Is it possible?

Let $A = \sqrt{4x}$ , $B = \sqrt{7x - 9}$ , $C = \sqrt{5x - 3}$ , $D = \sqrt{6x -6}$

The equation will become:

$\frac1A + \frac1B = \frac1C + \frac1D$

Simplifying this will be hard but upon inspection in the equation, I noticed that

$A^2 + B^2 = C^2 + D^2$

Further simplification will result to:

$((A^2 + B^2)(CD) + (A^2 + B^2)(AB) - 2ABCD)(CD - AB) = 0$

The first factor will result to imaginary roots but the second will have real roots.

$CD - AB = 0$

$CD = AB$

$\sqrt{5x - 3} \times \sqrt{6x -6} = \sqrt{4x} \times \sqrt{7x - 9}$

$4x(7x - 9) = (5x -3)(6x -6)$

$28x^2 - 36x = 30x^2 - 48x + 18$

$0 = 2x^2 - 12x + 18$

$(x - 3)^2 = 0$

$x = \boxed{3}$

This solution uses only intuition.

By some dumb guess/luck/genius (not so much the third), I set $6x-6=4x\Rightarrow x=3$

Checking this, I found that the expression became

$\frac{1}{12}+\frac{1}{12}=\frac{1}{12}+\frac{1}{12}$

However there is some logic behind this. Since there are all irrational denominators, one on the LHS will MOST LIKELY have to cancel out directly with another on the RHS. Thus setting two equal to each other, we find the answer.

SOME ONE PLEASE VARIFY THIS: The proof that there is only one answer is because the x coefficients add up to the same on both sides.