Solve $\frac{\left(x-7\right)}{\sqrt{x-3}+2}+\frac{\left(x-5\right)}{\sqrt{x-4}+1}=\sqrt{10}$ The answer in wolfram is 13. Any easier technique to solve this equation? 
 A: After the rationalization we get $$\sqrt{x-3}-2+\sqrt{x-4}-1=\sqrt{10}$$
so $$\sqrt{x-3}+\sqrt{x-4}=\sqrt{10}+3$$
Now the function on the left is strictly increasing so the equation has at most one solution and this is easy to guess...
A: $$\frac{x-7}{\sqrt{x-3}+2}+\frac{x-5}{\sqrt{x-4}+1}=\sqrt{10}$$
Substitute $x=y+4$
$$\frac{y-3}{\sqrt{y+1}+2}+\frac{y-1}{\sqrt{y}+1}=\sqrt{10}$$
$$\frac{y-3}{\sqrt{y+1}+2}+\frac{(\sqrt{y}+1)(\sqrt{y}-1)}{\sqrt{y}+1}=\sqrt{10}$$
$$\frac{y-3}{\sqrt{y+1}+2}+\sqrt{y}-1=\sqrt{10}$$
Substitute $y=z-1$
$$\frac{z-4}{\sqrt{z}+2}+\sqrt{z-1}-1=\sqrt{10}$$
$$\frac{(\sqrt{z}-2)(\sqrt{z}+2)}{\sqrt{z}+2}+\sqrt{z-1}-1=\sqrt{10}$$
$$\sqrt{z}-2+\sqrt{z-1}-1=\sqrt{10}$$
$$\sqrt{z-1}=3+\sqrt{10}-\sqrt{z}$$
$$(\sqrt{z-1})^2=\left(-\sqrt{z}+\sqrt{10}+3\right)^2$$
$$z-1=z-2  \sqrt{10 z}-6 \sqrt{z}+6 \sqrt{10}+19$$
$$z-\left(z-6 \sqrt{z}-2 \sqrt{10 z}+6 \sqrt{10}+19\right)-1=0$$
$$2 \sqrt{10} \sqrt{z}+6 \sqrt{z}-6 \sqrt{10}-20=0$$
Substitute $\sqrt{z}=w$
$$2 \sqrt{10} w+6 w-6 \sqrt{10}-20=0$$
$$\left(6+2 \sqrt{10}\right) w-6 \sqrt{10}-20=0$$
$$w=\frac{10+3 \sqrt{10}}{3+\sqrt{10}}\to w=\sqrt{10}$$
$z=w^2$ so $z=10$ and $y=z-1=9$ and $x=y+4\to \color{red}{x=13}$
Hope this is useful
