Find $x, y\in\mathbb Z$ such that $\frac{x}{2} + \frac{x}{y} - \frac{3}{2} = \frac{10}{y}$ Find $x, y\in\mathbb Z$ such that $\dfrac{x}{2} + \dfrac{x}{y} - \dfrac{3}{2} = \dfrac{10}{y}$
My try
$$\dfrac{x}{2} + \dfrac{x}{y} - \dfrac{3}{2} = \dfrac{10}{y} \iff \dfrac{y}{2} = \dfrac{-x + 10}{x-3} \ \ (x\neq 3,\ 10, \ y\neq 0, \ x, \ y\in\mathbb Z)$$
Let $t = \dfrac{y}{2} = \dfrac{-x + 10}{x-3}.$ We have $y = 2t\neq 0,$ and $x = \dfrac{3t + 10}{t + 1} = 3 + \dfrac{7}{t+1}.$
Then, $x\in\mathbb Z \Leftrightarrow (t+1)|7 \Leftrightarrow t\in \{-8, -2, 6\}.$
Hence, $$x\in \{-4, 2, 4\}, \mbox { and } y\in \{-16, -4, 12\}.$$

Note that listing possible values as sets is not concise enough as it suggests that, for example, x=4 and y=-16 work as a solution. Instead you should pair your answers.
Also, the method you used does give correct solutions, but it is not exhaustive. Rather, try a more standard Diophantine decomposition method.
$$ (y+2)(x-3) = 14  $$
