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Just got back from a week-long vacation. I had scheduled some things to be published on my blog during my vacation, but I did not realize that I had not finished the articles fully before leaving on the vacation. So, I came back and found some embarrassing oversights on my blog. Nice way to come back from vacation! Anyways, enough about me and my self-induced problems! Hope you people had a very merry Christmas and are looking forward to the 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 about the generalized form of merger operation on linear equations. We also learned in that lesson the formula for solution of the general form of the equation on which a merger operation can be performed so that we can solve the equation quickly without going through the process of solving the equation from first principles. In this lesson we will look at some equations that do not seem to fit the form of equation that can be handled by a merger, but which actually do after some simple transformations.

Consider an equation like the one below:

(2x – 4)/(x – 3) + (3x + 13)/(x + 2) = (5x + 34)/(x + 5)

We find that this equation does not have the standard form of an equation that can be solved using a merger operation. The numerators are not constants, and the sum of numerators on the left hand side of the equation is not the same as the sum of numerators on the right hand side of the equation.

However, consider the set of transformations below:

(2x – 4)/(x – 3) + (3x + 13)/(x + 2) = (5x + 34)/(x + 5) becomes
(2x – 6 + 2)/(x – 3) + (3x + 6 + 7)/(x + 2) = (5x + 25 + 9)/(x + 5) becomes
(2x – 6)/(x – 3) + 2/(x – 3) + (3x + 6)/(x + 2) + 7/(x + 2) = (5x + 25)/(x + 5) + 9/(x + 5) becomes
2 + 2/(x – 3) + 3 + 7/(x + 2) = 5 + 9/(x + 5) becomes
2/(x – 3) + 7/(x + 2) = 9/(x + 5)

Now we see that the equation does indeed have the structure required for it to be solved by the merger method since the sum of the numerators on the left hand side is equal to the numerator on the right hand side. We can then readily apply the formula and say that x = [2*2*(-3 – 5) – 3*7*(2 – 5)]/[2*(5 + 3) + 7*(5 – 2)] = 31/37. One can verify easily (OK, not that easily, to be perfectly honest) that this is indeed the correct solution to the given equation.

However, suppose we are given an equation of the general form (ax + b)/(cx + d) + (ex + f)/(cx + g) = (hx + j)/(cx + k). How are we to know that this equation can actually be transformed into a form that can be solved using the merger operation? There are two tests we can conduct to tell us whether the transformation will be successful.

The first test is very simple: we simply have to verify that a/c + e/c = h/c. In other words, after taking out the common factor c, we have to check if a + e = h. If this first test is passed, we then move on to the next test. For this test, we calculate the following three quantities:

Let us call p = (b – ad/c)
Let us call q = (f – eg/c)
Let us call r = (j – hk/c)

If p + q = r, then the second test is passed and the equation can be solved using merger. In other words, our next test is to see if (b – ad/c) + (f – eg/c) = (j – hk/c). Simplifying this, we can rewrite the condition as follows: If bc – ad + fc – eg = jc – hk, then the equation can be solved using merger.

You will notice that each of the terms in the equation above is simply a cross-multiplication between the coefficient of the unknown and constant term in the numerator and denominator of each term of the given equation. This makes the test quite easy to remember.

To solve the equation, we simply transform the equation as below, and then apply the standard formula for solving the equation using merger from the previous lesson:

(b – ad/c)/(cx + d) + (f – eg/c)/(cx + g) = (j – hk/c)/(cx + k)

Why do these tests and the subsequent transformation work? We can answer that question by taking the first term of the given general equation and doing a transformation on it as below:

(ax + b)/(cx + d) becomes
[(a/c)*(cx + d) + b – ad]/(cx + d) becomes
(a/c)*(cx + d)/(cx + d) + (b – ad)/(cx + d) becomes
a/c + (b – ad/c)/(cx + d)

This general procedure can then be applied to all the terms. This transformation applied to the second term will give us e/c + (f – eg/c)/(cx + g). The third term, similarly, can be transformed to h/c + (j – hk/c)/(cx + k). Thus the entire equation can be rewritten as below:

a/c + (b – ad/c)/(cx + d) + e/c + (f – eg/c)/(cx + g) = h/c + (j – hk/c)/(cx + k)

From the first test, we know that a + e = h (which implies that a/c + e/c = h/c). This means that the constant terms on both sides of the equation cancel out. This then leaves an equation with constant numerators as below:

(b – ad/c)/(cx + d) + (f – eg/c)/(cx + g) = (j – hk/c)/(cx + k)

The second test then tells us that the sum of the constant numerators on the left hand side is equal to the constant numerator on the right hand side. Thus, if the two tests are passed, we can perform a merger on the resulting equation and solve it using the formula we derived in the previous lesson for the merger method.

For examples of how to apply these tests and transformations to a real equation and solve it using the merger methodology, please read my full blog post here.

