Farhan's problem of absolute values

If $x < 0$ , then $|x|+x+y = - x+x + y = y = 10$ and $x+|y|-y = x + 10-10 = x = 12$ . But $x > 0$ which contradict with $x<0$ . This implies that there is no solution for $x<0$ . Therefore, $x \ge 0$ .

Similarly, if $y >0$ , then $x+|y|-y = x + y - y = x = 12$ and $|x|+x+y = 12+12 + y = 10$ , $\implies y = - 14 < 0$ . Again, contradict with $y > 0$ . Therefore, $y \le 0$ . Then we have:

$\begin{cases} |x|+x+y = 10 & \implies 2x + y = 10 &...(1) \\ x+|y|-y = 12 & \implies x - 2y = 12 & ...(2) \end{cases}$ Then from $2(1)+(2): \ 5x = 32$ $\implies x = \dfrac {32}5$ and from $(1): \ 2 \times \dfrac {32}5 + y = 10$ $\implies y = - \dfrac {14}5$ .

Therefore, $x+y = \dfrac {32}5-\dfrac {14}5 = \boxed{\dfrac {18}5}$ .

It is easy to figure out that x is not less than or equal to zero and y is not more than or equal to zero. Thus x>0 and y<0 which leads to x=32/5 and y=-14/5. So, x+y=18/5