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HW # 32 - Videos for section 3-4 AND p. 132 # 1-11 all

Week 9, Day Two. HW # 32 - Videos for section 3-4 AND p. 132 # 1-11 all Homework help online go.hrw.com keyword: MT8CA 3-4 Warm up 5n + 3n – n + 5 = 26 -81 = 7k + 19 + 3k 37 = 15a-5a-3

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HW # 32 - Videos for section 3-4 AND p. 132 # 1-11 all

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  1. Week 9, Day Two HW # 32- Videos for section 3-4 AND p. 132 # 1-11 all Homework help online go.hrw.com keyword: MT8CA 3-4 Warm up 5n + 3n – n + 5 = 26 -81 = 7k + 19 + 3k 37 = 15a-5a-3 Lydia rode 243 miles in a three-day bike trip. On the first day, Lydia rode 67 miles. On the second day, she rode 92 miles. How many miles per hour did she average on the third day if she rode for 7 hours?

  2. Warm Up Response 5n + 3n – n + 5 = 26 n=3 -81 = 7k + 19 + 3k k=-10 37 = 15a-5a-3 a=4 Lydia rod 243 miles in a three-day bike trip. On the first day, Lydia rode 67 miles. On the second day, she rode 92 miles. How many miles per hour did she average on the third day if she rode for 7 hours? 12 mi/h

  3. Goals for Today • Solving Equations with Variables on Both Sides (3-4) • Work on Simplifying Expressions from yesterday’s class work

  4. Slides for extra practice at home • Note, some of the formatting may be off and the ppt moves from Mac to PC.

  5. –3x = –3 6 –3 Additional Example 1A: Solving Equations with Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x To collect the variable terms on one side, subtract 4x from both sides. – 4x– 4x 6 = –3x Since x is multiplied by -3, divide both sides by –3. –2 = x

  6. Helpful Hint You can always check your solution by substituting the value back into the original equation.

  7. 4b 24 = 4 4 Additional Example 1B: Solving Equations with Variables on Both Sides Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 To collect the variable terms on one side, subtract 5b from both sides. – 5b– 5b 4b – 6 = 18 Since 6 is subtracted from 4b, add 6 to both sides. + 6+ 6 4b = 24 Since b is multiplied by 4, divide both sides by 4. b = 6

  8. 9w + 3 = 9w + 7 – 9w– 9w To collect the variable terms on one side, subtract 9w from both sides. Additional Example 1C: Solving Equations with Variables on Both Sides Solve. 9w + 3 = 9w + 7 3 ≠ 7 There is no solution. There is no number that can be substituted for the variable w to make the equation true.

  9. Helpful Hint if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution.

  10. –4x = –4 8 –4 Check It Out! Example 1A Solve. 5x + 8 = x 5x + 8 = x To collect the variable terms on one side, subtract 5x from both sides. – 5x– 5x 8 = –4x Since x is multiplied by –4, divide both sides by –4. –2 = x

  11. Check It Out! Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 To collect the variable terms on one side, subtract 2b from both sides. – 2b– 2b b – 2 = 12 Since 2 is subtracted from b, add 2 to both sides. + 2+ 2 b = 14

  12. 3w + 1 = 3w + 8 – 3w– 3w To collect the variable terms on one side, subtract 3w from both sides. Check It Out! Example 1C Solve. 3w + 1 = 3w + 8 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true.

  13. To solve more complicated equations, you may need to first simplify by combining like terms or clearing fractions. Then add or subtract to collect variable terms on one side of the equation. Finally, use properties of equality to isolate the variable.

