Giải hệ phương trình: $\left\{ \begin{array}{l}\dfrac{1}{{x - 2}} + \dfrac{4}{{3y + 1}} = 5\\\dfrac{2}{{x - 2}} - \dfrac{4}{{3y + 1}} =  - 2\end{array} \right.$ 

$\left\{ \begin{array}{l}y = 3x\\\sqrt x  + \sqrt y  = 1\end{array} \right.$

$\left\{ \begin{array}{l}x - 5y = 3\\y = 3x - 1\end{array} \right.$

$\left\{ \begin{array}{l}3 + 2y = 5x\\{x^3} = {y^2}\end{array} \right.$

$\left\{ \begin{array}{l}x + 2y + z = 10\\ - 2x + 3y + z = 2019\end{array} \right.$

$\left\{ \begin{array}{l}x - 2y =  - 1\\3x - y = 4\end{array} \right.$

$\left\{ \begin{array}{l}{x^2} - 3y =  - 4\\y - 3\sqrt 5 x = 4\end{array} \right.$

$\left\{ \begin{array}{l}5x + 6y = 19\\ - x - y =  - 3\end{array} \right.$

$\left\{ \begin{array}{l}x + \sqrt y  =  - 3\\4x + y = \sqrt 5 \end{array} \right.$

$\left( { - \dfrac{1}{6}; - \dfrac{5}{3}} \right)$     

$\left( {\dfrac{1}{3};\dfrac{5}{6}} \right)$

$\left( { - \dfrac{1}{3};\dfrac{5}{6}} \right)$    

$\left( { - \dfrac{1}{3}; - \dfrac{5}{6}} \right)$

$\left( { - 1;1} \right)$

$\left( { - 1; - 1} \right)$

$\left( {1; - 1} \right)$

$\left( {1;1} \right)$

$a = 1$ 

$a =  - 1$       

$a =  - 2$          

$a = 2$

$\left( {1;\dfrac{1}{2}} \right)$           

$\left( 1;-1 \right)$

$( - 1;1)$

$\left( {\dfrac{1}{2}; - 1} \right)$