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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Rearrange both equations to isolate the term with the square root, then square both sides.  This new system will likely add additional solutions, so we will need to check to make sure any result we get is really a solution.
Our new system is:
$$x=y^2-14y+49$$
$$y=x^2-22x+121$$
Substituting x from the first equation into the second equation gives the quartic equation: $$y=y^4-28y^3+98y^2+196y^2-1372y+2401-22y^2+308y-1078+121$$
$$0=y^4-28y^3+272y^2-1065y+1444=(y-4)(y^3-24y^2+176y-361)$$
We knew that we could factor out $(y-4)$ because we knew that we had a solution at $x=9,y=4$
Using the rational root theorem, because the leading coefficient is $1$, any rational roots must be factors of the constant term.  This results in possible roots $\pm1,\pm19,\pm361$.  Plugging in each of these possibilities shows that there are no other rational roots.  Using the cubic equation, we find that the other roots are approximately $y=12.8, y=7.9, and y=14.3$.
Our first original equation tells us that the only valid real solutions are for $y\leq7$, so we can reject all of these and we are left with the only solution being $x=9, y=4$.
