$\bf \begin{cases} 3x+2y=9\\ x-5y=4 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{solving the 2nd equation for "y"}}{x-5y = 4\implies x-4-5y=0}\implies x-4=5y\implies \cfrac{x-4}{5}=y \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 1st equation}}{3x+2\left(\cfrac{x-4}{5} \right) = 9}\implies 3x+\cfrac{2(x-4)}{5}=9 \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5\left( 3x+\cfrac{2(x-4)}{5} \right)=5(9)}\implies 15x+2(x-4)=45$

$\bf 15x+2x-8=45\implies 17x-8=45\implies 17x=53\implies \boxed{x=\cfrac{53}{17}} \\\\\\ \stackrel{\textit{we know that}}{\cfrac{x-4}{5}=y}\implies \cfrac{\left(\frac{53}{17} -4 \right)}{5}=y\implies \cfrac{\left(\frac{53-68}{17} \right)}{5}=y\implies \cfrac{~~\frac{-15}{17}~~}{5}=y \\\\\\ \cfrac{~~\frac{-15}{17}~~}{\frac{5}{1}}=y\implies \cfrac{-15}{17}\cdot \cfrac{1}{5}=y\implies \boxed{-\cfrac{3}{17}=y} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \left( \frac{53}{17}~~,~~-\frac{3}{17} \right)~\hfill$