Có 2 TH:
Nếu y \(\ge\) -2
<=> \(\left\{{}\begin{matrix}4x-\left(y+2\right)=3\\x+2\left(y+2\right)=3\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}4x-y-2=3\\x+2y+4=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-y-2=3\\4x+8y+16=3\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}9y+18=0\\4x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\left(TM\right)\\x=0,75\end{matrix}\right.\)
Nếu y < -2
<=> \(\left\{{}\begin{matrix}4x-\left[-\left(y+2\right)\right]=3\\x+\left[2.-\left(y+2\right)\right]=3\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}4x+y+2=3\\x+\left(-2y\right)-4=3\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}4x+y=1\\x-2y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+y=1\\4x-8y=28\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}-9y=27\\4x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\left(TM\right)\\x=1\end{matrix}\right.\)
Vậy Ta có 2 TH:
TH1: x = 0,75; y = -2
TH2: x = 1; y = -3
Ta có: \(\left\{{}\begin{matrix}4x-\left|y+2\right|=3\\x+2\left|y+2\right|=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-\left|y+2\right|=3\\4x+8\left|y+2\right|=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9\left|y+2\right|=-9\\x+2\left|y+2\right|=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y+2\right|=1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y+2=1\\x=1\end{matrix}\right.\\\left\{{}\begin{matrix}y+2=-1\\x=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-3\\x=1\end{matrix}\right.\end{matrix}\right.\)