Математика. 8 класс.

Решите задачу:

$\begin{array}{l} A(x;y).\\ \begin{dcases} x-2y=-10\\ x+y=-1 \end{dcases}\implies \begin{dcases} x=2y-10\\ 2y-10+y=-1 \end{dcases}\\ y=\frac{-1+10}{3} =3\implies \text{x=2\times 3-10=-4}.\\ A(-4;3).\\ B(x;y).\\ \begin{dcases} x-2y=-10\\ 11x+2y=34 \end{dcases}\implies \begin{dcases} x=2y-10\\ 11(2y-10)+2y=34 \end{dcases}\\ y=\frac{34+110}{24} =6. \implies \text{x=2\times 6-10=2.}\\ B(2;6).\\ C(x;y).\\ \begin{dcases} x+y=-1\\ 11x+2y=34 \end{dcases}\implies \begin{dcases} x=-(1+y)\\ -11(1+y)+2y=34 \end{dcases}\\ y=\frac{34+11}{-9} =-5\implies \text{x=-(1-5)=4.}\\ C(4;-5). \end{array}$