\(\left\{{}\begin{matrix}8y=8\\2x+3y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}2x+3y=8\\8y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3=8\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=1\end{matrix}\right.\\ b,\Leftrightarrow\left\{{}\begin{matrix}4\sqrt{3}x-2\sqrt{2}y=7\\\sqrt{3}x+2\sqrt{2}x=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{3}x=15\\\sqrt{3}x+2\sqrt{2}y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{15}{5\sqrt{3}}=\sqrt{3}\\3+2\sqrt{2}y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}\\y=\dfrac{5}{2\sqrt{2}}=\dfrac{5\sqrt{2}}{4}\end{matrix}\right.\)
\(b.\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{2}y=\dfrac{7}{2}\\\sqrt{3}x+2\sqrt{2}y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{2}y=\dfrac{7}{2}\\2\sqrt{3}x+4\sqrt{2}y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{2}y=\dfrac{25}{2}\\\sqrt{3}x+2\sqrt{2}y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5\sqrt{2}}{4}\\x=\sqrt{3}\end{matrix}\right.\)
