Explanation:
Given, ​DE∥AB
∴ $ \displaystyle \dfrac{{\text{CD}}}{{\text{AD}}}=\dfrac{{\text{CE}}}{{\text{BE}}}$​​
[by basic proportionality theorem] ⇒ $ \displaystyle \dfrac{{x+3}}{{3x+19}}=\dfrac{x}{{3x+4}}$
⇒(x + 3)(3x + 4) = x(3x + 19)
⇒ 3$ \displaystyle {{x}^{2}}$ + 4x + 9x + 12 = 3$ \displaystyle {{x}^{2}}$ + 19x
19x − 13x = 12
6x = 12
x = $ \displaystyle \dfrac{12}{6}$​ = 2​
Hence, the required value of $x$ is 2 .