Vedic Sutra #13:     The Ultimate and Twice the Penultimate

We want to solve for x from 1/AB + 1/AC = 1/AD + 1/BC

where D - B = 2(B - A)   ... so that the terms A, B, C and D are in arithmetic progression

Then D + 2C = 0

Example #1:
    1/(x2+7x+12) + 1/(x2+8x+15) = 1/(x2+9x+18) + 1/(x2+9x+20)

Rewrite this as:     1/(x+3)(x+4) + 1/(x+3)(x+5) = 1/(x+3)(x+6) + 1/(x+4)(x+5) ... if you can
and recognize the terms A = (x+3), B = (x+4), C = (x+5) and D = (x+6)
and note that A, B, C and D are in arithmetic progression ... the common difference being "1"

The solution is then obtained from The Ultimate and Twice the Penultimate, namely:
    D + 2C = 0   or   (x+6) + 2(x+5) = 0   or   x = - 5 1/3

Why?
  1. Write 1/AB + 1/AC = 1/AD + 1/BC   as   1/AB - 1/AD + 1/AC - 1/BC = 0
  2. This can also be written:   (1/A)(D - B)/(BD) + (1/C)(B - A)/(AB) = 0
  3. If (D - B) = 2(B - A) then we're left with:   2/D + 1/C = 0   or   D + 2C = 0.