You can find all the previous posts about Vedic Mathematics below:
Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
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.
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Straight Division I
Straight Division II
Solving Equations 1
Solving Equations 2
Solving Equations 3
Solving Equations 4
So, what is the special type of equation that is amenable to solution using the merger operation? Consider the equation below:
1/(x - 4) + 3/(x +2) = 4/(x - 5)
What is special about this equation? Looking at it gives one the impression that it will expand out into a quadratic equation that can be solved using the quadratic formula. However, this equation has a special property that guarantees that it is actually a linear equation whose solution does not involve cross-multiplication, collection of terms and use of the quadratic formula.
That special property is that the sum of numerators on the left hand side of the equation is equal to the numerator on the right hand side of the equation. How is this helpful to our cause? Let us look at the following manipulations of the equation:
1/(x - 4) + 3/(x + 2) = 4/(x - 5) becomes
1/(x - 4) + 3/(x + 2) = 1/(x - 5) + 3/(x - 5) becomes
1/(x - 4) - 1/(x - 5) + 3/(x + 2) - 3/(x - 5) = 0 becomes
[(x - 5) - (x - 4)]/[(x - 4)*(x - 5)] + [3(x - 5) - 3(x + 2)]/[(x + 2)*(x - 5)] = 0 becomes
-1/[(x - 4)*(x - 5)] - 21/[(x + 2)*(x - 5)] = 0 becomes
1/(x - 5) * [-1/(x - 4) - 21/(x + 2)] = 0 becomes
-1/(x - 4) - 21/(x + 2) = 0 becomes
1/(x - 4) + 21/(x + 2) = 0
Thus, the third term of the equation (which was on the right hand side), has been "merged" with the other two terms, giving us an equation with just two terms. Hence the term "merger" for this operation.
Also, we can recognize that the equation we have derived using this merger operation is actually an equation of the 4th type we identified in this earlier lesson on solving equations using the Paravartya Yojayet sutra. Applying the solution formula from that lesson, we can immediately solve the equation and say that x = [-1*2 - 21*(-4)]/(1 + 21), which gives us x = 41/11.
One can verify that the solution derived above is actually the correct solution for the original equation.
So, to formalize the methodology so that we can derive a formula for the solution of equations that can be solved by merger, let us take the general equation below:
p/(x + a) + q/(x + b) = (p + q)/(x + c)
Manipulating this equation using the same steps we used on the example equation above, we get the following series of transformations:
p/(x + a) + q/(x + b) = (p + q)/(x + c) becomes
p/(x + a) + q/(x + b) = p/(x + c) + q/(x + c) becomes
p/(x +a) - p/(x + c) + q/(x + b) - q/(x + c) = 0 becomes
p(x + c - x - a)/[(x + a)*(x + c)] + q(x + c - x - b)/[(x + b)*(x + c)] = 0
Removing the common factor (x + c) from the denominator, we get:
p(c - a)/(x + a) + q(c - b)/(x + b) = 0
We see that the numerators are purely constans, with no unknown terms in them. Therefore, we can now apply the formula for the fourth type of equation from this earlier lesson on deriving formulae for solving linear equations using the Paravartya Yojayet sutra. That gives us:
x = [bp(a - c) + aq(b - c)]/[p(c - a) + q(c - b)]
The formula looks long and complicated when written out using letters, but actually it is easy to apply once we get used to it through a few examples. So, let us work through a few of them now.
Let us take 2/(x + 2) + 3/(x + 3) = 5/(x + 5).
We see that 2 + 3 = 5. So, we deduce that the formula applies. Looking at the equation, we can tell that p = 2, q = 3, a = 2, b = 3 and c = 5. Substituting them in the formula gives us x = [3*2*(2 - 5) + 2*3*(3 - 5)]/[2*(5 - 2) + 3*(5 - 3)]. This gives us x = -5/2.
Take 3/(x + 5) + 5/(x - 6) = 8/(x - 4).
We see that 3 + 5 = 8, so we know the formula applies. We also see that p = 3, q = 5, a = 5, b = -6 and c = -4. Using these values in the formula above, we can immediately calculate x = [-6*3*(5 -(-4)) + 5*5*(-6 -(-4))]/[3*(-4 - 5) + 5*(-4 - (-6))], or x = 212/17. Substituting this value of x in the original equation will allow us to verify that this is indeed the correct solution of the equation.
Let us take another example:
4/(x + 5) + 1/(x - 6) = 5/(x + 5)
In this case, we find that a = c = 5. Therefore both (a - c) and (c - a) become zero, leaving us with a simplified formula for x. The simplified formula is x = aq(b - c)/q(c - b) = -a. Thus, this gives us x = -5. This appears to be a patently absurd solution since substituting x = -5 in the equation above would immediately give us two undefined terms. However, let us expand out the terms of the equation and then substitute x = -5 in the final equation that has no denominator terms. Expanding out the equation gives us:
4*(x - 6) + x + 5 = 5*(x + 5)*(x - 6)/(x + 5) which gives us
(x + 5)*(4x - 24 + x + 5) = 5*(x + 5)*(x - 6) which gives us
6x - 95 = -5x - 150
The solution to the equation above is obviously x = -5. So, our solution was correct even though at first glance it appeared not to be!
From this simple exercise, we derive another set of formulae for x when the equation has a specific structure. The special structure and the corresponding solutions are as below:
p/(x + a) + q/(x + b) = (p + q)/(x + a) => x = -a
p/(x + a) + q/(x + b) = (p + q)/(x + b) => x = -b
The solutions above may be counterintuitive, but they are in fact correct, as we verified above.
Now, let us consider the equation 1/(x + 3) - 2/(x + 4) = -1/(x + 5). We see that 1 - 2 = -1, so we know that the formula applies. However, we see that the denominator of the formula is p*(c - a) + q*(c - b), and substituting p = 1, q = -2, a = 3, b = 4, and c = 5 gives us zero for the denominator. Thus, the formula itself becomes undefined. Let us examine why by expanding out the equation. The series of manipulations is shown below:
1/(x + 3) - 2/(x + 4) = -1/(x + 5) becomes
x + 4 - 2*(x + 3) = -1*(x + 3)*(x + 4)/(x + 5) becomes
(-x - 2)*(x + 5) = -1*(x + 3)*(x + 4) becomes
-x^2 - 7x - 10 = -x^2 - 7x - 12 which gives us
10 = 12
This is obviously absurd and can not be true, so the equation itself was not a valid, meaningful equation for which a solution exists. Based on this, we can say that if the denominator of the formula becomes zero, then the equation does not have a valid solution. The actual algebraic proof of this is easy to derive, and the curious reader can verify that the above statement is true by deriving this algebraic proof.
In this lesson, we have learned about one more simple way of solving linear equations that have a specific structure using an operation called merging. This method allows us to solve such equations mentally and quickly rather than going through the process of expanding out the terms and solving the resulting equation from first principles. We also looked at a couple of corollaries that make the solution of equations that satisfy certain conditions even simpler. In the next lesson, we will look at some more applications of mergers. In the meantime, good luck, and happy computing!