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

Introduction to Vedic Mathematics

A Spectacular Illustration of Vedic Mathematics

10's Complements

Multiplication Part 1

Multiplication Part 2

Multiplication Part 3

Multiplication Part 4

Multiplication Part 5

Multiplication Special Case 1

Multiplication Special Case 2

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

Mergers 2

Mergers 3

Multiple Mergers

Consider the equation below:

4/(x + 3) - 16/(4x + 13) = 1/(2x + 5) - 1/(2x + 9)

This does not seem to be a prime candidate for a merger of any sort. Now, look at the series of manipulations we can perform on the above equation:

4/(x + 3) - 16/(4x + 13) = 1/(2x + 5) - 1/(2x + 9) becomes

[4*(4x + 13) - 16*(x + 3)]/[(x + 3)*(4x + 13)] = [(2x + 9) - (2x + 5)]/[(2x + 5)*(2x + 9)] becomes

4/[(x + 3)*(4x + 13)] = 4/[(2x + 5)*(2x + 9)]

Now, we see that the numerator is the same on both sides of the equation. We can then cancel out the common numerator, and invert the two terms to get a simple equation as below:

(x + 3)*(4x + 13) = (2x + 5)*(2x + 9)

We now recognize this to be a generalized form of the third type of equation we derived a formula for in the lesson on solving equations using the Paravartya Yojayet sutra. Applying the formula we derived then to the equation above, we can easily find the solution for the equation to be x = -2.

The operation we performed on the given equation to reduce it to a form that is easily amenable to solution using the formula we derived earlier is called a complex merger.

How do we identify when an equation is capable of being solved using complex merger? There are three tests that will tell us when the conditions are correct for the application of a complex merger operation on a given equation. Consider a general equation as below:

a/(bx + c) - d/(ex + f) = g/(hx + j) - k/(mx + n)

The first condition the equation has to satisfy is as below:

a/b = d/e and g/h = k/m

Thus, we have to make sure that the ratio of the numerator to the coefficient of the unknown is the same in both terms on the left hand side of the equation as well as in both terms on the right hand side of the equation. The ratios on the left and right hand side don't have to be the same, just the ratios of the two terms on either side should be the same.

What does this first condition ensure? Consider the equality a/b = d/e. It can be rewritten as ae = bd. On the left hand side of the given equation, when we take the LCM of (bx + c) and (ex + f), and then cross-multiply to combine the two terms into one, the coefficients of the unknown in the numerator will be ae and -bd. If ae = bd, then these coefficients will cancel out, leaving the numerator on the left hand side free of unknown terms. The condition g/h = k/m similarly ensures that the numerator on the right hand side contains no unknown terms either.

The second condition is that af - dc = gn - kj. The astute reader will have already surmised that the actual numerator on the left hand side of the equation (when we take the LCM of the two denominators and cross-multiply) is af - dc after the cancellation of the coefficients of the unknown term (according to the first condition). Similarly, gn - kj is the final numerator on the right hand side of the equation when that side is expanded out by cross-multiplication, after cancellation of the coefficients of the unknown terms.

When this second condition is satisfied, it enables us to cancel out the numerators on either side of the equation and invert the equation into the form below:

(bx + c)*(ex + f) = (hx + j)*(mx + n)

The third condition comes into play at this point. To ensure that this equation can be solved using the formula derived earlier, the third condition stipulates that be = hm. When this condition is satisfied, the two quadratic terms on either side of the equation cancel out, leaving us a linear equation that can be solved using the formula derived using the Paravartya Yojayet sutra. The final solution is then derived as x = (jn - cf)/(bf + ce - hn - mj)

If this third condition is not satisfied, all is not lost though. Because the first two conditions are satisfied, we have still managed to reduce a potentially cubic equation to just a quadratic equation which is easy to solve using the quadratic formula. Thus, if the first two conditions are satisfied, we should go ahead and take advantage of the complex merger to reduce the equation to a quadratic even if we are not able to reduce it to a linear form.

What happens if the second condition is not satisfied either? We are then left with an equation as below:

(af - dc)/[(bx + c)*(ex + f)] = (gn - kj)/[(hx + j)*(mx + n)]

When we cross-multiply, we get:

(af - dc)*(hx + j)*(mx + n) = (gn - kj)*(bx + c)*(ex + f)

If, at this point, we find that (af - dc)*hm = (gn - kj)*be, then the quadratic terms on the two sides of the equation will cancel out, and we will be left with a linear equation that can be solved by transposing and adjusting appropriately.

Let us solve a couple of equations using the principles we have outlined above so that we are clear on their application.

First let us solve the equation 2/(2x + 1) - 9/(9x + 5) = 2/(6x + 1) - 1/(3x + 1).

The first condition is satisfied since 2/2 = 9/9, and 2/6 = 1/3.

The second condition is also satisfied since 2*5 - 9*1 = 2*1 - 1*1.

We then simplify the equation to (2x + 1)*(9x + 5) = (6x + 1)*(3x + 1). We then verify that 2*9 = 6*3. So, we can apply the formula directly to solve the equation, and get the solution x = -2/5.

Now, let us solve the equation 2/(2x - 1) - 1/(x + 1) = 3/(2x + 1) - 3/(2x + 3).

We see that 2/2 = 1/1, and 3/2 = 3/2. Thus, the first condition is satisfied. However, 2*1 - 1*(-1) = 3, while 3*3 - 3*1 = 6. Thus, the second condition is not satisfied. However, we notice that (af - dc)*hm = 3*2*2 = 12, and (gn -kj)*be = 6*2*1= 12. Thus, we know that we will finally get a linear equation to solve.

Proceeding with the solution, we remove the denominators by cross-multiplying to the form:

(af - dc)*(hx + j)*(mx + n) = (gn - kj)*(bx + c)*(ex + f)

This gives us:

3*(2x + 1)*(2x + 3) = 6*(2x - 1)*(x + 1)

Expanding this and solving it by transposing and adjusting, we get the final solution of x = -5/6.

As we have noticed from previous lessons, there are various ways in which equations that can be solved using this method can undergo transformations that hide their true nature. In this case, once again, there are various ways in which it can become difficult to identify equations because of transformations they have undergone before being presented.

Many of these transformations have been dealt with in previous lessons. Most of them will probably make the equation too difficult to identify and/or solve on sight or mentally, so I will not be dealing with them in this lesson. However, if the equation has undergone some relatively minor transformation such that it is possible to unravel it on sight, then it is worth doing so to get back to a form in which solving the equation becomes much easier.

Identifying such transformations easily comes only with practice. So, it is important to practice working with equations to familiarize oneself with manipulations and transformations that can be performed on them.

We have extended the merger operation in this lesson to the case of complex mergers. We have identified the tests that a given equation has to satisfy, when presented in the standard form, for it to be solved using the complex merger operation. We have also dealt with some variations on the tests that may enable us to solve some equations that at first blush appear unsolvable using complex merger. Good luck, and happy computing!

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