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 \(xy\).

Using the suggested substitution, the first equation becomes \[\begin{equation} S+P=39.\label{eq:1} \end{equation}\] Also, \[S^2=x^2+y^2+2xy\] so the second equation becomes \[\begin{align} S^2-P^2&=741 \notag \\ (S-P)(S+P)&=741 \notag \\ S-P&=\frac{741}{39}=19. \label{eq:2} \end{align}\]

Solving \(\eqref{eq:1}\) and \(\eqref{eq:2}\) 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.