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Section 4.5

Chapter 4. Section 4.5. Chapter 4. Section 4.5. Exercise #3. Match the application with the appropriate system. Geometry . The length of a rectangle is 3 cm more than twice its width. If the perimeter of the rectangle is 36 cm, find the dimensions of the rectangle.

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Section 4.5

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  1. Chapter 4 Section 4.5

  2. Chapter 4 Section 4.5 Exercise #3

  3. Match the application with the appropriate system. Geometry. The length of a rectangle is 3 cm more than twice its width. If the perimeter of the rectangle is 36 cm, find the dimensions of the rectangle. L = length W = width 3 + L = 2W 36 = 2L + 2W

  4. (e) (a) (f) (b) (c) (g) (h) (d)

  5. The answer is (g). Geometry. The length of a rectangle is 3 cm more than twice its width. If the perimeter of the rectangle is 36 cm, find the dimensions of the rectangle. L = length W = width 3 + L = 2W 36 = 2L + 2W

  6. Chapter 4 Section 4.5 Exercise #43

  7. Solve the problem. Mixture problem. A garden store sold 8 bags of mulch and 3 bags of fertilizer for $24. The next purchase was for 5 bags of mulch and 5 bags of fertilizer. The cost of that purchase was $25. Find the cost of a single bag of mulch and a single bag of fertilizer.

  8. cost of mulch cost of fertilizer total cost of purchase = + M = cost of one bag of mulch F = cost of one bag of fertilizer 8M + 3F = 24 5M + 5F = 25 Eliminate the variable F: 40M  15F =120 15M + 15F = 75 25M =45 M = 1.80

  9. A bag of mulch costs $1.80 and a bag of fertilizer costs $3.20. Substitute 1.80 for M in 8M + 3F = 24. 8(1.80) + 3F = 24 14.40 + 3F = 24 3F = 9.60 F = 3.20

  10. Chapter 4 Section 4.5 Exercise #57

  11. Solve the problem. Coin problem. A cashier has 25 coins consisting of nickels, dimes, and quarters with a value of $4.90. If the number of dimes is 1 less than twice the number of nickels, how many of each type of coin does she have? n = # of nickels d = # of dimes q = # of quarters

  12. n + d + q = 25 (1) .05n + .1d + .25q = 4.90 (2) d = 2n  1 Rearrange equation above to 2n + d =1.(3) Multiply equation (1) by .25 to eliminate q from (1) and (2). 0.25n  .25d  0.25q =6.25 0.05n + 0.1d + 0.25q = 4.90 0.20n  0.15d =1.35 (4)

  13. n = 3 d = 5 Eliminate d from (3) and (4). Multiply (3) by .15. 0.30n + .15d =0.15 0.20n  .15d =1.35 0.50n =1.50 Substitute n = 3 in (3): 2(3) + d =1 6 + d =1

  14. q = 17 The cashier has 3 nickels, 5 dimes, and 17 quarters. Let n = 3 and d = 5 in (1): 3 + 5 + q = 25 8 + q = 25

  15. Chapter 4 Section 4.5 Exercise #61

  16. Solve the problem. Investments. Jovita divides $17,000 into three investments: a savings account paying 6% interest, a bond paying 9%, and a money market fund paying 11%. The annual interest from the three accounts is $1,540, and she has three times as much invested in the bond as in the savings account. What amount does she have invested in each account?

  17. s = amount invested in a savings account b = amount invested in a bond m = amount invested in a money market fund s + b + m = 17,000 (1) 0.06s + 0.09b + 0.11m = 1,540 (2) b = 3s Rearrange equation above to 3s + b = 0 (3). Eliminate m from (1) and (2). Multiply (1) by 0.11. 0.11s  0.11b  0.11m =1,870 0.06s + 0.09b + 0.11m = 1,540 0.05s  0.02b =330 (4)

  18. s = 3000 b = 9,000 Eliminate b from (3) and (4). Multiply (3) by 0.02. 0.06s + 0.02b = 0 0.05s  0.02b =330 0.11s =330 Substitute s = 3,000 into (3): 3(3,000) + b = 0 9,000 + b = 0 Substitute s = 3,000 and b = 9,000 into (1).

  19. m = 5,000 She has invested $3,000 in a savings account, $9,000 in a bond, and $5,000 in a money market fund. 3,000 + 9,000 + m = 17,000 12,000 + m = 17,000

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