Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$ Where do I start to solve a equation for x like the one below?
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
After squaring it, it's too complicated; but there's nothing to factor or to expand?
Ideas?

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
$$\implies \sqrt{7x-4}+\sqrt{4x-2}=\sqrt{4x-1}+\sqrt{7x-5} $$
Squaring we get, $$7x-4+4x-2+2\sqrt{(7x-4)(4x-2)}=4x-1+7x-5+2\sqrt{(4x-1)(7x-5)}$$
$$\implies  \sqrt{(7x-4)(4x-2)}= \sqrt{(4x-1)(7x-5)}$$
$$\text{Squaring we get, } (7x-4)(4x-2)=(4x-1)(7x-5)$$
$$\text{ On simplification we get, 8-(16+14)x=5-x(20+7)} $$
$$\implies x=1,\text{  which evidently satisfies the given equation.   }$$ 
Generalization: 
If $$\sqrt{ax+b}-\sqrt{ax+c}=\sqrt{dx+e}-\sqrt{dx+f} \text{ where }b-c=e-f$$
$$\implies \sqrt{ax+b}+\sqrt{dx+f}=\sqrt{dx+e}+\sqrt{ax+c} $$
Squaring we get, $$ax+b+dx+f+2\sqrt{(ax+b)(dx+f)}=dx+e+ax+c+2\sqrt{(dx+e)(ax+c)} $$
$$\implies \sqrt{(ax+b)(dx+f)}=\sqrt{(dx+e)(ax+c)}\text{ as } b-c=e-f\iff b+f=c+e $$
$$\text{Squaring we get, } (ax+b)(dx+f)=(dx+e)(ax+c)$$
$$\implies x(bd+af-cd-ae)=ec-bf $$
