Ta có: \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-8\\2x+4=3x-15y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+10y-3x=-8\\2x-3x+15y=-12-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-8\\-x+15y=-16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5y=-4\\-x+15y=-16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20y=-20\\x+5y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=-4-5y=-4-5\cdot\left(-1\right)=-4+5=1\end{matrix}\right.\)
Vậy: (x,y)=(1;-1)
$\begin{cases}5(x+2y)=3x-8\\2x+4=3x-15y-12\end{cases}$
`<=>` $\begin{cases}5x+10y=3x-8\\x-15y=16\end{cases}$
`<=>` $\begin{cases}2x+10y=-8\\x-15y=16\end{cases}$
`<=>` $\begin{cases}x+4y=-4\\x-15y=16\end{cases}$
`<=>` $\begin{cases}19y=-20\\x=15y+16\end{cases}$
`<=>` $\begin{cases}y=-\dfrac{20}{19}\\x=\dfrac{4}{19}\end{cases}$
