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.
$\hspace{2cm}$
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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$\ds{\bbox[10px,#ffd]{\left\{\begin{array}{ccccl}
\ds{\root{x}} & \ds{+} & \ds{y} & \ds{=} & \ds{7}
\\[2mm]
\ds{x} & \ds{+} & \ds{\root{y}} & \ds{=} & \ds{11}
\end{array}\right.}:\ {\Huge ?}}$.

Set $\ds{\pars{\vphantom{\LARGE A}\root{x} = 7\sin^{2}\pars{\theta} \implies x = 49\sin^{4}\pars{\theta}}}$ and
$\ds{y = 7\cos^{2}\pars{\theta}}$ such that the first equation is already satisfied.
\begin{align}
&\mbox{Then,}\qquad\qquad\left.\begin{array}{rcl}
\ds{49\sin^{4}\pars{\theta} + 7^{1/2}\cos\pars{\theta}} & \ds{=} & \ds{11}
\\[2mm]
\ds{49\bracks{1 - \cos^{2}\pars{\theta}}^{2} + 7^{1/2}\cos\pars{\theta}} &
\ds{=} & \ds{11}
\\[2mm]
\ds{\color{darkred}{49\cos^{4}\pars{\theta} - 98\cos^{2}\pars{\theta} + 7^{1/2}\cos\pars{\theta} + 38}} & \ds{\bf\color{black}{\large=}} & \ds{\color{darkred}{0}}
\end{array}\right\}
\\[1cm] & \implies
\pars{\cos\pars{\theta},\sin\pars{\theta}} =
\pars{{2 \over 7}\root{7},{\root{21} \over 7}}; 
\\[2mm] & \implies
\pars{x,y} = 
\pars{49\bracks{\root{21} \over 7}^{4},
7\bracks{{2 \over 7}\root{7}}^{2}} =
\pars{\color{red}{\Large9,4}}
\end{align}
See Quartic Function. In this approach, the first equation is satisfied identically, and what's more, the resulting quartic polynomial in $\cos\theta$ does not contain a cubic term. Consequently, one can use the Ferrari solution to solve this equation readily. The method used in Yves Daoust's answer has also this merit.
A: We assume $x, y \geq 0$.  Then let $u = \sqrt{x}, v = \sqrt{y}$, so that the equations become
$$
u+v^2 = 7 \\
u^2+v = 11
$$
Adding the two equations gives us
$$
u^2+u+v^2+v = 18 \\
u^2+u+\frac14+v^2+v+\frac14 = \frac{37}{2} \\
\left(u+\frac12\right)^2+\left(v+\frac12\right)^2 = \frac{37}{2}
$$
Conversely, subtracting the upper equation from the lower equation gives us
$$
(u^2-u)-(v^2-v) = 4 \\
\left(u^2-u+\frac14\right)-\left(v^2-v+\frac14\right) = 4 \\
\left(u-\frac12\right)^2-\left(v-\frac12\right)^2 = 4
$$
Plotting this circle and hyperbola on the first quadrant of the $u$-$v$ plane yields the following diagram:

It seems plausible that the sole point of intersection in this first quadrant is at $(3, 2)$; substitution quickly reveals this to be true.  This in turn yields $(9, 4)$ as the solution to the original equations.
A: Eliminate $\sqrt y$ to get
$$\sqrt x+(11-x)^2=7,$$ which can be rewritten as
$$((11-x)^2-7)^2-x=x^4-44x^3+712x^2-5017x+12996=0.$$
To make this more manageable, we deplete the polynomial with $x:=t+11$, giving
$$t^4-14t^2-t+38.$$
Now using the rational root theorem, we try $\pm2,\pm19$ and obtain the root $t=-2$,$$x=9$$ and from this $$y=4$$ make a valid solution.

Other solutions would be roots of
$$t^3-2t^2-10t+19=0$$ and by the rational root theorem, none are rational.
A: 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 $x=9, y=4$.
A: 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 $y=4$.
A: By subtracting the first equation from the second equation, we have
$$\sqrt{y}-\sqrt{x} + x-y = 4$$
and then $\sqrt{y}-\sqrt{x} + (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) = 4$. So, we get $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1) = 4$.
Put $t := \sqrt{x}-\sqrt{y}$. It follows from the above equation that $t(t+2\sqrt{y}-1) = 4$. By isolating $\sqrt{y}$, we obtain $\sqrt{y} = \frac{4+t-t^2}{2t}$. In additon, we also obtain $\sqrt{x} = t+\sqrt{y} = \frac{4+t+t^2}{2t}$.
Now, we have
\begin{align}
  \sqrt{x}+y &= \frac{4+t+t^2}{2t} + \frac{(4+t-t^2)^2}{4t^2} \\
&= \frac{(8t+2t^2+2t^3) + (16+t^2+t^4 + 8t-8t^2-2t^3)}{4t^2} \\
&= \frac{16 + 16t - 5t^2 + t^4}{4t^2}.
\end{align}
Therefore, the condition $\sqrt{x}+y = 7$ yields $t^4 - 33t^2 + 16t + 16 = 0$.
We can factorize the left hand side of this equation as below:
$$
  (t-1)(t^3 + t^2 - 32t - 16) = 0.
$$
There are three real roots of $t^3 + t^2 - 32t - 16$. However, in our case, $t$ must satisfy the condition $\sqrt{x} > 0$ and $\sqrt{y} > 0$. Unfortunately, none of them satisfies the condition.
Indeed, (i) $\sqrt{x} = \frac{(t+\frac{1}{2})^2+\frac{15}{4}}{4t} > 0$ requires at least $t > 0$, and (ii) under the assumption of $t > 0$, $\sqrt{y} = \frac{-(t-\frac{1}{2})^2+\frac{17}{4}}{4t} > 0$ requires at least $0 < t < 3$. However, (iii) from a sketch of the graph of $f(t) = t^3+t^2-32t-16$, it follows that $f(t) < 0$ for $0 \leq t \leq 3$.
Consequently, the only possible value of $t$ is $t = 1$, i.e. $(x,y) = (9,4)$ is the only real answer of the problem.
