Use a system of equations to solve each problem.

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Use a system of equations to solve each problem. PRACTICE 7-4 Use a system of equations to solve each problem.

Your teacher is giving you a test worth 100 points containing 40 questions. There are two point questions and four point questions on the test. How many of each type of question are on the test? Let x be the number of four point questions. What are you looking for? Let y be the number of two point questions. Write an equation relating the worth of each type of question and the worth of the test. 4x +2y = 100 Write an equation relating the number of each type of question and the total number of questions. x +y = 40 1

1 To solve by elimination: 1 Multiply the second equation By 4. 4x +2y = 100 2 x + y = 40 2 4x + 4y = 160 Subtract equation 1 from equation 2. 2 4x + 4y = 160 1 4x +2y = 100 2y = 60 Solve for y. y = 30 1

Substitute your answer for y back into one of the two original equations. 4x +2y = 100 I prefer this one. Do you know why? x + y = 40 Remember y = 30 x + y = 40 x + 30 = 40 x = 10 Solve for x. 1

1 Summary and explanation of results. There are 30 two-point questions and 10 four point questions on the test. 1

Suppose you are starting an office-cleaning service Suppose you are starting an office-cleaning service. You have spent $315 on equipment. To clean an office, you use $4 worth of supplies. You charge $25 per office. How many offices must you clean to break even? Let x be the number of offices you clean. What are you looking for? Write an equation relating your income (I) and the number of offices you clean. I = 25x Write an equation relating your expenses (E) and the number of offices you clean. E = 315 +4x The break even point is the value of x that makes income and expense the same. 2

2 To solve set I = E Substitute for I and E I = 25x E = 315 +4x 25x = 315 + 4x Substitute for I and E 21x = 315 Solve for x. x = 15 2

2 Summary and explanation of results. In order to break even, you must clean 15 offices. 2

The math club and science club had fundraisers to buy supplies for a hospice. The math club spent $135 buying six cases of juice and one case of bottled water. The science club spent $110 buying four cases of juice and two cases of bottled water. How much did a case of juice cost? How much did a case of bottled water cost? What are you looking for? Let x be the cost a case of of juice. Write an equation for the amount of money the Math club spent. Let y be the cost of a case of bottled water. 6x + y = 135 Write an equation for the amount of money the Science club spent. 4x + 2y = 110 Solve the System 3

3 6x + y = 135 The best method here might be subsittution. 4x + 2 (135 – 6x) = 110 4x + 270 – 12x = 110 -8x + 270 = 110 -8x = -160 x = 20 3

REMEMBER 3 The cost of a case of juice was $20 Let x be the cost a case of of juice. x = 20 Let y be the cost of a case of bottled water. y = 15 The cost of a case of juice was $20 and the cost of a case of bottled water was $15. 3