Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}
The answer is $x=9,\,y=4$. A geometrical investigation can give us better insights as depicted below.
$\hspace{2cm}$

Here is another elegant solution that one of my friends suggested. Although it is so simple and straight forward, it just works for these specific numbers $7$ and $11$! Let us start by rewriting the equations in the following forms
\begin{align*}
(\sqrt{x}-3)+(y-4)&=0,\\
(x-9)+(\sqrt{y}-2)&=0.
\end{align*}
Next, using the famous identity $a^2-b^2=(a-b)(a+b)$, we rearrange the above equations into the following form
\begin{align*}
(\sqrt{x}-3)+(\sqrt{y}-2)(\sqrt{y}+2)&=0,\\
(\sqrt{x}-3)(\sqrt{x}+3)+(\sqrt{y}-2)&=0.
\end{align*}
Solving for $\sqrt{y}-2$ from the second equation and substituting the result into the first leaves us with
\begin{align*}
(\sqrt{x}-3)-(\sqrt{x}-3)(\sqrt{x}+3)(\sqrt{y}+2)&=0.
\end{align*}
Now, it is easy to see what is going to happen. We just factor out the term $\sqrt{x}-3$, and then we will get a product which should be zero.
\begin{align*}
(\sqrt{x}-3)(1-(\sqrt{x}+3)(\sqrt{y}+2))&=0.
\end{align*}
At least one of these terms must vanish and it turns out that the only possibility is $\sqrt{x}-3=0$. This clearly implies that $x=9$ and substituting back into the original equations gives us $y=4$.
