$\begin{align}  & x-y=2----(a) \\ & {{x}^{2}}-{{y}^{2}}=8 \\ & (x-y)(x+y)=8---(b) \\ & 2(x+y)=8 \\ & x+y=4-----(c) \\ & \text{Adding }(a)\text{ and }(c)\text{ together} \\ & 2x=6 \\ & x=6 \\ & \text{Substitute }3\text{ for }x\text{ in (}a) \\ & y=1 \\ & (x,y)=(3,1) \\\end{align}$