jmc

We let $\sqrt{x}=y$. Then we have $$\begin{array}{rl}xy-5x-9y& =35\Rightarrow \\ xy-5x-9y+45& =35+45\Rightarrow \\ x(y-5)-9(y-5)& =80\Rightarrow \\ (x-9)(y-5)& =80.\end{array}$$ We know $y=\sqrt{x}$, so we resubstitute to find $(x-9)(\sqrt{x}-5)=80$. We construct a table of all factor pairs which multiply to $80$, and proceed to solve for $x$ and $\sqrt{x}$:

\begin{array}{c|c|c|c} $x-9%%DISP_1%%amp;$\sqrt{x}-5%%DISP_1%%amp;$x%%DISP_1%%amp;$\sqrt{x}$\\ \hline $1%%DISP_1%%amp;$80%%DISP_1%%amp;$10%%DISP_1%%amp;$85$\\ $2%%DISP_1%%amp;$40%%DISP_1%%amp;$11%%DISP_1%%amp;$45$\\ $4%%DISP_1%%amp;$20%%DISP_1%%amp;$13%%DISP_1%%amp;$25$\\ $5%%DISP_1%%amp;$16%%DISP_1%%amp;$14%%DISP_1%%amp;$21$\\ $8%%DISP_1%%amp;$10%%DISP_1%%amp;$17%%DISP_1%%amp;$15$\\ $10%%DISP_1%%amp;$8%%DISP_1%%amp;$19%%DISP_1%%amp;$13$\\ $16%%DISP_1%%amp;$5%%DISP_1%%amp;$25%%DISP_1%%amp;$10$\\ $20%%DISP_1%%amp;$4%%DISP_1%%amp;$29%%DISP_1%%amp;$9$\\ $40%%DISP_1%%amp;$2%%DISP_1%%amp;$49%%DISP_1%%amp;$7$\\ $80%%DISP_1%%amp;$1%%DISP_1%%amp;$89%%DISP_1%%amp;$6$ \end{array}

Of all solutions, only one satisfies the relationship ${\sqrt{x}}^2=x$, and that is $\sqrt{x}=7$ and $x=\boxed{49}$.