Solve the following system of equations by using the method of cross multiplication:

$\dfrac{5}{x} - \dfrac{4}{y} + 2 = 0, \dfrac{2}{x} + \dfrac{3}{y} = 13$, (x ≠ 0, y ≠ 0)

Substituting $\dfrac{1}{x}= a, \dfrac{1}{y} = b$ in $\dfrac{5}{x} - \dfrac{4}{y} + 2 = 0$, we get:

⇒ 5a - 4b + 2 = 0     ……….(1)

Substituting $\dfrac{1}{x} = a, \dfrac{1}{y} = b$ in $\dfrac{2}{x} + \dfrac{3}{y} = 13$, we get :

⇒ 2a + 3b = 13

⇒ 2a + 3b - 13 = 0     ……..(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(-4) \times (-13) - (3) \times (2)} = \dfrac{b}{(2) \times (2) - (-13) \times (5)} = \dfrac{1}{(5) \times (3) - (2) \times (-4)} \\[1em] \Rightarrow \dfrac{a}{52 - 6} = \dfrac{b}{4 + 65} = \dfrac{1}{15 + 8} \\[1em] \Rightarrow \dfrac{a}{46} = \dfrac{b}{69} = \dfrac{1}{23} \\[1em] \Rightarrow \dfrac{a}{46} = \dfrac{1}{23} \text{ and } \dfrac{b}{69} = \dfrac{1}{23} \\[1em] \Rightarrow a = \dfrac{46}{23} \text{ and } b = \dfrac{69}{23} \\[1em] \Rightarrow a = 2 \text{ and } b = 3.$

Now we have a = 2 and b = 3,

$\Rightarrow \dfrac{1}{x} = a \\[1em] \Rightarrow \dfrac{1}{x} = 2 \\[1em] \Rightarrow x = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{y} = b \\[1em] \Rightarrow \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

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