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I wish you all a very happy and prosperous new year!


This post is, once again, a summary of a longer post authored by me on my own blog. My blog covers a lot of areas, including Vedic Mathematics. If you are interested in reading my thoughts on other topics, please feel free to visit my blog and post comments on the other articles you find there also! Thank you!

In the previous lesson, we learned how to identify the most general form of an equation that can be solved using a merger operation. In general, such equations have two terms on the left hand side and one term on the right hand side. In this lesson, we will extend the methodology to some equations that have more than two terms on the left hand side.

To set the stage, consider an equation of the type below:

3/(x + 2) + 2/(x + 1) + 4/(x + 7) = 9/(x + 4)

We see that there are not two, but three terms on the left hand side of the equation, but there is only one term on the right hand side of the equation. Moreover, the sum of numerators on the left hand side (3 + 2 + 4) is equal to the numerator on the right hand side (9). Also, the coefficients of the unknown in all the terms is unity.

Now, look at the series of manipulations below:

3/(x + 2) + 2/(x + 1) + 4/(x + 7) = 9/(x + 4) becomes
3/(x + 2) – 3/(x + 4) + 2/(x + 1) – 2/(x + 4) + 4/(x + 7) – 4/(x + 4) = 0 becomes
3*(x + 4 – x – 2)/[(x + 2)*(x + 4) + 2*(x + 4 – x – 1)/[(x + 1)*(x + 4) + 4*(x + 4 – x – 7)/[(x + 7)*(x + 4)] = 0 becomes
[1/(x + 4)] * [3*2/(x + 2) + 2*3/(x +1) -4*3/(x + 7)] = 0 becomes
6/(x + 2) + 6/(x + 1) – 12/(x + 7) = 0 becomes
6/(x + 2) + 6/(x + 1) = 12/(x + 7)

We now see that the right hand side term has been merged with the left hand side, and we are left with an equation that is once again capable of undergoing another merger to get the solution (the numerators on the left hand side, 6 and 6, add up to the numerator on the right hand side, 12, and the coefficients of the unknown in all three terms are unity again).

Performing the merger operation on the resulting equation gives us a final equation of:

30/(x + 2) + 36/(x + 1) = 0

This then gives us a solution of x = -17/11. One can verify that this indeed is the correct solution to both the original given equation as well as the intermediate equations derived after the first and second mergers.

Since the solution procedure for the given equation involved performing two mergers one after the other, this procedure is referred to as a multiple merger. In fact, under the right conditions, it is possible to string together any number of mergers to solve an equation. The question then becomes: what are the right conditions?

To answer that question, let us consider a general equation as below:

p/(x + a) + q/(x + b) + r/(x + c) = s/(x + d)

Let us now assume that p + q + r = s. Thus we can perform a merger operation to begin with, and eliminate s and d, giving us:

p*(d – a)/(x + a) + q*(d – b)/(x + b) + r*(d – c)/(x + c) = 0

To be able to perform a second merger and solve the equation, the condition that has to be satisfied is as below:

Either p*(a – d) = q*(d – b) + r*(d – c) or
q*(b – d) = p*(d – a) + r*(d – c) or
r*(c – d) = p*(d – a) + q*(d – b)

If any of the above conditions are satisfied, then a second merger can be performed and the equation can be solved using merger. Assuming the third condition is satisfied, we can rewrite the equation after the first merger as below:

p*(d – a)/(x + a) + q*(d – b)/(x + b) = r*(c – d)/(x + c)

Because the third condition is true, we know that the sum of the numerators on the left hand side of the above equation is equal to the numerator on the right hand side of the equation. Therefore, we can perform another merger operation to eliminate r*(c – d) and c. This will give us:

p*(d – a)*(c – a)/(x + a) + q*(d – b)*(c – b)/(x + b) = 0

This can then be solved using the formula we derived in the first lesson on mergers here.

The extension of this methodology to more than 2 mergers must now be obvious to the reader. At each step, after the performance of a merger, the remaining terms must satisfy the conditions for another merger to be performed. The conditions imply that the sum of all but one of the terms must be equal to the negative of one of the terms. Depending on which term is thus isolated, a merger would once again be performed on the equation until the equation is reduced to two terms, at which point it can be solved using the formula we derived earlier.

You can read the rest of this lesson on my blog here. Good luck, and happy computing!

 

– The Vedic Maths Forum India


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Gaurav Tekriwal is the founder President of the Vedic Maths Forum India. Through television programs, workshops, DVDs, and Books he has taken the Vedic Maths System to over 4 million students in India, South Africa, United States, Australia, UAE, Ghana, and Colombia.

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