Answer:
[tex]\huge\boxed{x=\dfrac{19}{5};\ y=\dfrac{1}{5}\to\left(\dfrac{19}{5};\ \dfrac{1}{5}\right)}[/tex]
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}3x-2y=11&|\text{multiply both sides by (-2)}\\x-4y=3\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-6x+4y=-22\\x-4y=3\end{array}\right}\qquad|\text{add both sides of the equations}\\.\qquad-5x=-19\qquad|\text{divide both sides by (-5)}\\.\qquad\boxed{x=\dfrac{19}{5}}[/tex]
[tex]\text{Substitute it to the second equation}\\\\\dfrac{19}{5}-4y=3\qquad|\text{multiply both sides by 5}\\\\19-20y=15\qquad|\text{subtract 19 from both sides}\\\\-20y=-4\qquad|\text{divide both sides by (-20)}\\\\y=\dfrac{4}{20}\\\\\boxed{y=\dfrac{1}{5}}[/tex]
