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Wednesday, November 18, 2009

Vedic Mathematics Lesson 27: Solving Equations 2

In the previous lesson, we examined the derivation of different formulae for solving different types of equations so that we don't have to go through the process of solving for the unknown variable from first principles. But Vedic Mathematics provides another method of solving some equations on sight that we will examine in this lesson.

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

The Vedic method that we are going to examine in this lesson is based on a sutra that reads "Sunyam Samyasamuccaye". The literal translation of this sutra is that "if the Samuccaya is the same, then that Samuccaya must be zero".

What exactly does Samuccaya mean though? It turns out that the term means different things under different contexts. This may sound bad at first, but it actually turns out to be a good thing because the more contextual meanings Samuccaya has, the more contexts under which the sutra can be used. And the more contexts under which the sutra can be used, the more types of equations we may be able to solve on sight without expending any labor!

The first, and simplest, meaning of Samuccaya is an unknown quantity that occurs as a common factor throughout an equation. Take for instance, the equation 4x + 5x = 3x + x. Basic algebra skills would suffice for one to conclude that x = 0 is the solution to this equation. That is precisely what the first application of the sutra also says: since x is an unknown common factor throughout the equation, it is the sumuccaya and equating it to zero gives us the solution x = 0 right away.

However, the method can be applied in equations like the one below also:

7(x - 2) = 3(x - 2)

In this case, (x - 2) is the sumuccaya, and setting it to zero, gives us the solution x = 2. It is possible to expand the terms, collect like terms, transpose and adjust, and reach the same conclusion, but you can solve the equation on sight using this sutra.

The second meaning of Samuccaya is the product of independent terms in equations of the form (x + a)(x + b) = (x + c)(x + d). Suppose you are confronted with the equation (x + 2)(x + 3) = (x + 1)(x + 6). We notice right away that the product of the independent terms on the left hand side is 2*3 = 6, and the product of the independent terms on the right hand side is 1*6 = 6 also. Since they are the same, we can conclude that x = 0 is the solution to that equation!

The astute reader may have noticed that this type of equation was dealt with in the previous lesson as an equation of type 2. The formula for the solution of a type 2 equation has (cd - ab) as its numerator. The two products, cd and ab, represent the products of the independent terms on the right hand side and left hand side of the equation respectively. If they are equal, then the numerator becomes zero, thus making x equal to zero. So you could derive the same solution using the formula in that lesson instead of this sutra.

The third meaning and application of this sutra is in the solution of equations of the form m/(ax + b) + m/(cx + d) = 0. Notice that the numerators of the two terms are the same (m). The sutra says that the sum of the denominators is then the sumuccaya, which can be set to zero for the solution of the equation.

Let us take 1/(x - 1) + 1/(x - 2) = 0 as an example. Since the numerators are the same, calculate the sum of the denominators. It is x - 1 + (x - 2) = 2x - 3. Equating it to zero gives us x = 3/2 right away. We don't have to cross-multiply, expand terms, collect like terms, transpose or adjust. Simply by recognizing that the sum of the denominators is a samuccaya when the numerators are identical enables us to solve the equation on sight!

An extension of this meaning is that if the numerators of the two fractional expressions are not the same numerical constant, but the same unknown quantity, that unknown quantity is a samuccaya too (this can be considered an extension of the first meaning of samuccaya since the numerator becomes a common factor throughout the equation). Consider the following equation:

(x - 1)/(2x + 1) + (x - 1)/(3x - 4) = 0

The sum of the denominators is 2x + 1 + 3x - 4 = 5x - 3. Setting this to zero gives us x = 3/5 right away. But to solve the equation fully, we also need to set x - 1 (the common numerator) to zero, thus giving us x = 1 as another solution to the equation. By cross-multiplying, creating a quadratic equation and using the quadratic formula to solve it, you can verify that x = 1 and x = 3/5 are indeed the two solutions to the equation above.

The fourth meaning of samuccaya can be used to solve some equations that seemingly look very difficult to solve. They are equations of the form (ax + b)/(cx + d) = (ex + f)/(gx + h). In this context, the samuccaya means the sum of the numerators and the sum of the denominators. The sutra says that if the sum of the numerators is the same as the sum of the denominators, then that sum should be equated to zero.

Let us apply to an example to see how to use this interpretation of samuccaya. Suppose one needs to solve (3x + 1)/(x - 2) = (2x - 5)/(4x - 2). We immediately notice that the sum of the numerators is 3x + 1 + 2x - 5 = 5x - 4. The sum of the denominators is x - 2 + 4x - 2 = 5x - 4. Since the sum of the numerators is the same as the sum of the denominators, the sutra can be applied and 5x - 4 can be equated to zero. That immediately gives us one of the solutions to the equation, x = 4/5.

Obviously, once the cross-multiplication and other massaging of the equation is done, it will turn out to be a quadratic equation, so we still need to solve for the other solution, but being able to identify one solution with hardly any calculations is still a step in the right direction.

This application of the sutra is particularly useful when you have equations of the sort (ax + b)/(cx + d) = (cx + d)/(ax + b). These types of equations are easy to spot, and you can also tell right away that this sutra is applicable since the sum of the numerators is ax + b + cx + d, which is also the sum of the denominators. By setting this sum to zero as the sutra tells us to do for solving this equation, we get x = (-b - d)/(a + c).

