Question

помогите пожалуйста, очень срочно ​

Answer 1

$\left\{\begin{array}{ll}2(x+y)^2-7(x+y)+3=0\\2x-3y=-1\end{array}\right\\\\t=x+y\; ,\; \; \; 2t^2-7t+3=0\; \; ,\; \; D=25\; ,\; \; t_1=\frac{1}{2}\; ,\; t_2=3\\\\\left\{\begin{array}{ll}x+y=\frac{1}{2}\\2x-3y=-1\end{array}\right\; \; ili\; \; \; \left\{\begin{array}{ll}x+y=3\\2x-3y=-1\end{array}\right\\\\\\\left\{\begin{array}{ll}2x+2y=1\\2x-3x=-1\end{array}\right\; \; \; ili\; \; \; \left\{\begin{array}{ll}2x+2y=6\\2x-3y=-1\end{array}\right$

$\left\{\begin{array}{ll}-y=0\\2x=-1\end{array}\right\; \; \; ili\; \; \; \left\{\begin{array}{ll}-5y=5\\2x=3y-1\end{array}\right\\\\\\\left\{\begin{array}{ll}y=0\\x=-\frac{1}{2}\end{array}\right\; \; \; ili\; \; \; \left\{\begin{array}{ll}y=-1\\x=-2\end{array}\right\\\\\\Otvet:\; \; (-\frac{1}{2}\, ,\, 0)\; ,\; \; (-2,-1)\; .$