Setting these two equal, you can write

$\dfrac{a}{\sin x} = b - \dfrac{c}{\tan x} = b - \dfrac{c\cos x}{\sin x}$.

Multiplying both sides of the equation by $\sin x$ you get:

$a = b\sin x - c\cos x$

Next, going back to the first equation and rewriting, you get

$\sin x = \dfrac{a}{y} = \dfrac{\text{opp}}{\text{hyp}}$

So, $\cos x = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{\sqrt{y^2-a^2}}{y}$ (I applied the Pythagorean Theorem to find the length of the adjacent side).

Plugging in:

$a = \dfrac{ab}{y} + \dfrac{c\sqrt{y^2-a^2}}{y}$

Simplifying:

$ay - ab = c\sqrt{y^2-a^2}$

Squaring both sides:

$a^2y^2-2a^2by + a^2b^2 = c^2y^2-c^2a^2$

Now, you can get everything to one side of the equation:

$(a^2-c^2)y^2 - 2a^2by + a^2(b^2+c^2) = 0$

This gives you a quadratic equation. So, solving for $y$ and simplifying a bit, you get:

$y = \dfrac{a^2b \pm ac\sqrt{b^2 + 1 - a^2}}{(a^2-c^2)}$