Find the length of a triangle's side I have the following triangle and I'm supposed to find the value of x.

First thought that came to mind is to use the the following tangent equation
$$\tan(y)=\frac{x}{27}$$ and $$\tan(19+y) = \frac{2x}{27}$$ which implies that $$\tan(19+y) =2\tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.
I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).
Edit: Using the fact $\tan(a+b) = \dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$ yields the following quadratic equation $$2\tan(19)\tan^2(y)-\tan(y)+\tan(19) = 0$$which gives you that $\tan(y) = 0.890833$ or $\tan(y)= 0.561272$ which means $x = 27\times0.890833$ or $x= 27\times 0.561272$. 
So the question is now which $x$ should I pick and why? and is this more complicated than it should be?

I think you can just use the sinus rule and pythagorean theorem, then:
$$\frac{x}{\sin(19)} = \frac{\sqrt{x^2+27^2}}{\sin(71-y)}$$
And we see:
$$\sin(71-y)=\frac{27}{\sqrt{4x^2+27^2}}$$
Thus, if we substitute:
$$\frac{x}{\sin(19)} = \frac{\sqrt{x^2+27^2}\sqrt{4x^2+27^2}}{27}$$
Solving for $x$:
$$x=15.15...\lor x=24.05...$$
... witch is in compliance with your answer.
PS.
Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.
