Question

Решите систему уравнений

Answer 1

$\left \{ {{x+y+\frac{x}{y} =15 |*y} \atop {\frac{x(x+y)}{y} =56 |*y}} \right.$

$\left \{ x+y+x=15} \atop {{x(x+y)=56}} \right.$

$\left \{ 2x+y=15} \atop {{x^{2}+xy =56}} \right.$

$x^{2}+x(15-2x)=56\\x^{2}+15x-x^{2} =56\\-x^{2}+15x-56=0$

$D=15^{2}-4*(-56)*(-1)=225-224=1$

$x_{1}=\frac{-15-\sqrt{D}}{2*(-1)}=\frac{-15-\sqrt{1}}{-2}=\frac{-15-1}{-2}=\frac{-16}{-2}=8$

$x_{2}=\frac{-15+\sqrt{D}}{2*(-1)}=\frac{-15+\sqrt{1}}{-2}=\frac{-15+1}{-2}=\frac{-14}{-2}=7$

$x_{1}=8$

$\left \{ y=15-2x} \atop {{x^{2}+xy =56}} \right.$

$\left \{ y=15-2*8} \atop {{8^{2}+8y =56}} \right.$

$\left \{ y=15-16} \atop {{64+8y =56}} \right.$

$\left \{ y_{1}=-1} \atop {{64+8*(-1) =56}} \right.$

$\left \{ y_{1}=-1} \atop {{64-8=56}} \right.$

$56=56$

$x_{2}=7$

$\left \{ y=15-2x} \atop {{x^{2}+xy =56}} \right.$

$\left \{ y=15-2*7} \atop {{7^{2}+7y =56}} \right.$

$\left \{ y=15-14} \atop {{49+7y =56}} \right.$

$\left \{ y_{2}=1} \atop {{49+7*1 =56}} \right.$

$\left \{ y_{2}=1} \atop {{49+7=56}} \right.$

$56=56$

И вывод: $x_{1}=8\\x_{2}=7\\y_{1}=-1\\y_{2}=1$

Не благодари)