Biết hệ phương trình ​\(\left\{\begin{array}{l} \frac{x+1}{4}-\frac{y}{2}=x+y+1 \\ \frac{x-2}{2}+\frac{y-1}{3}=x+y-1 \end{array}\right.\) có nghiệm là:

* Hướng dẫn giải

 \(\begin{array}{l} \text { Ta có }\left\{\begin{array}{l} \frac{x+1}{4}-\frac{y}{2}=x+y+1 \\ \frac{x-2}{2}+\frac{y-1}{3}=x+y-1 \end{array}\right. \\ \Leftrightarrow\left\{\begin{array}{l} x+1-2 y=4 x+4 y+4 \\ 3 x-6+2 y-2=6 x+6 y-6 \end{array} \Leftrightarrow\left\{\begin{array}{l} 3 x+6 y=-3 \\ 3 x+4 y=-2 \end{array} \Leftrightarrow\left\{\begin{array}{l} y=-\frac{1}{2} \\ x=0 \end{array}\right.\right.\right. \end{array}\)