Solve the following pair of equations: $2x + y = \frac{7xy}{3}, x + 3y = \frac{11xy}{3}$

(N/A) Given equations are $2x + y = \frac{7xy}{3}$ and $x + 3y = \frac{11xy}{3}$.
It is clear that $x = 0$ and $y = 0$ satisfy both equations. Thus,$(0, 0)$ is a solution.
Assuming $x \neq 0$ and $y \neq 0$,we divide both equations by $xy$:
For the first equation: $\frac{2x}{xy} + \frac{y}{xy} = \frac{7}{3} \implies \frac{2}{y} + \frac{1}{x} = \frac{7}{3} \implies \frac{6}{y} + \frac{3}{x} = 7$ ... $(1)$
For the second equation: $\frac{x}{xy} + \frac{3y}{xy} = \frac{11}{3} \implies \frac{1}{y} + \frac{3}{x} = \frac{11}{3} \implies \frac{3}{y} + \frac{9}{x} = 11$ ... $(2)$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$3u + 6v = 7$ ... $(3)$
$9u + 3v = 11$ ... $(4)$
Multiply equation $(4)$ by $2$: $18u + 6v = 22$ ... $(5)$
Subtract equation $(3)$ from $(5)$: $(18u - 3u) + (6v - 6v) = 22 - 7 \implies 15u = 15 \implies u = 1$.
Substitute $u = 1$ into equation $(3)$: $3(1) + 6v = 7 \implies 6v = 4 \implies v = \frac{2}{3}$.
Since $u = \frac{1}{x} = 1 \implies x = 1$ and $v = \frac{1}{y} = \frac{2}{3} \implies y = \frac{3}{2}$.
The solutions are $(x, y) = (0, 0)$ and $(x, y) = (1, \frac{3}{2})$.