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In parallelogram DEFG, DH = x + 3, HF = 3y, GH = 4x – 5, and HE = 2y + 3. Find the values of x and y.

In parallelogram DEFG, DH = x + 3, HF = 3y, GH = 4x – 5, and HE = 2y + 3. Find the-example-1
User Enchantner
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2 Answers

5 votes
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
User Johnny Lim
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2 votes

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



User Object Object
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