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

$\dfrac{1}{x} + \dfrac{1}{y} = 7, \dfrac{2}{x} + \dfrac{3}{y} = 17$ (x ≠ 0, y ≠ 0)

Substituting $\dfrac{1}{x}= a, \dfrac{1}{y} = b$ in $\dfrac{1}{x} + \dfrac{1}{y} = 7$, we get :

⇒ a + b = 7

⇒ a + b - 7 = 0     ………(1)

Substituting $\dfrac{1}{x} = a, \dfrac{1}{y} = b$ in $\dfrac{2}{x} + \dfrac{3}{y} = 17$, we get :

⇒ 2a + 3b = 17

⇒ 2a + 3b - 17 = 0     ………(2)

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

$\Rightarrow \dfrac{a}{(1) \times (-17) - (3) \times (-7)} = \dfrac{b}{(-7) \times (2) - (-17) \times (1)} = \dfrac{1}{(1) \times (3) - (2) \times (1)} \\[1em] \Rightarrow \dfrac{a}{(-17) + 21} = \dfrac{b}{-14 + 17} = \dfrac{1}{3 - 2} \\[1em] \Rightarrow \dfrac{a}{4} = \dfrac{b}{3} = \dfrac{1}{1} \\[1em] \Rightarrow \dfrac{a}{4} = 1 \text{ and } \dfrac{b}{3} = \dfrac{1}{1} \\[1em] \Rightarrow a = \dfrac{4}{1} \text{ and } b = \dfrac{3}{1} \\[1em] \Rightarrow a = 4 \text{ and } b = 3.$

Now we have a = 4 and b = 3,

$\Rightarrow \dfrac{1}{x} = a \\[1em] \Rightarrow \dfrac{1}{x} = 4 \\[1em] \Rightarrow x = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{y} = b \\[1em] \Rightarrow \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

Hence, $x = \dfrac{1}{4}$ and $y = \dfrac{1}{3}$.