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Thread: CD's

  1. #1
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    CD's

    A DJ found a music store that had a great collection of used CD's. he purshased 38 CD's for a total of $235. The purchase included CD's priced at either $5.00 or $8.00. How many CD's in each price category did the DJ buy?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wild_flowr69
    A DJ found a music store that had a great collection of used CD's. he purshased 38 CD's for a total of $235. The purchase included CD's priced at either $5.00 or $8.00. How many CD's in each price category did the DJ buy?
    Call the number of CDs bought at $5.00 x. Call the number of CDs bought at $8.00 y. We know that the total cost was $235, so:
    $\displaystyle 5x+8y=235$.

    Additionally, we know that 38 CDs were bought, so:
    $\displaystyle x+y=38$.

    Two equations, two unknowns, so you can solve for x and y.

    -Dan
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  3. #3
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    Let me continue from topsquark's solution:
    The equations
    $\displaystyle \left\{\begin{array}{cc}5x+8y&=235\\x+y&=38\end{ar ray}\right$
    In the second equation solve for $\displaystyle y$:
    $\displaystyle y=38-x$
    Substitute that into the first:
    $\displaystyle 5x+8(38-x)=235$
    Now solve it,
    $\displaystyle 5x+304-8x=235$
    Thus,
    $\displaystyle -3x=-69$
    Thus,
    $\displaystyle x=23$
    Thus,
    $\displaystyle y=15$
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