Điều kiện xác định $x\ge \dfrac 1 2, y\ge -\dfrac 2 3$
$\begin{array}{l} a = \sqrt {2x - 1} ,b = \sqrt {3y + 2} \left( {a,b \ge 0} \right)\\ \Leftrightarrow \left\{ \begin{array}{l} a + b = 3\\ {a^2} - 2b = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = 3 - a\\ {a^2} - 2\left( {3 - a} \right) = 2 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} b = 3 - a\\ {a^2} + 2a - 8 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = 3 - a\\ \left[ \begin{array}{l} a = 2\\ a = - 4(L) \end{array} \right. \end{array} \right.\\ \Rightarrow a = 2 \Rightarrow b = 1\\ \Rightarrow \left\{ \begin{array}{l} \sqrt {2x - 1} = 2\\ \sqrt {3y + 2} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2x - 1 = 4\\ 3y + 2 = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{5}{2}\\ y = - \dfrac{1}{3} \end{array} \right. \end{array}$
