Problem:
If ∣x∣+x+y=10 and x+∣y∣−y=12, find x+y.
Answer Choices:
A. −2
B. 2
C. 518
D. 322
E. 22
Solution:
Label the equations
(1) ∣x∣+x+y=10,(2) x+∣y∣−y=12
If x≤0, then (1)⇒y=10. Then (2)⇒x=12, a contradiction. Thus x>0 and (1) becomes
(3) 2x+y=10.
If y≥0,(2)⇒x=12. Then (3)⇒y=−14, a contradiction. Therefore y<0 and equation (2) becomes
(4) x−2y=12.
Solving (3) and (4) simultaneously, one finds x=532,y=−514 and x+y=518.
Alternatively, graph equations (1) and (2). The graph of (1) has two pieces: y=10 for x≤0 and 2x+y=10 for x≥0. Each piece is easy to sketch. Similarly, (2) has two pieces: x−2y=12 for y≤0 and x=12 for y≥0. A quick sketch of both equations shows that the only solution is in Quadrant IV, where the problem reduces to solving the linear equations (3) and (4).