Решить систему уравнений

$\left \{ {{ \frac{1}{x+y}- \frac{6}{x-y} =-2} \atop {\frac{1}{x+y}+ \frac{3}{x-y} = \frac{1}{4} }} \right. \\ \\ x \neq y \\ x \neq -y \\ \\$
отнимем
$\frac{3}{x-y}+\frac{6}{x-y} = \frac{1}{4} +2 \\ \\ \frac{9}{x-y} = \frac{9}{4} \\ \\ x-y=4 \\ \\ x=4+y \\ \\ \frac{1}{4+y+y} - \frac{6}{4+y-y} =-2 \\ \\ \frac{1}{4+2y} =- \frac{1}{2} \\ \\ 2y=-6 \\ \\ y=-3 \\ \\ x=4-3=1 \\ \\ OTVET: (1;-3)$