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A. In a box of letters, there are 3x's, 4y's, and 13 z's. In a second box, there are 7 x's, 15 y's, and 2 z's. The contents of the two boxes are put together in a third box which already contains 2 x's, 3 y's, and 8 z's. What are the final contents of the third box?
SOLUTION:The contents of the first box can be expressed as
3x + 4y + 13z
The contents of the second box can be expressed as
7x + 15y + 2z
The contents of the third box can be expressed as
2x + 3y + 8z
If the contents of the first two boxes are put together in the third box, the final contents of the third box can be expressed as the sum of the three expressions above. So, we have
3x + 4y + 13z
7x + 15y + 2z
+2x + 3y + 8z
__________________ 12x + 22y + 23z
Therefore, the final contents of the third box are 12 x's, 22 y's, and 23 z's.
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