  14. 8z8 = 8 8 Additional Example 2A: Solving Multi-Step Equations with Variables on Both Sides Solve. 10z – 15 – 4z = 8 – 2z – 15 10z – 15 – 4z = 8 – 2z – 15 6z – 15 = –2z – 7 Combine like terms. + 2z+ 2z Add 2z to both sides. 8z – 15 = – 7 + 15+15 Add 15 to both sides. 8z = 8 Divide both sides by 8. z = 1

  15. 7 10 7 10 7 10 7 10 3y 5 3y 5 3y 5 3y 5 y 5 y 5 y 5 y 5 3 4 3 4 3 4 3 4 + – = y – 20( ) = 20( ) + – y – 20() + 20( ) – 20( )= 20(y) – 20( ) Additional Example 2B: Solving Multi-Step Equations with Variables on Both Sides + – = y – Multiply by the LCD, 20. 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14 Combine like terms.

  16. 4y 4 –1 –1 4 = y = 4 Additional Example 2B Continued 16y – 15 = 20y – 14 – 16y– 16y Subtract 16y from both sides. –15 = 4y – 14 + 14+ 14 Add 14 to both sides. –1 = 4y Divide both sides by 4.

  17. 10z50 = 10 10 Check It Out! Example 2A Solve. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 Combine like terms. + 2z+ 2z Add 2z to both sides. 10z – 12 = 38 + 12+12 Add 12 to both sides. 10z = 50 Divide both sides by 10. z = 5

  18. 6 8 6 8 6 8 6 8 5y 6 5y 6 5y 6 5y 6 y 4 y 4 y 4 y 4 3 4 3 4 3 4 3 4 + + = y – 24( ) = 24( ) + + y – 24() + 24( )+ 24( )= 24(y) – 24( ) Check It Out! Example 2B + + = y – Multiply by the LCD, 24. 6y + 20y + 18 = 24y – 18 26y + 18 = 24y – 18 Combine like terms.

  19. 2y 2 –36 2 = Check It Out! Example 2B Continued 26y + 18 = 24y – 18 – 24y– 24y Subtract 24y from both sides. 2y + 18 = – 18 – 18– 18 Subtract 18 from both sides. 2y = –36 Divide both sides by 2. y = –18

  20. Additional Example 3: Business Application Daisy’s Flowers sells a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florists' bouquets cost the same price. What is the price? Write an equation for each service. Let c represent the total cost and r represent the number of roses. total cost is flat fee plus cost for each rose Daisy’s: c = 39.95 + 2.95 r Other: c = 26.00 + 4.50 r

  21. 13.95 1.55r1.55 = 1.55 Additional Example 3 Continued Now write an equation showing that the costs are equal. 39.95 + 2.95r = 26.00 + 4.50r Subtract 2.95r from both sides. – 2.95r– 2.95r 39.95 = 26.00 + 1.55r Subtract 26.00 from both sides. – 26.00– 26.00 13.95 = 1.55r Divide both sides by 1.55. 9 = r The two bouquets from either florist would cost the same when purchasing 9 roses.

  22. Additional Example 3 Continued To find the cost, substitute 9 for r into either equation. Daisy’s: Other florist: c = 39.95 + 2.95r c = 26.00 + 4.50r c = 39.95 + 2.95(9) c = 26.00 + 4.50(9) c = 39.95 + 26.55 c = 26.00 + 40.50 c = 66.5 c = 66.5 The cost for a bouquet with 9 roses at either florist is $66.50.

  23. Check It Out! Example 3 Marla’s Gift Baskets sells a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both baskets cost the same price. Write an equation for each service. Let c represent the total cost and b represent the number of balloons. total cost is flat fee plus cost for each balloon Marla’s: c = 22.00 + 2.25 b Other: c = 16.00 + 3.00 b

  24. 6.00 0.75b0.75 = 0.75 Check It Out! Example 3 Continued Now write an equation showing that the costs are equal. 22.00 + 2.25b = 16.00 + 3.00b Subtract 2.25b from both sides. – 2.25b– 2.25b 22.00 = 16.00 + 0.75b Subtract 16.00 from both sides. – 16.00– 16.00 6.00 = 0.75b Divide both sides by 0.75. 8 = b The two services would cost the same when purchasing a muffin basket with 8 balloons.

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