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

In this lesson, we will concentrate on one meaning of the term that is very powerful. We will devote this entire lesson to this meaning since there are several variations of this meaning that can take time to identify correctly for application of this sutra.

The basic equation we will start with has the form 1/(x - a) + 1/(x - b) = 1/(x - c) + 1/(x - d). As you can tell, this does not seem to be an easy type of equation to deal with. In fact, at first glance, it looks as if it is likely to become a cubic equation. Actually, if you take the LCM of the denominators, add up the terms on either side, and simplify the equation, you will get an equation of this sort:

[2x - (a + b)][x^2 - (c + d)x + cd] = [2x - (c + d)][x^2 - (a + b)x + ab]

As you can tell, the x^3 terms on the left and right hand side will cancel out, leaving you with a quadratic equation at best, not a cubic equation. But, the prospect of expanding out the terms further, collecting them to create a standard quadratic equation, and then applying the quadratic formula on it to get the final solution is daunting at best.

The meaning of Samuccaya that we are dealing with in this lesson can make problems like this much simpler under some special circumstances. The special circumstance, in this case, happens to be as follows:

If the sum of the denominators on either side of the equation is the same, that is the Samuccaya. Set it equal to zero to solve the equation. Thus, if 2x - (a + b) = 2x - (c + d), then x = (a + b)/2 (or equivalently, x = (c + d)/2) is the solution. That is all there is to it!

Let us see this meaning of Samuccaya in action with a couple of examples. First, let us consider the equation 1/(x + 2) + 1/(x + 3) = 1/(x + 1) + 1/(x + 4). The sum of the denominators on the left hand side is 2x + 5. We find that the sum of the denominators on the right hand side is also 2x + 5. The sutra then says that 2x + 5 = 0, giving us the solution x = -2.5.

Now consider the equation 1/(x - 9) + 1/(x - 7) = 1/(x - 5) + 1/(x - 11). Once again, the sum of denominators on the left hand side is 2x - 16, which is the same as the sum of the denominators on the right hand side. Thus, x = 8.

Note that the numerator need not be 1 in all the terms above. It can be any constant without affecting the outcome. Thus, it is easy to verify that the solution to 3/(x + 2) + 3/(x - 7) = 3/(x - 1) + 3/(x - 4) is also x = 2.5. That is obviously because the equation could be divided by the constant throughout to make all the numerators equal to 1. In fact, the numerators can be any quantity (known or unknown). The sutra is applicable as long as each term has the same numerator.

The reason this meaning can be hard to apply is because there are many ways in which equations can morph themselves so that it is not readily obvious that this meaning is applicable. So, we will now consider some variations of this type of equation that one may encounter. Some of these variations are hard to spot, but once we do, their solution using this meaning of Samuccaya is practically instantaneous. That not only makes this meaning of Samuccaya very powerful, it also makes it very advantageous for us to spot these variations so that we can convert the equations to standard form and get their solution quickly.

The first variation is quite easy to spot. Consider the equation 1/(x - 4) - 1/(x - 3) = 1/(x - 6) - 1/(x - 5). We only have to realize that two of the terms have been transposed to the opposite sides of the equation for us to change the equation back to the standard form that can be solved by the application of this sutra. Thus, we rewrite the equation as 1/(x - 4) + 1/(x - 5) = 1/(x - 6) + 1/(x - 3). We then see that the sum of the denominators on both sides of the equation is 2x - 9, and therefore x = 4.5.

Thus, the first variation of the standard form of the equation simply involves transposition of the terms. The transposition could involve just one term from each side being moved t0 the other side, or it could be more radical. Consider, for instance the equation below:

1/(x - 3) - 1/(5 - x) = 1/(x - 6) + 1/(x - 2)

In this case, we have to realize that -1/(5 - x) = 1/(x - 5). Thus rewriting that term gives us 1/(x - 3) + 1/(x - 5) = 1/(x - 6) + 1/(x - 2), giving us the solution x = 4.

Thus, the first step in identifying equations to which the sutra is applicable is to make sure that every term is of the form 1/(x + a), and the terms are added to each other, not subtracted from each other (note that I have written the term as 1/(x + a), but a could be negative without affecting the general form into which we are trying to eventually get the terms. Thus the more general way to write the step would be to make sure that every term is either of the form 1/(x + a) or 1/(x - b), where a and b are positive numbers). Once again, if there is some other quantity in the numerator instead of 1, it does not make a difference to the eventual outcome, as long as the numerators are all the same.

Now, consider the equation below:

(x + 2)/(x + 4) + (x + 3)/(x + 5) = (x + 1)/(x + 3) + (x + 4)/(x + 6)

The equation can be rewritten as:

(x + 4 - 2)/(x + 4) + (x + 5 - 2)/(x + 5) = (x + 3 - 2)/(x + 3) + (x + 6 - 2)/(x + 6)

This can then be written as:

(x + 4)/(x + 4) - 2/(x + 4) + (x + 5)/(x + 5) - 2/(x + 5) = (x + 3)/(x + 3) - 2/(x + 3) + (x + 6)/(x + 6) - 2/(x + 6)

Notice that we now have two terms on each side of the equation that simplify to 1. These can be cancelled out, leaving us with:

-2/(x + 4) - 2/(x + 5) = -2/(x + 3) - 2/(x + 6)

Dividing throughout by -2 gives us a standard form of the equation which then leads to the solution x = -4.5.

