Question

Giai hệ phương trình:\(\left\{{}\begin{matrix}\left|x+2\right|+\left|y-3\right|=8\\\left|x+2\right|-5y=1\end{matrix}\right.\)

Answer 1

Lời giải:

Lấy PT(1) trừ PT(2) ta thu được:

\(|y-3|+5y=8-1=7\)

\(\Leftrightarrow |y-3|=7-5y\)

\(\Rightarrow \left\{\begin{matrix} 7-5y\geq 0\\ \left[\begin{matrix} y-3=7-5y\\ y-3=5y-7\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y\leq \frac{7}{5}\\ \left[\begin{matrix} y=\frac{5}{3}\\ y=1\end{matrix}\right.\end{matrix}\right.\Rightarrow y=1\)

Thay vào PT(2) suy ra:

\(|x+2|=5y+1=5.1+1=6\)

\(\Rightarrow \left[\begin{matrix} x+2=6\\ x+2=-6\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=4\\ x=-8\end{matrix}\right.\)

Vậy \((x,y)\in\left\{(4,1); (-8,1)\right\}\)