Помогите решить номер1562

Решите задачу:

$\left \{ {{ \frac{x+y}{8} +\frac{y}{2} =4} \atop { \frac{x}{10} -\frac{x-y}{5} =1}} \right. \\ \\ \left \{ {{x+y+4y=32} \atop {x-2x+2y=10}} \right. \\ \\ + \left \{ {{x+5y=32} \atop {2y-x=10}} \right. \\ \\ \left \{ {{7y=42} \atop {x=2y-10}} \right. \\ \\ \left \{ {{y=6} \atop {x=2}} \right. \\ \\ OTBET:(2;6)$

$\left \{ {{ \frac{x-y}{2} -\frac{x+y}{4} =-3} \atop { \frac{y}{2} -\frac{x+y}{5} =0}} \right. \\ \\ \left \{ {{2x-2y-x-y=-12} \atop {5y-2x-2y=0}} \right. \\ \\ +\left \{ {{x-3y=-12} \atop {3y-2x=0}} \right. \\ \\ \left \{ {{-x=-12} \atop {y= \frac{2x}{3} }} \right. \\ \\ \left \{ {{x=12} \atop {y= 8 }} \right. \\ \\ OTBET:(12;8)$