If $\dfrac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, find the value of k.

Since, $\dfrac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, x = $\dfrac{2}{3}$ satisfies the given equation.

Substituting x = $-\dfrac{2}{3}$ and solving for k we get:

$\Rightarrow 7\Big(\dfrac{2}{3}\Big)^2 + k \times \dfrac{2}{3} -3 = 0 \\[1em] \Rightarrow 7 \times \dfrac{4}{9} + \dfrac{2k}{3} -3 = 0 \\[1em] \Rightarrow \dfrac{28}{9} - 3 + \dfrac{2k}{3} = 0 \\[1em] \Rightarrow \dfrac{28}{9} - \dfrac{27}{9} + \dfrac{6k}{9} = 0 \text{ (Taking L.C.M.) } \\[1em] \Rightarrow \dfrac{28 - 27 + 6k}{9} = 0 \\[1em] \Rightarrow 1 + 6k = 0 \\[1em] \Rightarrow 6k = -1 \\[1em] \Rightarrow k = -\dfrac{1}{6}$

Hence, the value of k is $-\dfrac{1}{6}.$