Alternative methods to solving system of equations

Question

Let $$ \begin{cases} \sqrt{x} = y -1 \\ \sqrt{y} = 11 - x\end{cases}$$ Solve $x$ and $y$ in real numbers.

After a long calculation, I get to this equation: $(y−4)(y^3−16y−25)=0$. I found $x = 9$ and $y = 4$ are the only answers. What's the other methods for solving that maybe using derivation or inequalities ?

Answer 1

Rewrite like this:

$$\underbrace {\sqrt{\sqrt{x}+1}}_{f(x)} = \underbrace {11-x}_{g(x)}$$

Since $f$ is increasing and $g$ is decreasing this equation has at most one solution. After some ''playing'' with numbers in $[0,11]$ we guess that $x=9$ is a solution and thus unique.

Answer 2

$$y\ge1,\quad x\le 11.\tag1$$

$$ \begin{cases} (y-1)^2 = x\\ (11-x)^2 = y \end{cases}\rightarrow \begin{cases} (y-1)^2 - 9 = x-9\\ (x-11)^2 - 4 = y-4 \end{cases}\rightarrow \begin{cases} (y-4)(y+2) = x - 9\\ y-4 = (x-9)(x-13). \end{cases} $$ $y=-2$ do not lead to the solution.

$x=13$ do not lead to solution.

So $$ \begin{cases} (y+2)(x-9)(x-13) = x-9\\ (y-4)(y+2)(x-13) = y-4 \end{cases}\rightarrow \begin{cases} (x-9)((y+2)(x-13) -1 ) = 0\\ (y-4)((y+2)(x-13) - 1) = 0. \end{cases} $$ Taking in account $(1),$ $$(y+2)(x-13) - 1 < 0,$$ so $$(x,y) = (9,4).$$