Can we solve the simultaneous equations $x + y + \sqrt{xy} = 39$ and $x^2 + y^2 + xy = 741$?

Can we solve the simultaneous equations $x + y + \sqrt{xy} = 39$ and $x^2 + y^2 + xy = 741$?

Question

By substituting

$x + y = S$

and

$xy = P^2$

, or otherwise, solve the simultaneous equations

$$\begin{align*} x + y + \sqrt{\vphantom{1}xy} &= 39, \\ x^2 + y^2 + xy &= 741. \end{align*}$$

The expression $\sqrt{\vphantom{1}xy}$ denotes the positive square root of $xy$.

Solution

Using the suggested substitution, the first equation becomes

$$\begin{equation} S+P=39. \end{equation}$$

Also,

$$S^2=x^2+y^2+2xy$$

so the second equation becomes

$$\begin{align} S^2-P^2&=741 \not ag \\ (S-P)(S+P)&=741 \not ag \\ S-P&=\frac{741}{39}=19. \end{align}$$

Solving $(1)$ and $(2)$ simultaneously, we obtain $S=29$ and $P=10$.

Thus $x + y = 29$ and $xy = 100$.

Combining these we have

$$\begin{align*} y&=29-x \\ x(29-x)&=100\\ x^2-29x+100&=0 \\ (x-25)(x-4)&=0\\ x&=25\text{ or }4 \end{align*}$$

The original equations are symmetrical in $x$ and $y$ so the values of $x$ and $y$ are $25$ and $4$ in either order.