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
Sep 01, 18 04:07 PM
These heart of algebra questions will help you prepare to take the math portion of the SAT
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Sep 01, 18 04:07 PM
These heart of algebra questions will help you prepare to take the math portion of the SAT
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