Xy+2(x-y)=10 xy=z z+2t=10 z+2t=10
5xy-3(x-y)=11 (x-y)=t ⇔ 5z-3t=11 -13t =-39 t=3
z=4
xy=4 (3+y)y=4 y²+3y-4=0 ⇔ y1=-4 y2=1
(x-y)=3 ⇔x=3+y ⇔ x=3+y x1=-1 x2=4
проверка
x1=-1 у1=-4
xy+2(x-y)=10 (-1)(-4)+2(-1-(-4))=10 верно
5xy-3(x-y)=11 5(-1)(-4)-3(-1-(-4))=11 верно
x2=4 у2=1
xy+2(x-y)=10 (4)(1)+2(4-1)=10 верно
5xy-3(x-y)=11 5(4)(1)-3(4-(1))=11 верно