1,
$\begin{cases}x+2y=4\\x-2y=-4\end{cases}$
`<=>` $\begin{cases}2x=0\\x-2y=-4\end{cases}$
`<=>` $\begin{cases}x=0\\0-2y=-4\end{cases}$
`<=>` $\begin{cases}x=0\\y=2\end{cases}$
Vậy `(x;y)=(0;2)`
2,
$\begin{cases}x+2y=-3\\x+3y=-4\end{cases}$
`<=>` $\begin{cases}y=-1\\x+3.(-1)=-4\end{cases}$
`<=>` $\begin{cases}y=-1\\x=-1\end{cases}$
`<=>` $\begin{cases}x=-1\\y=-1\end{cases}$
Vậy `(x;y)=(-1;-1)`
1) Ta có: \(\left\{{}\begin{matrix}x+2y=4\\x-2y=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}x+2y=-3\\x+3y=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=1\\x+3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=-4-3y=-4-3\cdot\left(-1\right)=-4+3=-1\end{matrix}\right.\)
3) Ta có: \(\left\{{}\begin{matrix}2\left(x+2\right)=3\left(y-1\right)\\3x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3-4\\3x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-7\\3x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-9y=-21\\6x+2y=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-33\\3x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\3x=6-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Giúp mình với ạ, mình cần gấp lắm ạ!
