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5.4 Problem Solving Using Systems of Equations

5.4 Problem Solving Using Systems of Equations. Steps. Read, read, read Assign a variable to each unknown (need 2 variables) Translate so you have 2 eqns. Solve system Check answer in original wording of problem.

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5.4 Problem Solving Using Systems of Equations

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  1. 5.4 Problem Solving Using Systems of Equations

  2. Steps • Read, read, read • Assign a variable to each unknown (need 2 variables) • Translate so you have 2 eqns. • Solve system • Check answer in original wording of problem

  3. Ex. The sum of two numbers is 17. If one number is subtracted from the other, their difference is -3. Find the numbers. x = 1st # y = 2nd # system: x + y = 17 x – y = -3 Add. Method: x + y = 17 x – y = -3 2x + 0 = 14 2x = 14 2x = 14 2 2 x = 7 Find y (x = 7): x + y = 17 7 + y = 17 7 + y – 7 = 17 – 7 y = 10 answer: 7, 10

  4. Ex. Each day, the sum of the average times spent on grooming for 20- to 24-year-old women and men is 86 minutes. The difference between grooming times for 20- to 24-year-old women and men is 12 minutes. How many minutes per day do 20- to 24-year old women and men spend grooming? x = ave. time wom. spend groom. y = ave. time men spend groom system: x + y = 86 x – y = 12 Add. Method: x + y = 86 x – y = 12 2x + 0 = 98 2x = 98 2x = 98 2 2 x = 49 Find y (x = 49): x + y = 86 49 + y = 86 49 + y – 49 = 86 – 49 y = 37 ans: women spend 49 min. men spend 37 min.

  5. Ex. A collection of Halloween candy contains a total of five Mr. Goodbars and Mounds bars. Each Mr. Goodbar contains 16.3 grams of fat. Each Mounds bar contains 14.1 grams of fat. The grams of fat in these candy bars exceeds the daily maximum desirable fat intake of 70 grams by 7.1 grams. How many bars of each kind of candy are contained in the Halloween collection? x = # Mr. Goodbars y = # Mounds system: x + y = 5 16.3x + 14.1y = 70 + 7.1 Add. Method: x + y = 5 -16.3(x + y) = -16.3(5) 16.3x + 14.1y = 77.1 16.3x + 14.1y = 77.1 -16.3x – 16.3y = -81.5 • 16.3x + 14.1y = 77.1 0 – 2.2y = -4.4 -2.2y = -4.4 -2.2y = -4.4 -2.2 -2.2 y = 2

  6. Find x (y = 2): x + y = 5 x + 2 = 5 x + 2 – 2 = 5 – 2 x = 3 ans: 3 Mr. Goodbars 2 Mounds

  7. Ex. The perimeter of a badminton court is 128 feet. After a game of badminton, a player’s coach estimates that the athlete has run a total of 444 feet, which is equivalent to six times the court’s length plus nine times its width. What are the dimensions of a standard badminton court? W L L = length W = width P= 2L + 2W system: 2L + 2W = 128 6L + 9W = 444 Add. Method: 2L + 2W = 128  -3(2L + 2W) = -3(128) 6L + 9W = 444  6L + 9W = 444  -6L – 6W = -384 • 6L + 9W = 444 0 + 3W = 60 3W = 60 3W = 60 3 3 W = 20

  8. ans: Width is 20 feet Length is 44 feet Find L (W = 20): 2L + 2W = 128 2L + 2(20) = 128 2L + 40 = 128 2L + 40 – 40 = 128 – 40 2L = 88 2L = 88 2 2 L = 44

  9. Ex. You are choosing between two long-distance telephone plans. Plan A has a monthly fee of $20 with a charge of $0.05 per minute for long distance calls. Plan B has a monthly fee of $5 with a charge of $0.10 per minute for long distance calls. (a) For how many minutes of long-distance calls will the cost for the two plans be the same? What will be the cost for each plan? (b) If you make approximately 10 long-distance calls per month, each averaging 20 minutes, which plan should you select? Explain your answer.

  10. Ex. (a) For how many minutes of long-distance calls will the cost for the two plans be the same? What will be the cost for each plan? x = # minutes y = cost system: Plan A: y = 20 + 0.05x Plan B: y = 5 + 0.10x How many minutes for same cost? Sub. Method: y = 20 + 0.05x y = 5 + 0.10x 20 + 0.05x = 5 + 0.10x 20 + 0.05x – 0.10x = 5 + 0.10x – 0.10x 20 – 0.05x = 5 20 – 0.05x – 20 = 5 – 20 -0.05x = -15 -0.05x = -15 -0.05 -0.05 x = 300 Ans: 300 minutes at same cost

  11. Cost for 300 minutes? Find y (x = 300): y = 5 +0.10x y = 5 +0.10(300) y = 5 + 30 y = 35 ans: $35 for 300 minutes

  12. Ex. (b) If you make approximately 10 long-distance calls per month, each averaging 20 minutes, which plan should you select? Explain your answer. 10 calls averaging 20 minutes = 10 calls x 20 min = 200 min Plan A: y = 20 + 0.05x y = 20 + 0.05(200) y = 20 + 10 y = 30 $30 for 200 min Plan B: y = 5 + 0.10x y = 5 + 0.10(200) y = 5 + 20 y = 25 $25 for 200 min Ans: Plan B is better because it costs $5 less.

  13. Ex. A community center sells a total of 301 tickets for a basketball game. An adult tickets costs $3. A student ticket costs $1. The sponsors collect $487 in ticket sales. Find the number of each type of ticket sold. x = # adult tickets y = # student tickets total # of tickets = 301 x + y = 301 cost of adult ticket = $3 cost of student ticket = $1 total cost = $487 3x + 1y = 487 system: x + y = 301 3x + y = 487 Add. Method: x + y = 301  -1(x + y) = -1(301) 3x + y = 487  3x + y = 487  -x – y = -301 • 3x + y = 487 2x + 0 = 186 2x = 186 2x = 186 2 2 x = 93

  14. ans: 93 adult tickets 208 student tickets Find y (x = 93): x + y = 301 93 + y = 301 93 + y – 93 = 301 – 93 y = 208

  15. Ex. When a crew rows with the current, it travels 16 miles in 2 hours. Against the current, the crew rows 8 miles in 2 hours. Find the rate of rowing in still water and the rate of the current. x = crew’s rowing rate in still water y = rate of the current system: 2(x + y) = 16 2(x – y) = 8 Add. Method: 2(x + y) = 16  2x + 2y = 16 2(x – y) = 8  2x – 2y = 8 4x + 0 = 24 4x = 24 4x = 24 4 4 x = 6

  16. ans: rate of crew is 6 mph rate of current is 2 mph Find y (x = 6): 2x + 2y = 16 2(6) + 2y = 16 12 + 2y = 16 12 + 2y – 12 = 16 – 12 2y = 4 2y = 4 2 2 y = 2

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