If LM ∥ AB, AL=x-3, AC=2x, BM=x-2, BC=2x+3. What is value of AC?
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In â–³ABC,
It's given LM ∥ AB,So, ALCA=BMCB   [By BPT](x−3)2x=(x−2)(2x+3)
(x−3)×(2x+3) = (x−2)×2x
2x2 - 3x - 9 =Â 2x2 -Â 4x
x=9So AC=2x=18
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