Question

In triangle ABC, DE is parallel to AB, AD=2x , DC=x+3, BE=2x-1, CE=x then find value of x

Answer 1

hi! here is your answer !
here.....

DE || AB


By the Basic Proportionality Theorem

$\frac{cd }{ad } = \: \frac{ce}{be}$

$\frac{x + 3}{2x} = \frac{x}{2x - 1}$

$(x + 3)(2x - 1) =(2x)(x)$

$2 {x}^{2} - x + 6x - 3 = 2 {x}^{2}$
$- x + 6x - 3 = 0$

$5x - 3 = 0$

$5x = 3$

$x = \frac{3}{5}$

$x = 0.6$
hope this helps ! ! !#