In ∆ABC, DE ∥ BC so that AD = (7x − 4) cm, AE = (5x − 2) cm, DB

(c) x = 4

It is given that $D E \| B C$.

Applying Thales' theorem, we get:

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

$\Rightarrow \frac{7 x-4}{3 x+4}=\frac{5 x-2}{3 x}$

$\Rightarrow 3 x(7 x-4)=(5 x-2)(3 x+4)$

$\Rightarrow 21 x^{2}-12 x=15 x^{2}+20 x-6 x-8$

$\Rightarrow 21 x^{2}-12 x=15 x^{2}+14 x-8$

$\Rightarrow 21 x^{2}-12 x-15 x^{2}-14 x+8=0$

$\Rightarrow 6 x^{2}-26 x+8=0$

$\Rightarrow 2\left(3 x^{2}-13 x+4\right)=0$

$\Rightarrow 3 x^{2}-13 x+4=0$

$\Rightarrow 3 x^{2}-12 x-x+4=0$

$\Rightarrow 3 x(x-4)-1(x-4)=0$

$\Rightarrow(x-4)(3 x-1)=0$

$\Rightarrow x-4=0$ or $3 x-1=0$

$\Rightarrow x=4$ or $x=\frac{1}{3}$

If $x=\frac{1}{3}, 7 x-4=-\frac{5}{3}<0$; it is not possible.

Therefore, $x=4$