# cost of cd

by Mackenzie

Andrew had \$93.00. With this money, he could buy 3 CDs and 4 books. However, he bought only 2 CDs and 3 books and had \$27.00 left. What was the cost of each CD?

Solution

This problem will lead to a system of linear equations

Let x be the cost of 1 cd

Let y be the cost of 1 book

Cost of 3 CDs + cost of 4 books must equal to 93 dollars.

3 × x + 4 × y = 93

3x + 4y = 93

When he bought 2 CDs and 3 books he had 27 dollars left.

The statement above means that

Cost of 2 CDs + cost of 3 books = 93 - 27

Cost of 2 CDs + cost of 3 books = 66

2x + 3y = 66

You now have a system of equations to solve

3x + 4y = 93 equation 1

2x + 3y = 66 equation 2

Multiply equation 1 by -2 and equation 2 by 3

We get

-6x + -8y = -186 equation 3

6x + 9y = 198 equation 4

Add the left sides and right sides of equation 3 and equation 4.

-6x + 6x + -8y + 9y = -186 + 198

0 + -8y + 9y = -186 + 198

-8y + 9y = -186 + 198

y = -186 + 198

y = 12

Use equation 2 to find y

2x + 3y = 66

2x + 3 × 12 = 66

2x + 36 = 66

2x + 36 - 36 = 66 - 36

2x = 66 - 36

2x = 30

Divide both sides by 2

2x / 2 = 30 / 2

x = 15

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