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Verified Answer :

SJ2006

Here, DH = HF 

x+3 = 3y

x = 3y-3   ----(I)

GH = HE

4x-5 = 2y+3

4x = 2y+8

Substitute value of x,

4(3y-3) = 2y+8

12y-12 = 2y+8

12y-2y = 12+8

10y = 20

y = 2

Now, substitute it in equation 1,

x = 3(2)-3

x = 6-3 = 3

So, your final answer is      x=3  & y=2

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

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