Question

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

Answer 1

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