a: Ta có: \(\left\{{}\begin{matrix}2x-y=3\\x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=3\\2x+4y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=-5\\x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4-2y=2\end{matrix}\right.\)
b: Ta có: \(\left\{{}\begin{matrix}4x-5y=3\\3x-y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x-15y=9\\12x-4y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-15\\3x-y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{15}{11}\\3x=y+6=\dfrac{81}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{27}{11}\\y=\dfrac{15}{11}\end{matrix}\right.\)
c: Ta có: \(\left\{{}\begin{matrix}4x-2y=-6\\-2x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-2y=-6\\-4x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=0\\-2x+y=3\end{matrix}\right.\left(luônđúng\right)\)
d: Ta có: \(\left\{{}\begin{matrix}4x+3y=6\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\4x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x=4-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
