Word problems involving polynomials with solution?
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 as3x + 4y + 13zThe contents of the second box can be expressed as7x + 15y + 2zThe contents of the third box can be expressed as2x + 3y + 8zIf 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 have3x + 4y + 13z7x + 15y + 2z+2x + 3y + 8z__________________ 12x + 22y + 23zTherefore, the final contents of the third box are 12 x's, 22 y's, and 23 z's.