$MID$
$có$ $thể:$ \(\left[ \begin{array}{l}3x-5y=3\\3x+2y+1=3\end{array} \right.\)
$=>$ \(\left[ \begin{array}{l}3x-5y-3=0\\3x+2y+1-3=0\end{array} \right.\)
$=>$\(\left[ \begin{array}{l}(3x-3)-5y=0\\(3x-3)+2y+1=0\end{array} \right.\)
$gọi:(3x-3)=a$
$=>$\(\left[ \begin{array}{l}a-5y=0\\a+2y+1=0\end{array} \right.\)
$=>$\(\left[ \begin{array}{l}a-y=0\\a+y=\frac{-1}{2}\end{array} \right.\)
$=>do$$\left \{ {{x-y=0} \atop {x+y=\frac{-1}{2}}} \right.$
$=>x=y$ $do:x+y=\frac{-1}{2}$
$=>x=y=\frac{-1}{4}$
$\frac{chúc-bạn-học-tốt}{mình-xin-hay-nhất-nha}$