since DH=HF you set x+3=3y
GH=HE so 4x-5=2y+3
This gives you a system and I solved by substitution so I got x alone in the first equation-> x=3y-3
then I plugged this x into the second equation so
4(3y-3)-5=2y+3
12y-12-5=2y+3
12y-17=2y+3
10y=20
y=2
plug y into the first equation to find x
x=3(2)-3
x=6-3
x=3
c. x=3, y=2