In parallelogram defg, dh = x + 3, hf = 3y, gh = 4x – 5, and he = 2y + 3. find the values of x and y.

Ince 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 

therefore x=3, y=2