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**Solving Multi-Step Equations**

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

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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

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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

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**Learn to solve multi-step equations.**

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

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**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

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**Additional Example 1 Continued**

Check 8x x – 2 = 37 8(3) (3) – 2 = 37 ? Substitute 3 for x. – 2 = 37 ? 37 = 37 ?

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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

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**Check It Out: Example 1 Continued**

9x x – 2 = 42 9(3) (3) – 2 = 42 ? Substitute 3 for x. – 2 = 42 ? 42 = 42 ?

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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.

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**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

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**Additional Example 2A Continued**

– 7 –7 Subtract 7 from both sides. 5n = –10 5n 5 –10 = Divide both sides by 5 n = –2

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

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**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.

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**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

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**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 = ?

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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

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**Check It Out: Example 2A Continued**

– 5 – Subtract 5 from both sides. 3n = –6 3n 3 –6 = Divide both sides by 3. n = –2

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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.

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**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

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**Check It Out: Example 2B Continued**

3 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 = ?

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**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

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**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.

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**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.

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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

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**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.

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**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.

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**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

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