X в 2 - ху =10 у-х = -5 решите систему

Решите задачу:

$\left \{ {{ x^{2}-xy=10 } \atop {y-x=5}} \right. \left \{ {{x^{2}-xy=10} \atop {x=5+y}} \right. \left \{ {{(x+y)^2-(5+y)*y=10} \atop {x=5+y}} \right. (x+y)^2-(5+y)*y=10$
$25+10y+y^2-(-5y+y^2)=10; 25+10y+y^2+5y-y^2=10;$
$25+15y=10; 15y=10-25; 15y=-15; y=-1.$
$\left \{ {{y=-1} \atop {x=5-1}} \right. \left \{ {{y=-1} \atop {x=4}} \right.$