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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