Now, if a = c in the above equation, then it turns out to be a simple linear equation without any quadratic terms when expanded, and transposed and adjusted. Then, we can conclude that x = (-b - d)/2a is the only solution to the equation. But if a is not equal to c, then the equation will expand out to a quadratic equation that will require us to use a different method to obtain the other solution.

The next meaning of samuccaya may be able to help in that regard though. In this context, samuccaya also means the difference between the numerator and denominator on both sides of the equation. Consider the equation (5x + 3)/(3x + 2) = (3x + 6)/(x + 5). On the left hand side, the difference between the numerator and denominator is 5x + 3 - 3x - 2 = 2x + 1. The difference between the numerator and denominator on the right hand side is 3x + 6 - x - 5 = 2x + 1. The sutra then says that this common difference, being the samuccaya, can be equated to zero for a solution to the equation. Thus, we get x = -1/2.

The important thing to note in this context is that by "difference", I literally mean difference (without any particular order of terms in the subtraction). I am not talking about any order between the numerator and denominator when I do the subtraction on each side of the equation. If the equation were N1/D1 = N2/D2, I am calculating either N1 - D1 or D1 - N1 (depending on my convenience), and I am either calculating N2 - D2 or D2 - N2 (again depending on my convenience). I can compare N1 - D1 against N2 - D2 or against D2 - N2. Similarly, I can compare D1 - N1 against either D2 - N2 or N2 - D2.

As an aside, this is actually quite easy to understand since D2 - N2 is simply -1*(N2 - D2). Since -1 can never be equal to zero, it is the difference that has to be equal to zero.

But, why is this important? There are several reasons why this could be important: First of all, it expands the applicability of this sutra. If N1 - D1 ≠ N2 - D2, you can still try and see whether N1 - D1 = D2 - N2, and if it is, the sutra still applies, with its attendant labor-savings in solving the equation.

More importantly, let us now revisit equations of the sort (ax + b)/(cx + d) = (cx + d)/(ax + b). We already saw that we could use the previous meaning of sumuccaya to get one of the solutions to this equation as (-b -d)/(a + c). Now, consider the difference between the numerators and denominators. On the left hand side, we get ax + b - cx - d when we do N1 - D1. On the right hand side, we are guaranteed to get the same difference when we do D2 - N2 because the actual terms on the numerator and denominator are the same on both sides of the equation, only their positions have been changed.

Thus, according to this sutra, ax + b - cx - d should be set equal to zero for another solution to this equation. This gives us x = (d - b)/(a - c). Notice that if a = c, then the equation is not really a quadratic equation and will expand out to be a linear equation. This second formula is undefined in that circumstance, as it should be!

Consider the equation (5x + 1)/(4x - 2) = (4x - 2)/(5x + 1). Using the last two meanings of the sutra, we can say that the two solutions to this quadratic equation are x = 1/9 and -3. We did not do any cross-multiplication, expanding of terms, collecting of terms, application of the quadratic formula or anything of the sort. We simply looked at the equation, recognized it as something that can be tackled with this sutra, applied the appropriate meaning of the sutra, and got both solutions of the equation within seconds! That is the power of this sutra, especially the last two meanings we have been examining!!

Now, notice that this makes it trivial to solve equations such as (4x + 5)^2 = (3x + 2)^2. All we have to do is rewrite it in a slightly different form as follows, and we are pretty much done: (4x + 5)/(3x + 2) = (3x + 2)/(4x + 5). The solutions are then x = -1 and x = -3! Thus, if you have a quadratic equation that is of the form (ax + b)^2 = (cx + d)^2, the two solutions to it are x = (-b - d)/(a + c) and x = (d - b)/(a - c). Obviously, if a = c in the equation above, it is not a quadratic equation, so we can ignore the second solution. No squares, square roots, discriminants and all the rest of the quadratic paraphernalia! I don't know of any other system anywhere that formalizes this property of quadratic equations of the above sort!!

This brings up an even more powerful property of this sutra to solve quadratic equations that are not expressed readily in the form (ax + b)^2 = (cx + d)^2. But we will examine that in a later lesson.

As you may have guessed by now, this word samuccaya has some more meanings also. Correspondingly, this sutra has a lot more applications also. We will examine them in future lessons. In the meantime, here is a table summarizing the results we have covered in this lesson.

Type Of Equation

At-sight solution technique

Unknown quantity is a common factor throughout the equation

Set the unknown quantity equal to zero

(x + a)(x + b) = (x + c)(x +d) and ab = cd

x = 0

m/(ax + b) + m/(cx + d) = 0

Set the sum of the denominators equal to zero. Thus, x = (–b – d)/(a + c)

Instead of m above, numerator is an unknown quantity

In addition to above solution, set the unknown quantity equal to zero for another solution

(ax + b)/(cx + d) = (ex + f)/(gx + h) and ax + b + ex + f = cx + d + gx + h (sum of the numerators = sum of denominators)

Equate the sum to zero for one solution

(ax + b)/(cx + d) = (ex + f)/(gx + h) and |ax + b – cx – d| = |ex + f – gx – h| (difference between numerator and denominator is the same on both sides of the equation)

Equate the difference to zero for one solution

(ax + b)^2 = (cx + d)^2

Combine the two solution strategies above to derive x = (–b – d)/(a + c) and (if a ≠ c), x = (d – b)/(a – c)


As always, practice is key to success in the application of these methods to the solution of equations. Identifying the samuccaya and applying the right interpretation of the sutra to any given equation takes time and practice. But the rewards can be quite substantial. Good luck, and happy computing!

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