1988 AHSME Problems/Problem 17

Problem

If $|x| + x + y = 10$ and $x + |y| - y = 12$, find $x + y$

$\textbf{(A)}\ -2\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ \frac{18}{5}\qquad \textbf{(D)}\ \frac{22}{3}\qquad \textbf{(E)}\ 22$

Solution

We proceed by casework:

Case 1: x, y >= 0

Thus we have 2x + y = 10, x = 12. This makes y < 0, a contradiction.

Case 2: x >= 0 > y

2x + y = 10, x - 2y = 12. By multiplying the first equation by two and adding, we have 4x + x = 2*10 + 12 thus x = 32/5. Since the question only specifies 1 solution, plugging in x = 32/5 into the first equation yields y = -14/5 thus x + y = 18/5 $C$.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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