This problem can be solved graphically, but can it be solved mathematically? 


I can solve it graphically, but not mathematically. Graphically, I found X and Y to be equal, X = Y = 1.33

Let $ t = \angle ABC $
then
$ x = (y + 2) \sin t  $
$ x = 1 \cdot \sin t + y \cos 45^\circ = \sin t + y \cos 45^\circ $
$ \dfrac{1}{x} = \dfrac{(y+1)}{(y + 2) \cos t } $
The last equation becomes
$ x(y + 1) = (y + 2) \cos t $
Use the first equation,
$  (y+1) (y+2) \sin t = (y+2) \cos t $
So
$\tan t = \dfrac{1}{y+1} $
It follows that
$ \cos t = \dfrac{y+1}{\sqrt{y^2 + 2 y + 2 }} $
$ \sin t = \dfrac{1}{\sqrt{y^2 + 2 y + 2 }} $
Substitute into the second equation,
$\dfrac{y + 2}{\sqrt{y^2 + 2 y + 2}} = \dfrac{1}{\sqrt{y^2+ 2 y + 2}} + \dfrac{y}{\sqrt{2}} $
And this simplifies to
$ y + 2 = 1 + \dfrac{y}{\sqrt{2}} \sqrt{y^2 + 2 y + 2} $
And further to
$ \sqrt{2} (y + 1) = y \sqrt{ y^2 + 2 y + 2} $
So that
$ 2 (y^2 + 2 y + 1) = y^2 (y^2 + 2 y + 2 ) $
And finally,
$ y^4 + 2 y^3 - 4 y - 2 = 0 $
whose solution is (from wolframalpha.com)
$ y = \dfrac{1}{2} (-1 + \sqrt{2} \sqrt[4]{3} + \sqrt{3} ) = 1.29663026289$
It follows that
$ t = \tan^{-1} \bigg( \dfrac{1}{y+1} \bigg) = 23.5292985676^\circ $
And
$ x = (y + 2) \sin t = 1.31607401295 $
