Giải hệ phương trình sau: |x - 1| + |y + 2| = 2 và 4|x - 1| + 3|y + 2| = 7

* Hướng dẫn giải

$$\begin{gathered} \left\{\begin{array}{l}\left|x-1\right|+\left|y+2\right|=2\\ 4\left|x-1\right|+3\left|y+2\right|=7\end{array}\right.\iff \left\{\begin{array}{l}\left|x-1\right|=1\\ \left|y+2\right|=1\end{array}\right. \\ \iff \left\{\begin{array}{l}\left[\begin{array}{l}x-1=1\\ x-1=-1\end{array}\right.\\ \left[\begin{array}{l}y+2=1\\ y+2=-1\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}\left[\begin{array}{l}x=2\\ x=0\end{array}\right.\\ \left[\begin{array}{l}y=-1\\ y=-3\end{array}\right.\end{array}\right.\Rightarrow \left(x;y\right)\in \left\{(2;-1);(2;-3);(0;-1);(0;-3)\right\} \end{gathered}$$