Giải hệ phương trình sau: \(\left\{\begin{array}{l} \frac{y}{5}-\frac{x-y}{2}=\frac{1}{10} \\ \frac{y}{2}-\frac{x+y}{5}=\frac{1}{5} \end{array}\right.\)

* Hướng dẫn giải

Ta có:

\(\left\{\begin{array}{l} \frac{y}{5}-\frac{x-y}{2}=\frac{1}{10} \\ \frac{y}{2}-\frac{x+y}{5}=\frac{1}{5} \end{array} \Leftrightarrow\left\{\begin{array}{l} 2 y-5(x-y)=1 \\ 5 y-2(x+y)=2 \end{array}\right.\right.\)

\(\begin{aligned} &\Leftrightarrow\left\{\begin{array}{l} 2 y-5 x+5 y=1 \\ 5 y-2 x-2 y=2 \end{array} \Leftrightarrow\left\{\begin{array}{l} -5 x+7 y=1 \\ -2 x+3 y=2 \end{array} \Leftrightarrow\left\{\begin{array}{l} y=\frac{5}{7} x+\frac{1}{7} \\ -2 x+3\left(\frac{5}{7} x+\frac{1}{7}\right)=2 \end{array}\right.\right.\right.\\ &\Leftrightarrow\left\{\begin{array}{l} y=\frac{5}{7} x+\frac{1}{7} \\ -2 x+\frac{15}{7} x+\frac{3}{7}=2 \end{array} \Leftrightarrow\left\{\begin{array}{l} y=\frac{5}{7} x+\frac{1}{7} \\ \frac{1}{7} x=\frac{11}{7} \end{array} \Leftrightarrow\left\{\begin{array}{l} x=11 \\ y=8 \end{array}\right.\right.\right. \end{aligned}\)