Solve for x and y When (x+$\sqrt {x^2+1})\times (y+\sqrt {y^2+4})=7$

[Albert1]

$if \,\, (x+\sqrt {x^2+1})\times (y+\sqrt {y^2+4})=7$

$find: \,\, x\sqrt {y^2+4}+ y\sqrt {x^2+1}=?$

 

[lfdahl]

My attempt:

Spoiler

Let $x + \sqrt{x^2+1}$ take on the value $\alpha$ (note that $\alpha > 0$ for all $x$) and solve the given equation for $y$: \[\alpha (y+\sqrt{y^2+4})=7 \Rightarrow y = \frac{7}{2\alpha }-\frac{2\alpha }{7}\]In order to facilitate the algebra let $\beta = \frac{7}{2\alpha } > 0$, for all $x$.We are looking for the value of the sum: $x\sqrt{y^2+4}+y\sqrt{x^2+1}$. Elaborating on each term:(i). \[x\sqrt{y^2+4} = x\sqrt{\left ( \beta -\frac{1}{\beta } \right )^2+4}=x\sqrt{\left ( \beta +\frac{1}{\beta } \right )^2}= x\left ( \beta +\frac{1}{\beta } \right )\]
(ii). \[y\sqrt{x^2+1} = y(\alpha -x)=\left (\beta -\frac{1}{\beta } \right )\left ( \alpha -x \right )\]Summing the terms:

\[x\sqrt{y^2+4}+y\sqrt{x^2+1} =x\left ( \beta +\frac{1}{\beta} \right )+\left (\beta -\frac{1}{\beta} \right )\left ( \alpha -x \right )\]\[=\alpha \beta +\frac{1}{\beta }\left ( 2x-\alpha \right )= \frac{7}{2}+\frac{2}{7}(x+\sqrt{x^2+1})(x-\sqrt{x^2+1}) = \frac{7}{2}-\frac{2}{7} = \frac{45}{14}.\]