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.
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We have
$$\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}$$
Under the constraints
$$0\le x \le 11,\quad  0\le y \le 7$$
A contour plot shows a single point of intersection

We can isolate the square root of $x$ in first equation, square both sides (this causes us to have extraneous roots that we have to eliminate) and then do the same in second equation, substitute and arrive at
$$y=(11−(49−14y+y^2))^2$$
Solving this, we find four roots for $y$, including the only valid root at $y = 4$ and solving for $x$, we arrive at $x = 9$.
To verify this, we can use a Groebner Basis and eliminate either variable and if we first eliminate $x$
$$y^4-28 y^3+272 y^2-1065 y+1444 = (y-4) \left(y^3-24 y^2+176 y-361\right) = 0$$
We could have also chosen to eliminate $y$
$$x^4-44 x^3+712 x^2-5017 x+12996 = (x-9) \left(x^3-35 x^2+397 x-1444\right) = 0$$
Using either of these, we have four real roots, but only one meets the constraints and original equations $(x, y) = (9, 4)$.
We can verify this result using Wolfram Alpha.