How do we identify this variation on sight though? The actual transformation of the given equation to the standard form seems somewhat contrived, and we may not want to embark on a series of such transformations without any assurance that the final outcome is going to be the standard form.

There are actually three tests that we can perform to make sure that the equation can in fact be transformed to the standard form. If the equation passes these tests, we can be assured that the equation can indeed be transformed to the standard form.

The first test is performed as follows: Let the given equation be:

(ax + b)/(cx + d) + (ex + f)/(gx + h) = (jx + k)/(mx + n) + (px + q)/(rx + s)

If a/c + e/g = j/m + p/r, then the equation passes the first test. In other words, if the sum of the ratios of the coefficients of the unknown in the numerator and denominator of each term on the left is equal to the sum of the ratios of the coefficients of the unknown in the numerator and denominator of each term on the right, then the equation passes the first test. That is all there is to it! We can then move on and perform the second test.

The second test is performed as follows: Take the same given equation as before. If |b -d(a/c)| = |f - h(e/g)| = |k - n(j/m)| = |q - s(p/r)|, then the equation passes the second test also.

For the third test, look at the signs of each of the four calculated values in the second test without taking their absolute values: b - d(a/c), f - h(e/g), k - n(j/m) and q - s(p/r). If all four of them have the same sign, the equation passes the third test. If there is one negative and one positive term on each side of the equation, then also, the equation passes the third test.

If the equation passes all three tests, rewrite the equation above as below:

(b - d(a/c))/(cx + d) + (f - h(e/g))/(gx + h) = (k - n(j/m))/(mx + n) + (q - s(p/r))/(rx + s)

The resulting equation will either be in standard form or will require a simple transposition of terms from one side of the equation to the other to produce the standard form (the latter occurs if there is one positive and one negative term on each side when performing the third test).

After the equation is brought to standard form, check to make sure that the sum of the denominators is the same on both sides of the equation. If it is, the sutra is applicable for solving the equation. Then, it should be easy to apply the sutra to get the final solution.

Essentially, all we are doing is transforming each term in the given equation as below:

(ax + b)/(cx + d) is rewritten as:

[(a/c)(cx + d) + b - d(a/c)]/(cx + d)

You can expand out the terms to convince yourself that the two forms of the first terms are indeed equivalent.

We then transform the term further as below by expanding the numerator into two terms as below:

(a/c)(cx + d)/(cx + d) + (b -d(a/c))/(cx + d)

This then leaves us with (a/c) + (b - d(a/c))/(cx + d) because the (cx + d)'s in the numerator and denominator of the first term cancel out, leaving us the constant a/c as the first term. When we do a similar transformation of all the terms in the given equation, we get a/c and e/g on the left hand side, and j/m and p/r on the right hand side as constants.

The first test tells us that a/c + e/g = j/m + p/r. So, the constant terms on both sides of the equation cancel out, leaving us with the transformed equation mentioned after the third test.

Because of the second test, we now know that all the numerators have the same absolute value. Because of the third test, we know that either all four of them have the same sign (in which case, the equation is in standard form), or there are terms of opposite signs on each side of the equation (in which case, we need to transpose the negative terms to the opposite sides of the equation to get the standard form).

Let us work through this with a couple of examples so that we can be sure we know how to apply the tests and do the transformations.

Take the equation (2x + 7)/(x + 5) + (4x + 5)/(x + 2) = (3x + 9)/(x + 4) + (3x + 6)/(x + 3) as an example. We perform the first test and verify that 2/1 + 4/1 = 3/1 + 3/1. So, the equation passes the first test.

We perform the second test and verify that |7 - (2/1)*5| = |5 - (4/1)*2| = |9 - (3/1)*4| = |6 - (3/1)*3|. Thus the second test is also passed.

We notice that all four of the terms above, before we take their absolute value, are of the same sign (they are all equal to -3). This tells us that the equation passes the third test also. We now rewrite the equation in its transformed form as below:

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

We now verify that the sum of the denominators on both sides of the equation is 2x + 7, therefore, the sutra is applicable. We set 2x + 7 = 0, giving us the solution x = -3.5.

Consider the next equation as below:

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

Applying the first test, we find that 2/1 + 1/2 = 3/2 + 1/1. So, the equation passes the first test.

Now, apply the second test. We can verify that |3 - (2/1)*3| = |4 - (1/2)*2| = |0 - (3/2)*-2| = |-4 - (1/1)*(-1)|. So, the equation passes the second test.

We also notice that the terms in the second test before taking their absolute values are -3, 3, 3 and -3. Since there is one positive and one negative term on each side of the equation, it passes the third test. So, we rewrite the equation in its transformed state as below:

-3/(x + 3) + 3/(2x + 2) = 3/(2x -2) - 3/(x - 1)

Transposing the negative terms to the opposite sides of the equation gives us an equation in standard form:

3/(2x + 2) + 3/(x - 1) = 3/(2x - 2) + 3/(x + 3)

We now see that the sum of the denominators on both sides of this equation is 3x + 1. Therefore the sutra tells us that the solution to this equation (and the original equation from which this is derived) is x = -1/3.

Obviously, the multitude of ways in which equations can be transformed so that it becomes difficult to identify whether this sutra applies or not, makes it hard to use this meaning of samuccaya in many circumstances. But by being able to perform the tests listed in this lesson mentally and on sight, one can identify equations that are susceptible to this quick and easy solution technique. That can result in large savings of time and effort in actually solving such equations.

Therefore, I hope you will take the time to practice performing these tests on various equations so that you can do them quickly and efficiently. Good luck, and happy computing!

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