Ху=5 х-2у=3 Система уравнения

$\left \{ {{xy=5} \atop {x-2y=3}} \right. ; \left \{ {{x= \frac{5}{y} } \atop {x=3+2y}} \right. . \\ \\ \frac{5}{y} =3+2y \\ \\ y(3+2y)=5 \\ \\ 3y+2y ^{2} =5 \\ 2y ^{2} +3y-5=0 \\ D=3 ^{2} -4*2*(-5)=49=7 ^{2} \\ \\ y_{1} = \frac{-3+7}{2*2} = \frac{4}{4} =1; \\ \\ y_{2}= \frac{-3-7}{2*2}= \frac{-10}{4} =-2.5 \\ \\ x_{1} =3+2*1=5; \\ x_{2} =3+2*(-2.5)=-2.$
Ответ: х₁ = 5; у₁ = 1.
х₂ = - 2; у₂ = - 2,5.