## Wednesday, December 16, 2009

### Vedic Mathematics Lesson 31: Mergers 2

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.

You can find all the previous posts about Vedic Mathematics below:

Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I
Straight Division II
Vinculums
Divisibility Rules
Simple Osculation
Multiplex Osculation
Solving Equations 1
Solving Equations 2
Solving Equations 3
Solving Equations 4
Mergers 1

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.

Let us solve a few equations of this sort to familiarize ourselves with the mechanics of the application of the above formula.

Take the equation below, for instance:

5/(3x + 7) + 4/(3x + 2) = 9/(3x + 5)

In this case, p = 5, q = 4, a = 3, b = 7, c = 2 and d = 5. Substituting these values in the formula above, we get x = [(5*7 - 5*5)*2 - (4*2 - 4*5)*7]/[3*(5*5 - 5*7 + 4*5 - 4*2)] = -32/3. This solution can be verified to be true.

Take another equation as shown below:

9/(2x - 7) -7/(2x + 3) = 2/(2x - 11)

In this case, p = 9, q = -7, a = 2, b = -7, c = 3 and d = -11. Substituting these values in the formula above, we get x = [(9*(-7) - 9*(-11))*3 + ((-7)*3 - (-7)*(-11))*(-7)]/[2*(9*(-11) - 9*(-7) + (-7)*(-11) - (-7)*(3))] = 397/62.

Now, consider an equation such as the one below:

1/(3x + 5) + 3/(3x + 2) = 4/(3x + 5)

We can instantly recognize that in this case, b = d = 5. What happens to our formula when b = d? We notice that pd - pb becomes zero, thus the numerator and denominator can be simplified by removing those terms from it. This leaves us with x = [(qc - qd)*b]/[(qd - qc)*a], which can be simplified to x = -b/a. Thus, the solution to the above equation would be -5/3.

We saw in the previous lesson that this may seem counterintuitive at first since substituting x = -5/3 in the above equation would seemingly lead to two divisions by zero. But if we go through the process of cross-multiplying and collecting like terms so that there are no terms in the denominator, we do finally get a linear equation whose solution is actually -5/3.

Thus, just like we did in the previous lesson, we can derive a set of corollary results as below:

p/(ax + b) + q/(ax + c) = (p + q)/(ax + b) => x = -b/a
p/(ax + b) + q/(ax + c) = (p + q)/(ax + c) => x = -c/a

Notice, once again, that these formulae become equivalent to the formulae derived in the previous lesson for similar cases when "a" is equal to one. Thus, this lesson is just a derivation of the formulae for the more general case of merger operations we dealt with in the previous lesson.

We will also carry over from the previous lesson the result that if the entire denominator of our original formula becomes zero because of the values of various terms in it, then the equation is not a valid, meaningful equation, and it has no solution. We will not go into proving this result again in this lesson.

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. We then found out that there are some special equation forms that result in much more simplified solutions. We have derived these special forms and their solutions as corollaries. In the next lesson, we will deal with some more extensions of the merger operation.

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