SOLUTION: EF is bisected by AC at D. Find AC if EF = 2x + 22, AC = 3x-5, and DE = 2x-4.

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Question 1204153: EF is bisected by AC at D. Find AC if EF = 2x + 22, AC = 3x-5, and DE = 2x-4.
Found 2 solutions by math_tutor2020, mananth:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

EF is bisected at D, which means EF is split into two equal pieces ED and DF

EF = ED+DF
EF = ED+ED
EF = 2*ED
EF = 2*DE

EF = 2*DE
2x+22 = 2*(2x-4)
2x+22 = 4x-8
2x-4x = -8-22
-2x = -30
x = -30/(-2)
x = 15

Then,
AC = 3x-5
AC = 3*15-5
AC = 45-5
AC = 40

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
DE + DF = EF
DE=DF
2x - 4 + 2x - 4 = 2x + 22


4x - 8 = 2x + 22
2x - 8 = 22
2x = 30
(2x)/2 = 30/2
x = 15
we can find the value of AC:
AC = 3x - 5
AC = 3(15) - 5
AC = 45 - 5
AC = 40
So, AC = 40.