Giải hệ phương trình: $\left\{ \begin{array}{l}\sqrt {2x + 3}  + \sqrt {4 - y}  = 4{\rm{    }}\left( 1 \right)\\\sqrt {2y + 3}  + \sqrt {4 - x}  = 4{\rm{    }}\left( 2 \right)\end{array} \right.$

Điều kiện:  $\left\{ \begin{array}{l} - \dfrac{3}{2} \le x \le 4\\ - \dfrac{3}{2} \le x \le 4\end{array} \right.$

$\begin{array}{l}\left( 1 \right) - \left( 2 \right) \Leftrightarrow \left\{ \begin{array}{l}\sqrt {2x + 3}  + \sqrt {4 - y}  = 4\\\left( {\sqrt {2x + 3}  - \sqrt {2y + 3} } \right) + \left( {\sqrt {4 - y}  - \sqrt {4 - x} } \right) = 0\end{array} \right.\\{\rm{    }} \Leftrightarrow \left\{ \begin{array}{l}\sqrt {2x + 3}  + \sqrt {4 - y}  = 4\\\dfrac{{2\left( {x - y} \right)}}{{\sqrt {2x + 3}  + \sqrt {2y + 3} }} + \dfrac{{x - y}}{{\sqrt {4 - x}  + \sqrt {4 - y} }} = 0\end{array} \right.\\{\rm{    }} \Leftrightarrow \left\{ \begin{array}{l}\sqrt {2x + 3}  + \sqrt {4 - y}  = 4\\\left( {x - y} \right)\left( {\dfrac{2}{{\sqrt {2x + 3}  + \sqrt {2y + 3} }} + \dfrac{1}{{\sqrt {4 - x}  + \sqrt {4 - y} }}} \right) = 0\end{array} \right.\\{\rm{    }} \Leftrightarrow \left\{ \begin{array}{l}\sqrt {2x + 3}  + \sqrt {4 - y}  = 4\\x - y = 0\end{array} \right.{\rm{ }}\left( {do:\dfrac{2}{{\sqrt {2x + 3}  + \sqrt {2y + 3} }} + \dfrac{1}{{\sqrt {4 - x}  + \sqrt {4 - y} }} > 0} \right)\\{\rm{    }} \Leftrightarrow \left\{ \begin{array}{l}\sqrt {2x + 3}  + \sqrt {4 - x}  = 4\\x = y\end{array} \right.\\{\rm{    }} \Leftrightarrow \left\{ \begin{array}{l}x + 7 + 2\sqrt {\left( {2x + 3} \right)\left( {4 - x} \right)}  = 16\\x = y\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = y = 3\\x = y = \dfrac{{11}}{9}\end{array} \right.\end{array}$

So với điều kiện, hệ có hai nghiệm: $S = \left( {x;y} \right) = \left\{ {\left( {3;3} \right),\left( {\dfrac{{11}}{9};\dfrac{{11}}{9}} \right)} \right\}$.