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

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Presentation on theme: "Course 3 11-2 Solving Multi-Step Equations 11-2 Solving Multi-Step Equations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation."— Presentation transcript:

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

2 Course Solving Multi-Step Equations Warm Up Solve. 1. 3x = = z – 100 = w = 98.6 x = 34 y = 225 z = 121 w = 19.5 y 15

3 Course Solving Multi-Step Equations 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 Course Solving Multi-Step Equations Learn to solve multi-step equations.

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

6 Course Solving Multi-Step Equations Solve. 8x x – 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 x 11 =

7 Course Solving Multi-Step Equations Check Additional Example 1 Continued 8x x – 2 = 37 8(3) (3) – 2 = 37 ? – 2 = 37 ? 37 = 37 ? Substitute 3 for x.

8 Course Solving Multi-Step Equations Solve. 9x x – 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 x 13 =

9 Course Solving Multi-Step Equations Check Check It Out: Example 1 Continued 9x x – 2 = 42 9(3) (3) – 2 = 42 ? – 2 = 42 ? 42 = 42 ? Substitute 3 for x.

10 Course Solving Multi-Step Equations 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 Course Solving Multi-Step Equations Solve. + = – Additional Example 2A: Solving Equations That Contain Fractions n5n 4 Multiply both sides by 4 to clear fractions, and then solve. 7 4 –3 4 5n5n = 4 ( ) ( ) ( ) ( ) ( ) 5n – = 4 5n + 7 = –3 Distributive Property.

12 Course Solving Multi-Step Equations Additional Example 2A Continued 5n + 7 = –3 – 7 –7 Subtract 7 from both sides. 5n = –10 5n5n 5 –10 5 = Divide both sides by 5 n = –2

13 Course Solving Multi-Step Equations The least common denominator (LCD) is the smallest number that each of the denominators will divide into. Remember!

14 Course Solving Multi-Step Equations Solve. + – = Additional Example 2B: Solving Equations That Contain Fractions 2 3 The LCD is 18. x 2 7x7x ( ) + 18 ( ) – 18 ( ) = 18 ( ) 7x7x 9 x x + 9x – 34 = 12 23x – 34 = 12 Combine like terms. ( ) ( ) x x7x – = 18 Distributive Property. Multiply both sides by 18.

15 Course Solving Multi-Step Equations Additional Example 2B Continued 23x = 46 = 23x Divide both sides by 23. x = Add 34 to both sides. 23x – 34 = 12 Combine like terms.

16 Course Solving Multi-Step Equations Additional Example 2B Continued = ? Check x 2 7x7x – = Substitute 2 for x. 7(2) 9 + – = (2) ? – = ? – = 17 9 ? 1 The LCD is – = ?

17 Course Solving Multi-Step Equations Solve. + = – Check It Out: Example 2A n3n 4 Multiply both sides by 4 to clear fractions, and then solve. ( ) ( ) 5 4 –1 4 3n3n = 4 ( ) ( ) ( ) 3n – = 4 3n + 5 = –1 Distributive Property.

18 Course Solving Multi-Step Equations Check It Out: Example 2A Continued 3n + 5 = –1 – 5 –5 Subtract 5 from both sides. 3n = –6 3n3n 3 –6 3 = Divide both sides by 3. n = –2

19 Course Solving Multi-Step Equations Solve. + – = Check It Out: Example 2B 1 3 The LCD is 9. x 3 5x5x ( ) + 9 ( ) – 9 ( ) = 9 ( ) 5x5x 9 x x + 3x – 13 = 3 8x – 13 = 3 Combine like terms. ( ) x x5x – = 9 ( ) Distributive Property. Multiply both sides by 9.

20 Course Solving Multi-Step Equations 8x = 16 = 8x8x Divide both sides by 8. x = Add 13 to both sides. 8x – 13 = 3 Combine like terms. Check It Out: Example 2B Continued

21 Course Solving Multi-Step Equations = ? Check x 3 5x5x – = Substitute 2 for x. 5(2) 9 + – = (2) ? – = ? The LCD is – = ? Check It Out: Example 2B Continued

22 Course Solving Multi-Step Equations 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 Davids average speed is his combined speeds for the two days divided by 2. average speed = Day 1 speedDay 2 speed + 2

23 Course Solving Multi-Step Equations Additional Example 3 Continued Multiply both sides by the LCD Multiply both sides by = 2(12) 3m 5 m2m2 m2m2 10 = 10(24) 3m 5 + m2m2 Substitute for Day 1 speed and for Day 2 speed. 3m 5 m2m2 2 = 12 3m53m

24 Course Solving Multi-Step Equations Additional Example 3 Continued 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 Simplify m 11 =

25 Course Solving Multi-Step Equations 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 Penelopes average speed is her combined speeds for the two days divided by 2. average speed = Day 1 speedDay 2 speed + 2

26 Course Solving Multi-Step Equations Check It Out: Example 3 Continued Multiply both sides by the LCD Multiply both sides by = 2(20) 2m 7 m3m3 m3m3 Substitute for Day 1 speed and for Day 2 speed. 2m 7 m3m3 21 = 21(40) 2m 7 + m3m3 2 = 20 2m72m

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

28 Course Solving Multi-Step Equations Solve. 1. 6x + 3x – x + 9 = –9 = 5x 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? Lesson Quiz x = –3.75 x = 3 x = x x6x 7 4. – = 2x2x $8.50 x =


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