để hệ phương trình có nghiệm là \(\left(3;-2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-2b=3\\6a+6y=36\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9a-6b=9\\6a+6b=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15a=45\\6a+6b=36\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\18+6b=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=\dfrac{36-18}{6}=3\end{matrix}\right.\)
vậy \(a=b=3\)