a) \(\left\{{}\begin{matrix}3\left(x+2\right)+4\left(y-1\right)-7=0\\2x-3y-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+6+4y-4-7=0\\2x-3y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=5\\2x-3y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+8y=10\\6x-9y=27\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}17y=-17\\2x-3y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x-3.\left(-1\right)=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{\left(3;-1\right)\right\}\)
b) \(\left\{{}\begin{matrix}x\sqrt{2}-3y=1\\2x+y\sqrt{2}=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\sqrt{2}y=\sqrt{2}\\2x+\sqrt{2}y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-4\sqrt{2}y=\sqrt{2}+2\\2x+\sqrt{2}y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1-\sqrt{2}}{4}\\2x+\sqrt{2}.\left(\dfrac{-1-\sqrt{2}}{4}\right)=-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=\dfrac{-1-\sqrt{2}}{4}\\2x-\dfrac{2+\sqrt{2}}{4}=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1-\sqrt{2}}{4}\\2x=-2+\dfrac{2+\sqrt{2}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1-\sqrt{2}}{4}\\2x=\dfrac{-6+\sqrt{2}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1-\sqrt{2}}{4}\\x=\dfrac{-6+\sqrt{2}}{8}\end{matrix}\right.\)
Vậy \(S=\left\{\left(\dfrac{-6+\sqrt{2}}{4};\dfrac{-1-\sqrt{2}}{4}\right)\right\}\)
