Question

помогите решить, пожалуйста
система:
x^2 + 10xy + 25y^2 = 49
x - 5y = -3​

Answer 1

(x+5y)²=49

x-5y=-3

получаем две системы

1) x+5y=7

x-5y= -3

2x=4

x=2

y=1

2) x+5y= -7

x-5y= -3

2x= -10

x= -5

y= -2/5

Ответ: (2;1), (-5; -0,4)

Answer 2

$\left\{\begin{array}{l}x^2+10xy+25y^2=49\\x-5y=-3\end{array}\right\; \; \left\{\begin{array}{l}(x+5y)^2=7^2\\x-5y=-3\end{array}\right\; \; \left\{\begin{array}{l}(x+5y-7)(x+5y+7)=0\\x=5y-3\end{array}\right\\\\\\\left\{\begin{array}{l}(10y-10)(10y+4)=0\\x=5y-3\end{array}\right\; \; \left\{\begin{array}{l}y_1=1\; ,\; y_2=-\frac{2}{5}\\x_1=2\; ,\; x_2=-5\end{array}\right\\\\\\Otvet:\; \; (2,1)\; ,\; (-5,-\frac{2}{5})\; .$