keeping in mind that in a parallelogram the diagonals bisect each other, namely cut each other into two equal halves. Check the picture below.
[tex]\stackrel{GH}{3x-4}~~ = ~~\stackrel{HE}{5y+1}\implies 3x=5y+5\implies x=\cfrac{5y+5}{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{DH}{x+1}~~ = ~~\stackrel{HF}{3y}\implies \stackrel{\textit{substituting "x" in the equation}}{\cfrac{5y+5}{3}+1~~ = ~~3y}[/tex]
[tex]\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3\left( \cfrac{5y+5}{3}+1 \right)=3(3y)}\implies 5y+5+3=9y\implies 5+3=4y\implies 8=4y \\\\\\ \cfrac{8}{4}=y\implies \boxed{2=y}~\hfill x=\cfrac{5y+5}{3}\implies x=\cfrac{5(2)+5}{3}\implies \boxed{x=5}[/tex]
