This 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 solve an equation by using an operation called a merger. In that lesson, we applied the merger to a standard form of equation in which the coefficients of the unknown in each term of the equation was unity. In this lesson, we will extend the merger operation to deal with a bigger set of equations: equations in which the coefficients of the unknown terms are not necessarily unity.
In particular, in the previous lesson, we considered equations of the sort below:
p/(x + a) + q/(x + b) = (p + q)/(x + c)
Now, consider an equation like the one below:
3/(2x + 3) + 4/(2x + 5) = 7/(2x -7)
We notice that the coefficients of the unknowns in the equation above are not unity. But we also notice that they are the same (2) in all terms of the equation. Moreover, the numerators on the left hand side add up to the numerator on the right. The question therefore arises as to whether a merger is possible under this scenario. Is it?
It turns out that it is indeed possible to merge the term on the right hand side into the terms on the left hand side using the series of steps below:
3/(2x + 3) + 4/(2x + 5) = 7/(2x – 7) becomes
3/(2x + 3) + 4/(2x + 5) = 3/(2x – 7) + 4/(2x – 7) becomes
3/(2x + 3) – 3/(2x – 7) + 4/(2x + 5) – 4/(2x – 7) = 0 becomes
[3*(2x – 7) – 3*(2x + 3)]/[(2x + 3)*(2x – 7)] + [4*(2x – 7) – 4*(2x + 5)]/[(2x + 5)*(2x – 7)] = 0 becomes
[1/(2x – 7)]*[(6x – 21 – 6x – 9)/(2x + 3) + (8x – 28 – 8x -20)/(2x + 5)] = 0 becomes
-30/(2x + 3) – 48/(2x + 5) = 0 becomes
30/(2x + 3) + 48/(2x + 5) = 0.
At this point, we recognize that the equation is actually a general form of the fourth type of equation we identified in an earlier lesson on solving equations using the Paravartya Yojayet sutra. We then apply the formula we derived for such equations directly from that lesson, and solve the equation to get x = (-30*5 – 48*3)/(30*2 + 48*2) = -49/26. We can verify that this is indeed the correct solution of the given equation.
Thus, we can generalize the form of equation that can be solved using merger to the equation below:
p/(ax + b) + q/(ax + c) = (p + q)/(ax + d)
What is the solution of the above general form of equation? By going through the merger operation step by step as illustrated with the example above, we can derive the solution to it as below:
p/(ax + b) + q/(ax + c) = p/(ax + d) + q/(ax + d) becomes
p/(ax + b) – p/(ax + d) + q/(ax + c) – q/(ax + d) = 0 becomes
[p(ax + d) – p(ax + b)]/[(ax + b)(ax + d)] + [q(ax + d) – q(ax + c)]/[(ax + c)(ax + d)] = 0 becomes
[1/(ax + d)]*[(pd – pb)/(ax + b) + (qd – qc)/(ax + c)] = 0 becomes
(pd – pb)/(ax + b) + (qd – qc)/(ax + c) = 0
Since the numerators are now constants without any unknown quantities in them, the formula from the lesson on solving equations using the Paravartya Yojayet sutra is applicable. We get the solution as:
x = [p*(b – d)*c + q*(c – d)*b]/[a*(p*(d – b) + q*(d – c))]
Once again, the equation looks quite complicated when written down using letters, but once we start applying it, we can recognize the pattern of appearance of various terms in the equation, and after some practice, quickly become able to substitute the appropriate values in the appropriate places to solve equations quickly and efficiently.
In fact, the formula above is actually the same as the formula we derived in the previous lesson except for an extra division by “a” (the coefficient of the unknown quantity in each of the terms of the equation). Thus, this formula becomes the same as the previous formula if in fact the unknown quantities have a coefficient of unity in all the terms of the equation.
For some examples of how to apply the formula above to some real equations, as well as some corollary results that give rise to simpler formulae for the solution provided certain conditions are met, please visit my blog and read the full lesson here. You can also find links to all my previous posts on Vedic Mathematics in the full lesson in my blog.
Thus, in this lesson, we have extended the merger operation to cover a more general case. We have derived a formula for solving equations of the type that can be solved using this more general form of merger. In the next lesson, we will deal with some more extensions of the merger operation.