$\begin{array}{l} \left\{ \begin{array}{l} \left| x \right| + 3y = 7\\ 2x + 2\left| {y - 1} \right| = 3 \end{array} \right.\\ + x \ge 0,y \ge 1\\ \Leftrightarrow \left\{ \begin{array}{l} x + 3y = 7\\ 2x + 2y - 2 = 3 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x + 3y = 7\\ 2x + 2y = 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{1}{4}\\ y = \dfrac{9}{4} \end{array} \right.(TM)\\ + x < 0,y < 1\\ \left\{ \begin{array}{l} - x + 3y = 7\\ 2x + 2\left( {1 - y} \right) = 3 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - x + 3y = 7\\ 2x - 2y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{{17}}{4}\\ y = \dfrac{{15}}{4} \end{array} \right.(L)\\ x \ge 0,y < 1\\ \left\{ \begin{array}{l} x + 3y = 7\\ 2x + 2\left( {1 - y} \right) = 3 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x + 3y = 7\\ 2x - 2y = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{{17}}{8}\\ y = \dfrac{{13}}{8} \end{array} \right.(L)\\ x < 0,y \ge 1\\ \left\{ \begin{array}{l} - x + 3y = 7\\ 2x + 2\left( {y - 1} \right) = 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - x + 3y = 7\\ 2x + 2y = 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{1}{8}\\ y = \dfrac{{19}}{8} \end{array} \right.(L)\\ \to \left( {x;y} \right) = \left( {\dfrac{1}{4};\dfrac{9}{4}} \right) \end{array}$
