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# Find the value of $ x $ for which $ \mathrm{DE} \| \mathrm{AB} $ in figure below.

"

To do:

We have to find the value of \( x \) for which \( \mathrm{DE} \| \mathrm{AB} \).

Solution:

$D E \| AB$

This implies, by Basic Proportionality Theorem,

$\frac{C D}{A D} =\frac{C E}{B E}$

$\frac{x+3}{3 x+19} =\frac{x}{3 x+4}$

$(x+3)(3 x+4) =x(3 x+19)$

$3 x^{2}+4 x+9 x+12=3 x^{2}+19 x$

$19 x-13 x =12 $

$6 x =12$

$x=\frac{12}{6}$

$x=2$

Therefore, the value of \( x \) for which \( \mathrm{DE} \| \mathrm{AB} \) is $2$.

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