Solve $\dfrac{1}{x + y} - \dfrac{1}{2x} = \dfrac{1}{30}, \dfrac{5}{x + y} + \dfrac{1}{x} = \dfrac{4}{3}.$ Hence, find the value of 2x² - y².

Given,

$\dfrac{1}{x + y} - \dfrac{1}{2x} = \dfrac{1}{30}$

$\dfrac{5}{x + y} + \dfrac{1}{x} = \dfrac{4}{3}$

Substituting $\dfrac{1}{x + y} = a \text{ and } \dfrac{1}{x} = b$ in above equations,

$a - \dfrac{b}{2} = \dfrac{1}{30}$ ……..(i)

$5a + b = \dfrac{4}{3}$ ……..(ii)

Multiplying eq. (i) by 5 we get,

$5a - \dfrac{5b}{2} = \dfrac{1}{6}$ …….(iii)

Subtracting eq. (iii) from (ii) we get,

$\Rightarrow 5a + b - \Big(5a - \dfrac{5b}{2}\Big) = \dfrac{4}{3} - \dfrac{1}{6} \\[1em] \Rightarrow b + \dfrac{5b}{2} = \dfrac{8 - 1}{6} \\[1em] \Rightarrow \dfrac{2b + 5b}{2} = \dfrac{7}{6} \\[1em] \Rightarrow \dfrac{7b}{2} = \dfrac{7}{6} \\[1em] \Rightarrow b = \dfrac{1}{3} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 3.$

Substituting value of b in eq. (i) we get,

$\Rightarrow a - \dfrac{\dfrac{1}{3}}{2} = \dfrac{1}{30} \\[1em] \Rightarrow a - \dfrac{1}{6} = \dfrac{1}{30} \\[1em] \Rightarrow a = \dfrac{1}{30} + \dfrac{1}{6} \\[1em] \Rightarrow a = \dfrac{1 + 5}{30} \\[1em] \Rightarrow a = \dfrac{6}{30} \\[1em] \Rightarrow a = \dfrac{1}{5} \\[1em] \therefore \dfrac{1}{x + y} = \dfrac{1}{5} \\[1em] \Rightarrow x + y = 5 \\[1em] \Rightarrow 3 + y = 5 \\[1em] \Rightarrow y = 2.$

Substituting values of x and y in 2x² - y² we get,

⇒ 2x² - y²

= 2(3)² - (2)²

= 2(9) - 4

= 18 - 4 = 14.

Hence, x = 3, y = 2 and 2x² - y² = 14.