4. If $x$ and $y$ are real numbers for which $|x|+x+5 y=2$ and $|y|-y+x=7$, find the value of $x+y$.
(A) -3
(B) -1
(C) 1
(D) 3
(E) 5
4. Answer: (D)
If $x \leq 0$, then $|x|=-x$, and we obtain from $|x|+x+5 y=2$ that $y=\frac{2}{5}$. Thus $y$ is positive, so $|y|-y+x=7$ gives $x=7$, which is a contradiction since $x \leq 0$. Therefore we must have $x>0$. Consequently, $|x|+x+5 y=2$ gives the equation
$$
2 x+5 y=2 \text {. }
$$

If $y \geq 0$, then $|y|-y+x=7$ gives $x=7$. Substituting $x=7$ into $|x|+x+5 y=2$, we get $y=-\frac{12}{5}$, which contradicts $y \geq 0$. Hence we must have $y<0$, and it follows from the equation $|y|-y+x=7$ that
$$
x-2 y=7 \text {. }
$$

Solving equations (1) and (2) gives $x=\frac{13}{3}, y=-\frac{4}{3}$. Therefore $x+y=3$.