# Solve the following system of equations - (6).

Solve the following system of equations \large \left \{ \begin{aligned} x^2 + y^2 &= 8\\ \sqrtx - \sqrty = (\sqrty - \sqrtx)&(xy + x + y + 2020)\end{aligned} \right.

This was a question in an exam I recently took and I was stumped. Seriously, I can't think of any approach to solve this problem.

Looking at the domain of the equations, we can conclude that $$x \ge 0$$ and $$y \ge 0$$. Also, if $$x \ne y$$, the second equation does not have solutions as both sides will have different signs. ( If $$x>y$$, $$\sqrtx - \sqrty >0$$ and $$\sqrty - \sqrtx<0$$ and vice versa)
Thus, $$x=y=2$$ is the only solution.
I'm going to assume that $$x$$ and $$y$$ are nonnegative, or the LHS of the second equation doesn't make sense. Note that $$x=y$$ is a solution, which yields $$x=y=2$$.
Assume for contradiction there exists some other solution. Then we have $$xy + x + y + 2020 = \frac{\sqrtx-\sqrty}{\sqrty-\sqrtx}.$$ Note that the LHS is positive, and since $$x \neq y$$, the RHS is negative (since either $$x>y$$ in which the numerator is positive and the denominator is negative, or $$x in which it's the other way around). So there can exist no solutions where $$x\neq y$$.