a) Ta có:
\(2x-y=8\)⇒\(y=2x+8\)
Theo đề ra \(3x+5y=1\)
⇒\(3x+5\left(2x+8\right)=1\)
⇒\(13x+40=1\)
⇒\(x=-3\)
Thay \(x=-3\) vào \(2x-y=-8\) ta có:
\(2.-3-y=-8\)
⇒\(y=2\)
a. \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=-39\\3x+5y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}-x-\sqrt{2}y=\sqrt{3}\\\sqrt{2}x+2y=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{2}y=-\sqrt{3}\\\sqrt{2}x+2y=-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x+2y=\sqrt{6}\\\sqrt{2}x+2y=-\sqrt{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0=2\sqrt{6}\left(VLí\right)\\-x-\sqrt{2}y=\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\varnothing\\y=\varnothing\end{matrix}\right.\)
