In ΔABC, DE ∥ BC so that AD = (7x − 4) cm, AE = (5x − 2) cm, DB = (3x + 4) cm and EC = (3x) cm.
Then x equals:

1. 2.5

2. 3

3. 4

4. 5

By basic proportionality theorem,

If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those two sides in the same ratio.

Since DE ∥ BC in ΔABC,

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC} \\[1em] \Rightarrow \dfrac{(7x - 4)}{(3x + 4)} = \dfrac{(5x - 2)}{3x} \\[1em] \Rightarrow 3x \times (7x - 4) = (5x - 2) \times (3x + 4) \\[1em] \Rightarrow 21x^2 - 12x = 15x^2 + 20x -6x -8 \\[1em] \Rightarrow 21x^2 - 12x - 15x^2 - 20x + 6x + 8 = 0\\[1em] \Rightarrow 21x^2 - 15x^2 - 12x - 20x + 6x + 8 = 0 \\[1em] \Rightarrow 6x^2 - 26x + 8 = 0 \\[1em] \Rightarrow 6x^2 - 24x - 2x + 8 = 0 \\[1em] \Rightarrow 6x(x - 4) - 2(x - 4) = 0 \\[1em] \Rightarrow (6x - 2)(x - 4) = 0 \\[1em] \Rightarrow (6x - 2) = 0 \text{ or } (x - 4) = 0 \text{[Using Zero-product rule]}\\[1em] \Rightarrow 6x= 2 \text{ or } x = 4 \\[1em] \Rightarrow x= \dfrac{2}{6} \text{ or } x = 4 \\[1em] \Rightarrow x= \dfrac{1}{3} \text{ or } x = 4 \\[1em]$

x cannot be $\dfrac{1}{3}$ as it yields negative AD and AE values. Therefore x = 4.

Hence, option 3 is the correct option.