Question

I challenge all the brainly members to solve this question. Please don't put useless answers, best answer marked as brainliest ,the question is :- In triangle ABC, DE parallel to BC, find the value of x, if AD = 3x + 19, BE = 3x + 4, CD = x + 3, CE= x, This question is totally right there is no figure.

Answer 1

GivenAD=4x-3, BD=3x-1 , AE=8x-7 and EC=5x-3
SinceDE is parllel to BC,by basic proportionality theorem
(AD/DB)=(AE/EC)
(4x-3/3x-1)=(8x-7/5x-3)
By cross multiplication we get
20x2-27x+9=24x2-29x+7
2x2-x-1=0
2x2-2x+x-1=0
2x(x-1)+1(x-1)=0
(2x+1)(x-1)=0
so x=-1/2,1
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