Решить линейное упавнение 2х+3(х+у)=11 7(х+3у)-4у=-23

Решите задачу:

$\begin{cases}2x+3(x+y)=11\\7(x+3y)-4y=-23\end{cases}\Rightarrow\begin{cases}2x+3x+3y=11\\7x+21y-4y=-23\end{cases}\Rightarrow\\\Rightarrow\begin{cases}5x+3y=11\\7x+17y=-23\end{cases}\Rightarrow\begin{cases}x=\frac{11-3y}5\\7\cdot\frac{11-3y}5+17y=-23\end{cases}\\7\cdot\frac{11-3y}5+17y=-23\\\frac{77-21y}5+17y=-23\\\frac{77-21y+85y}5=-23\\64y+77=-115\\64y=-192\\y=-3\\\begin{cases}x=\frac{11-3\cdot(-3)}5=4\\y=-3\end{cases}$