Vedic Sutra #5:     If the Samuccaya is the Same it is Zero

We want to solve for x, from: (Ax +B) / (Cx + D) = (Ex + F) / (Gx + H)

Here, Samuccaya refers to some combination of terms.
If the combination is the same ... we set it to zero to find x.

Example #1:
Solve for x from (2x+9) / (2x+7) = (2x+7) / (2x+9)

  1. Note that the sum of the numerators are the same as the sum of denominators, namely 4x + 16.
  2. Since they are the same, it is equal to zero, so : 4x + 16 = 0 hence x = -4 is the solution.
This can be extended to the case where the sum of numerators is constant times the sum of denominators.

Example #2:
Solve for x from (3x+4) / (6x+7) = (x+1) / (2x+3)

  1. Note that the sum of the numerators, namely 4x +5, is a 0.5 times the sum of denominators (which is 8x +10).
  2. Hence (according to this sutra) the sum is equal to zero. That is:   4x + 5 = 0 hence x = -5/4.
Why?
Suppose that   N1 / D1 = N2 / D2
where the sum of numerators is a constant, k, times the sum of denominators. That is: N1+N2 = k(D1+D2) .

Cross multiplying we get:
[1]       N1D2 = N2D1

Adding N2D2 to each side gives:
[2]       (N1+N2)D2 = (D1+D2)N2

If [2] is satisfied, then so it our original equation [1].

Now assume that the sum of the numerators (that's N1+N2) is a constant k times the sum of denominators ... so N1+N2 = k(D1+D2).
Hence, if N1+N2 = 0 then D1+D2 = 0 as well ... and [2] is satisfied as is [1].

Then a solution for x can be found from:
[4]       N1+N2 = 0   or, equivalently,   D1+D2 = 0

In the above examples, if we were to solve for x by cross multiplying, we'd find the coefficients of x2 on each side of the resultant equation are the same. The result is then a simple linear equation for x.
In example #1, cross-multplication would give:     4x2+36x+81 = 4x2+28x+49   and the x2 term cancels
In example #2, cross-multplication would give:     6x2+17x+12 = 6x2+13x+7   and the x2 term cancels

If the x2 term did not cancel, then cross-multiplying would result in a quadratic equation to solve for x
... so there should be a second solution !.

Note, however, that [2] hence [1] is also satisfied when D2 = N2 ... and that'll give a second solution.

Example #3:
Solve for x from (3x+4) / (6x+7) = (5x+6) / (2x+3)

  1. Although the sum of the numerators = the sum of denominators, namely 8x + 10, the x2 term does not cancel.
  2. Nevertheless, we get one solution via setting 8x + 10 = 0, namely x = -5/4 ... as in Example #2
  3. For the second solution we set D2 = N2 giving 2x+3 = 5x+6 so (solving) we get the solution x = -1 as well.
Observation:
If we have that D2 = N2, then (from [2]) we'd also have D1 = N1 namely 6x+7 = 3x+4... so that would also give that second solution