Question

Решите систему уравнения подстановки: 15+2(х+3у)=3(4х+у), 2(5х-у)-3у=2+3(2х-у);

Answer 1

$\begin{cases}15+2(x+3y)=3(4x+y)\\2(5x-y)-3y=2+3(2x-y)\end{cases}\Rightarrow\begin{cases}15+2x+6y=12x+3y\\10x-2y-3y=2+6x-3y\end{cases}\\\Rightarrow\begin{cases}10x=15-3y\\2y=4x-2\end{cases}\Rightarrow\begin{cases}10x=15-3(1-2x)\\y=2x-1\end{cases}\\10x=15-3(1-2x)\\10x=15-3+6x\\4x=12\\x=3\\\begin{cases}x=3\\y=5\end{cases}$