If $$\sqrt{2\frac{23}{49} + 73\frac{23}{49}} - \sqrt{4\frac{29}{49} + 1\frac{15}{49}} = \frac{a}{b}$$ where $$a$$ and $$b$$ are coprime natural numbers, then $$a - b = $$ ______.

The first radicand is $$\frac{121}{49} + \frac{3600}{49} = \frac{3721}{49} = \left(\frac{61}{7}\right)^2$$ and the second is $$\frac{225}{49} + \frac{64}{49} = \frac{289}{49} = \left(\frac{17}{7}\right)^2$$. So the expression equals $$\frac{61}{7} - \frac{17}{7} = \frac{44}{7}$$, where 44 and 7 are coprime. Hence $$a - b = 44 - 7 = 37$$.