Xy+y(квадрат)=10 x-y=1 РОЗВ‘ЯЖІТЬ СИСТЕМУ

$\left \{ {{xy+y^{2}=10} \atop {x-y=1}} \right.$
$\left \{ {{y(x+y)=10} \atop {x=1+y}} \right.$
$\left \{ {{y+y^2+y^2=10} \atop {x=1+y}} \right.$
$\left \{ {{2y^2+y-10=0} \atop {x=1+y}} \right.$
$2y^2+y-10=0$
$D=b^2-4ac$
D=1+80=81 > 0
$x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{-1+9}{4} =2$
$x_{2}= \frac{-b- \sqrt{D} }{2a} = \frac{-1-9}{4}= -2,5$
$y_{1}=x_{1}-1=2-1=1$
$y_{2}=x_{2}-1=-2.5-1=-3.5$