
18( ) + 18( ) – 18( ) = 18( ) 7x 9 x 2 17 3
Distributive Property.
14x + 9x – 34 = 12
23x – 34 = 12 Combine like terms.

15. Additional Example 2B Continued
23x – 34 = Combine like terms.
Add 34 to both sides.
23x = 46 =
23x 23 46
Divide both sides by 23.
x = 2

16. Additional Example 2B Continued
Check x 2 7x 9 17 2 3
+ – = 2 3
Substitute 2 for x.
7(2) 9
– =
(2) 17 ? 2 3 14 9
+ – = 17 ? 2 3 14 9
+ – = 17 ? 1
The LCD is 9. 6 9 14
+ – = 17 ? 6 9 = ? 

17. Check It Out: Example 2A
Solve.
+ = – 3n 4 5 4 1 4
Multiply both sides by 4 to clear fractions, and then solve.
( ) ( ) 5 4 –1 3n
= 4
( ) ( ) ( ) 3n 4 5 –1
= 4
Distributive Property.
3n + 5 = –1

18. Check It Out: Example 2A Continued
– 5 – Subtract 5 from both sides.
3n = –6 3n 3 –6 =
Divide both sides by 3.
n = –2

19. Check It Out: Example 2B
Solve.
+ – = x 3 5x 9 13 1 3
The LCD is 9.
( ) x 3 1 5x 9 13
– = 9( )
Multiply both sides by 9.
9( ) + 9( )– 9( ) = 9( ) 5x 9 x 3 13 1
Distributive Property.
5x + 3x – 13 = 3
8x – 13 = 3 Combine like terms.

20. Check It Out: Example 2B Continued
8x – 13 = Combine like terms.
Add 13 to both sides.
8x = 16 = 8x 8 16
Divide both sides by 8.
x = 2

21. Check It Out: Example 2B Continued3 5x 9 13 1 3
+ – = 1 3
Substitute 2 for x.
5(2) 9
– =
(2) 13 ? 1 3 10 9
+ – = 2 13 ?
The LCD is 9. 3 9 10
+ – = 6 13 ? 3 9 = ? 

22. Additional Example 3: Travel Application
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.
David’s average speed is his combined speeds for the two days divided by 2.
Day 1 speed
Day 2 speed + 2 =
average speed

23. Additional Example 3 Continued
Substitute for Day 1 speed
and for Day 2 speed.
3m 5 m2 2
= 12
3m5 + +
Multiply both sides by 2. 2
= 2(12)
3m5 m2 1 1 m2
= 10(24)
3m5 +
Multiply both sides by the LCD 10.

24. Additional Example 3 Continued
Simplify.
5m + 6m = 240
Combine like terms. Divide both sides by 11.
240 11
11m11 =
m ≈ 21.82
On the second day David rode 3 times m (3m) or approximately 65.5 miles.

25. Check It Out: Example 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.
Penelope’s average speed is her combined speeds for the two days divided by 2.
Day 1 speed
Day 2 speed + 2 =
average speed

26. Check It Out: Example 3 Continued
Substitute for Day 1 speed
and for Day 2 speed.
2m 7 m3 2
= 20
2m7 + +
Multiply both sides by 2. 2
= 2(20)
2m7 m3 1 1 m3
= 21(40)
2m7 +
Multiply both sides by the LCD 21.

27. Check It Out: Example 3 Continued
Simplify.
7m + 6m = 840
Combine like terms. Divide both sides by 13.
840 13
13m13 =
m ≈ 64.62
On Sunday Penelope rode 2 times m, (2m), or approximately miles.

28. Lesson Quiz Solve. 1. 6x + 3x – x + 9 = 33 2. –9 = 5x + 21 + 3x x = 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?
x = 3
x = –3.75 5 8 x 8 33 8
x = 28
x = 1 9 16
– = 6x 7 2x 21 25 21
$8.50