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Solving Multi-Step Equations. 11-2. Course 3. Warm Up. Problem of the Day. Lesson Presentation. y 15. Warm Up Solve. 1. 3 x = 102 2. = 15 3. z – 100 = 21 4. 1.1 + 5 w = 98.6. x = 34. y = 225. z = 121. w = 19.5. Problem of the Day

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11-2

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

  2. y 15 Warm Up Solve. 1.3x = 102 2. = 15 3.z – 100 = 21 4. 1.1 + 5w = 98.6 x = 34 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. 33 11x = 11 11 Additional Example 1: Solving Equations That Contain Like Terms Solve. 8x + 6 + 3x – 2 = 37 11x + 4 = 37 Combine like terms. – 4– 4Subtract 4 from both sides. 11x = 33 Divide both sides by 11. x = 3

  7. ? 8(3) + 6 + 3(3) – 2 = 37 ? 24 + 6 + 9 – 2 = 37 ? 37 = 37 Additional Example 1 Continued Check 8x + 6 + 3x – 2 = 37 Substitute 3 for x. 

  8. 39 13x = 13 13 Check It Out: Example 1 Solve. 9x + 5 + 4x – 2 = 42 13x + 3 = 42 Combine like terms. – 3– 3Subtract 3 from both sides. 13x = 39 Divide both sides by 13. x = 3

  9. ? 9(3) + 5 + 4(3) – 2 = 42 ? 27 + 5 + 12 – 2 = 42 ? 42 = 42 Check It Out: Example 1 Continued Check 9x + 5 + 4x – 2 = 42 Substitute 3 for x. 

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

  12. –10 Divide both sides by 5 5 5n = 5 Additional Example 2A Continued 5n + 7 = –3 – 7–7Subtract 7 from both sides. 5n = –10 n = –2

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

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

  15. 46 = Divide both sides by 23. 23 23x 23 Additional Example 2B Continued 23x – 34 = 12 Combine like terms. + 34+ 34Add 34 to both sides. 23x = 46 x = 2

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

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

  18. –6 Divide both sides by 3. 3 3n = 3 Check It Out: Example 2A Continued 3n + 5 = –1 – 5–5Subtract 5 from both sides. 3n = –6 n = –2

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

  20. 16 = Divide both sides by 8. 8 8x 8 Check It Out: Example 2B Continued 8x – 13 = 3 Combine like terms. + 13+ 13Add 13 to both sides. 8x = 16 x = 2

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

  22. + Day 1 speed Day 2 speed 2 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. = average speed

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

  24. Combine like terms. Divide both sides by 11. Simplify. 11m11 24011 = Additional Example 3 Continued 5m + 6m = 240 m ≈ 21.82 On the second day David rode 3 times m (3m) or approximately 65.5 miles.

  25. + Day 1 speed Day 2 speed 2 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. = average speed

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

  27. Combine like terms. Divide both sides by 13. Simplify. 13m13 84013 = Check It Out: Example 3 Continued 7m + 6m = 840 m ≈ 64.62 On Sunday Penelope rode 2 times m, (2m), or approximately 129.2 miles.

  28. 9 16 25 2x 5 x 6x 33 8 8 8 7 21 21 x = 1 Lesson Quiz 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? x = 3 x = –3.75 x = 28 4. – = $8.50

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