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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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