[tex]\bf \begin{array}{lcrrrrr} 2x-5y=11&\stackrel{\stackrel{\textit{\large first step}}{\downarrow }}{\times -3}\implies &-6x&+&15y&=&-33\\ 3x+2y=7&\times 2\implies &6x&+&4y&=&14\\\cline{3-7} &&0&+&19y&=&-19 \end{array} \\\\\\ y = \cfrac{-19}{19}\implies \boxed{y=-1} \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{2x-5(-1)=11\implies 2x+5=11}\implies 2x=6\implies x = \cfrac{6}{2}\implies \boxed{x = 3} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (3~~,~~-1)~\hfill[/tex]