what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respectively (note since this is a geometry question the lengths of a side cannot be negative so the negative solution sets do not matter)

Maybe try this way:
$b^2+16=1+(a+b)^2\Rightarrow 15=a(a+2b)$ and $a^2+9=1+(a+b)^2\Rightarrow 8=b(2a+b)$
$\frac{8}{15}=\frac{b(2a+b)}{a(a+2b)}=\frac{x(2+x)}{1+2x}$, where $x=\frac{b}{a}$ (Assuming $a$ and $b$ are positive).
On solving, we get $x=\frac{2}{5}=\frac{b}{a}$.
Thus, $5b=2a$. 
As $a^2+9=b^2+16=(\frac{2a}{5})^2+16$. This gives $a=\frac{5}{\sqrt{3}}$. Thus, $b=\frac{2}{\sqrt{3}}$. (Considering only positive numbers)
