practice 7-4 use a system of equations to solve each problem
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PRACTICE 7-4PRACTICE 7-4
Use a system of equations to Use a system of equations to solve each problem.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?
What are you looking for?
Write an equation relating the worth of each type ofquestion and the worth of the test.
4x +2y = 100
Let x be the number of four point questions.
Let y be the number of two point questions.
Write an equation relating the number of each type of question and the total number of questions.
x +y = 40
4x +2y = 100
x + y = 40
To solve by elimination:To solve by elimination:
Multiply the second equationBy 4. 4x + 4y = 160
Subtract equation 1 from equation 2.
1
2 2
4x + 4y = 1602
4x +2y = 1001
2y = 60
y = 30Solve for y.
4x +2y = 100
x + y = 40
Substitute your answer for y back into one of the two original equations.
I prefer this one.
Do you know why?
x + y = 40Remember y = 30
x + 30 = 40
Solve for x. x = 10
Summary and explanation of results.
There are 30 two-point questions and10 four point questions on the test.
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?
What are you looking for? Let x be the number of offices you clean.
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 xthat makes income and expense the same.
To solve set I = ETo solve set I = E
I = 25x E = 315 +4x
25x = 315 + 4x
Substitute for I and E
Solve for x.21x = 315
x = 15
Summary and explanation of results.
In order to break even,
you must clean 15 offices.
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 ofof juice.
Let y be the cost of a case of bottled water.
Write an equation for the amountof money the Math club spent.
Write an equation for the amountof money the Science club spent.
6x + y = 135
4x + 2y = 110
6x + y = 135
4x + 2y = 110
The best method here might be subsittution.
y = 135 – 6x
4x + 2 (135 – 6x) = 110
4x + 270 – 12x = 110 -8x + 270 = 110
-8x = -160 x = 20
y = 135 – 6(20)y = 135 – 120y = 15
Let x be the cost a case ofof juice.
Let y be the cost of a case of bottled water.
x = 20
y = 15
The cost of a case of juice was $20and the cost of a case of bottled water was $15.