Now, consider a different equation as below:

1/(2x + 1) + 2/(3x + 1) = 7/(6x + 7)

In this case we see that the coefficients of the unknown terms are not the same in each of the three terms. Moreover, the sum of numerators on the left hand side does not match the numerator on the right hand side. Thus, this equation does not seem to fall under the merger type of equation. However, notice that the equation above can be rewritten as below:

3/(6x + 3) + 4/(6x + 2) = 7/(6x + 7)

Now, we notice that the sum of the numerators on the left hand side is equal to the numerator on the right hand side (3 + 4 = 7), and also, the coefficient of the unknown quantity in all terms of the equation is the same (6). So, the equation actually is capable of being solved using merger.

How do we then look at any equation of the general form a/(bx + c) + d/(ex + f) = g/(hx + j), and decide whether a merger is possible. In this case, there is just one simple test that will tell us all we need to know.

For this test, calculate a/b, d/e and g/h. If a/b + d/e = g/h, then the test is satisfied.

To solve the equation, transform the equation as below:

(La/b)/(Lx + Lc/b)] + (Ld/e)/(Lx + Lf/e) = (Lg/h)/(Lx + Lj/h), where L is the LCM of b, e, and h.

Because a/b + d/e = g/h, we know that La/b + Ld/e = Lg/h. Thus, the equation is now in the standard form that is required for solution by merger. Just apply the formula derived in the previous lesson on this transformed equation, and we are done.

For examples of how to apply these tests and transformations to a real equation and solve it using the merger methodology, please read my full blog post here.

The third type of equation we may encounter, as you may have guessed by now involves a combination of both transformations we have dealt with so far. Take an equation like the one below:

(2x – 1)/(x – 1) + (3x + 7)/(3x + 4) = (6x + 7)/(2x + 1)

This can be transformed as follows:

(2x – 2 + 1)/(x – 1) + (3x + 4 + 3)/(3x + 4) = (6x + 3 + 4)/(2x + 1) which then becomes
(2x – 2)/(x – 1) + 1/(x – 1) + (3x + 4)/(3x + 4) + 3/(3x + 4) = (6x + 3)/(2x + 1) + 4/(2x + 1) becomes
2 + 1/(x – 1) + 1 + 3/(3x + 4) = 3 + 4/(2x + 1) becomes
1/(x – 1) + 3/(3x + 4) = 4/(2x + 1) becomes
6/(6x – 6) + 6/(6x + 8) = 12/(6x + 6)

We now recognize that the equation is suitable for solution using merger. We get the solution as x = -17/4. But how are we to recognize that the equation is even suitable for solution using merger?

Once again, let us take the generalized form of the equation as below:

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

The tests we need to apply are going to be a combination of the tests we devised earlier. The first test is to check whether a/c + e/g = j/m. If this test is satisfied, we then proceed to the second test. The second test is also very straightforward. Calculate the following quantities:

p = (bc – ad)/c^2
q = (fg – eh)/g^2
r = (km – jn)/m^2

If p + q = r, then the second test is also satisfied. To solve the equation, simply transform it as below, and then apply the formula derived in the previous lesson for solving an equation using merger:

[L*(bc – ad)/c^2]/(Lx + dL/c) + [L*(fg – eh)/g^2]/(Lx + hL/g) = [L*(km – jn)/m^2]/(Lx + nL/m)

In the above equation, L is the LCM of c, g and m.

Why does the above transformation work?

Let us take the term (ax + b)/(cx + d). This can be transformed as below:

(ax + b)/(cx + d) becomes
[(a/c)*(cx + d) + b – ad/c]/(cx + d) becomes
a/c + (b – ad/c)/(cx + d) becomes
a/c + [(L/c)*(bc – ad)/c]/(Lx + dL/c) becomes
a/c + [L*(bc – ad)/c^2]/(Lx + dL/c)

Applying a similar transformation to the other two terms gives us:

e/g + [L*(fg – eh)/g^2]/(Lx + hL/g) for the second term, and
k/m + [L*(km – jn)/m^2]/(Lx + nL/m) for the third term.

Now, we know from the first test that a/c + e/g = k/m. Therefore, the constant terms that are added to each of these transformed terms cancel out on both sides of the equation. We are left with:

[L*(bc – ad)/c^2]/(Lx + dL/c) + [L*(fg – eh)/g^2]/(Lx + hL/g) = [L*(km – jn)/m^2]/(Lx + nL/m)

We know from the second test that (bc – ad)/c^2 + (fg – eh)/g^2 = (km – jn)/m^2. In the equation above, these are numerators of the three terms (they are all multiplied by L, which does not make a difference to the equality). Also, the coefficient of the unknown in all three denominators is L. This then tells us that the equation is now in a form on which the merger formula can be applied directly to find the answer.

For examples of how to apply these tests and transformations to a real equation and solve it using the merger methodology, please read my full blog post here.

This lesson has involved a lot of algebraic manipulation and transformation of equations. This can be confusing at first, but the principles are quite basic, and practice will enable one to perform the tests and transformations on the fly without hesitation. Good luck, and happy computing! And a very happy and prosperous New Year to all too!

– 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.

1 COMMENT

  1. What is the significance of numbers 153
    which follows the rule XYZ=X CUBE + Y CUBE +Z CUBE. IT IS SAID THERE ARE ONLY 4 SUCH NUMBERS FROM ZERO TO INFINITY WHICH FOLLOW THIS RULE.
    EXAMPLE 1 CUBE + 5 CUBE + 3 CUBE= 153

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