Solve : $4x + \dfrac{6}{y} = 15 \text{ and } 3x - \dfrac{4}{y} = 7$.

Hence, find 'a' if y = ax - 2.

Given, equations :

$\Rightarrow 4x + \dfrac{6}{y} = 15$ ……….(1)

$\Rightarrow 3x - \dfrac{4}{y} = 7$ ……….(2)

Multiplying equation (1) by 2, we get :

$\Rightarrow 2\Big(4x + \dfrac{6}{y}\Big) = 2 \times 15 \\[1em] \Rightarrow 8x + \dfrac{12}{y} = 30 \text{ ………(3)}$

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(3x - \dfrac{4}{y}\Big) = 3 \times 7 \\[1em] \Rightarrow 9x - \dfrac{12}{y} = 21 \text{ ………(4)}$

Adding equation (3) and (4), we get :

$\Rightarrow 8x + \dfrac{12}{y} + 9x - \dfrac{12}{y} = 30 + 21 \\[1em] \Rightarrow 17x = 51 \\[1em] \Rightarrow x = \dfrac{51}{17} = 3.$

Substituting value of x in equation (2), we get :

$\Rightarrow 3 \times 3 - \dfrac{4}{y} = 7 \\[1em] \Rightarrow 9 - \dfrac{4}{y} = 7 \\[1em] \Rightarrow \dfrac{4}{y} = 9 - 7 \\[1em] \Rightarrow \dfrac{4}{y} = 2 \\[1em] \Rightarrow y = \dfrac{4}{2} = 2.$

Given,

⇒ y = ax - 2

⇒ 2 = 3a - 2

⇒ 3a = 2 + 2

⇒ 3a = 4

⇒ a = $\dfrac{4}{3} = 1\dfrac{1}{3}$.

Hence, x = 3, y = 2 and a = $1\dfrac{1}{3}$.