CAT 2021 Slot 3 · Quant · Question 13

Modulus equations are handled by splitting into sign cases and discarding the ones that contradict themselves.

Suppose x<0. Then x+\left| x \right|=0, so the second equation gives 3y=1 and y=\frac{1}{3}. With y>0 the first becomes 3x+3y=7, so 3x=6 and x=2 — which contradicts x<0.

So x\ge 0, and the second equation becomes 2x+3y=1.

If y\ge 0, the first is 3x+3y=7. Together with 2x+3y=1 this gives x=6 and y=-\frac{11}{3}, contradicting y\ge 0.

So y<0, and 2\left| y \right|+y=-2y+y=-y, making the first equation 3x-y=7. With 2x+3y=1: substituting y=3x-7 gives 2x+9x-21=1, so x=2 and y=-1.

Both conditions hold (x\ge 0, y<0), and x+2y=2-2=0.