Question

Даю 30 баллов! Прошу помочь решить систему уравнений.

Answer 1

Ответ:

$\displaystyle \tt \left \{ {{x_1=\frac{\boxed{\displaystyle \tt 1}}{\boxed{\displaystyle \tt 3}}} \atop {y_1=\boxed{\displaystyle \tt 2}}} \displaystyle \tt \right. \left \{ {{x_2=-\frac{\boxed{\displaystyle \tt 1}}{\boxed{\displaystyle \tt 3}}} \atop {y_2=-\boxed{\displaystyle \tt 2}}} \right.$

Решение:

$\displaystyle \tt \left \{ {{(2x+y)(x+5y)=\displaystyle \tt \frac{248}{9}} \atop {\displaystyle \tt \frac{2x+y}{x+5y}=\frac{8}{31}}} \right. \: \to \: \left \{ {{(2x+y)(x+5y)=\displaystyle \tt \frac{248}{9}} \atop {54x-9y=0}} \right. \: \to\\\\\\ \displaystyle \tt \to \: \left \{ {{(2x+y)(x+5y)=\displaystyle \tt \frac{248}{9}} \atop {y=6x}} \right. \\\\\\\displaystyle \tt (2x+6x)(x+5\cdot6x)=\frac{248}{9}\\\\\displaystyle \tt \bold{x=-\frac{1}{3}} \: \: \: \: u \: \: \: \: \bold{x=\frac{1}{3}}$

$\displaystyle \tt y=6\cdot\bigg(-\frac{1}{3}\bigg) \: \to \: \bold{y=-2}\\\\ \displaystyle \tt y=6\cdot\frac{1}{3} \: \to \: \bold{y=2}$