$\dfrac{2}{3}$ and 1 are the solutions of equation mx² + nx + 6 = 0. Find the values of m and n.

Since, $\dfrac{2}{3}$ is a solution of equation mx² + nx + 6 = 0.

Substituting $\dfrac{2}{3}$ in mx² + nx + 6 = 0,

$\Rightarrow m\Big(\dfrac{2}{3}\Big)^2 + n \times \dfrac{2}{3} + 6 = 0 \\[1em] \Rightarrow m \times \dfrac{4}{9} + \dfrac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \dfrac{4m}{9} + \dfrac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \dfrac{4m + 6n + 54}{9} = 0 \\[1em] \Rightarrow 4m + 6n + 54 = 0 \\[1em] \Rightarrow 2(2m + 3n + 27) = 0 \\[1em] \Rightarrow 2m + 3n + 27 = 0 \\[1em] \Rightarrow 2m + 3n = -27 …….(i)$

Since, 1 is a solution of equation mx² + nx + 6 = 0.

Substituting 1 in mx² + nx + 6 = 0,

$\Rightarrow m(1)^2 + n(1) + 6 = 0 \\[1em] \Rightarrow m + n + 6 = 0 \\[1em] \Rightarrow m = -(n + 6) ……..(ii)$

Sustituting above value of m in eq. 1 we get,

$\Rightarrow 2.-(n + 6) + 3n = -27 \\[1em] \Rightarrow -2(n + 6) + 3n = -27 \\[1em] \Rightarrow -2n - 12 + 3n = -27 \\[1em] \Rightarrow n - 12 = -27 \\[1em] \Rightarrow n = -27 + 12 \\[1em] \Rightarrow n = -15.$

Substituting value of n in (ii) we get,

$\Rightarrow m = -(-15 + 6) \\[1em] \Rightarrow m = -(-9) \\[1em] \Rightarrow m = 9.$

Hence, the value of m = 9 and n = -15.