<s>[INST] 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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 [/INST] 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.
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