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Wednesday, December 2, 2009

Vedic Mathematics Lesson 29: Solving Equations 4

In the previous lesson, we learned about a powerful application of the Sunyam Samyasamuccaye sutra. We also learned about some ways in which equations could be transformed such that it becomes difficult to identify that the sutra does actually apply to the equation at hand. We then learned how to perform a set of three tests that would reveal whether the given equation can be transformed into an equation of the standard form so that the sutra can then be applied to solve the equation.

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

In this lesson, we will study another transformation that can hide equations so that it becomes hard to recognize that the sutra is applicable to solve the equation. We will once again start with a simple example.

Consider the equation written below:

4/(4x + 1) + 6/(6x + 3) = 3/(3x + 2) + 12/(12x + 1)

By making the numerator equal to 12 in all the terms (because 12 is the least common multiple of the the current numerators, 3, 4, 6, and 12), we can rewrite the equation as below:

12/(12x + 3) + 12/(12x + 6) = 12/(12x + 8) + 12/(12x + 1)

Now, we recognize that not only are the numerators all identical, but also the sum of the denominators on each side of the equation is 24x + 9. At this point, we recognize that the sutra is applicable, and its application gives us the answer x = -9/24.

Once again, we face the conundrum of whether it would be easier to solve the given equation by expanding the terms or whether we should try the LCM approach without any certainty that the sutra would actually be applicable once we go through the trouble of actually making the numerator the same in all four terms.

It turns out the conundrum is not as bad as it sounds. Once again, there is a test one can perform mentally on the given equation to verify whether it can be transformed into the standard form for application of the sutra.

Consider the given equation to be as below:

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

To see if the equation can be solved using the samuccaya sutra, calculate the following quantities:

a*(ex + f) + d*(bx + c) and
g*(mx + n) + k*(hx + j)

Obviously, these are the sums of cross-multiplications of the numerators with the denominators of the other term on each side of the equation. If the two sums calculated above are exactly equal to each other, then the second test is passed, and the equation can be transformed into the standard form. Notice that if you take an equation in standard form for which the sutra is applicable, then this test will always be satisfied since bx + c + ex + f = hx + j + mx + n (note that in the standard form, the numerators of the terms are all 1).

Even if the two sums calculated above are not equal to each other, if they are equal to each other after they are reduced to their lowest terms by taking out common multiples, then the test is passed, and the equation can be transformed into the standard form.

If the test is satisfied, then the solution can actually be found even without going through the steps required to convert the given equation into standard form. In fact, once the test is satisfied, the two quantities calculated above become another meaning of Samuccaya. Therefore, one can equate either of the two quantities calculated in the test to zero to find the solution to the given equation.

Let us consider some simple examples to make sure the test and the final solution are fully understood. Take the case of 2/(2x + 3) + 6/(6x + 5) = 3/(3x + 1) + 4/(4x + 8)

To perform the test, we crossmultiply the numerators and denominators on the left hand side, and we get 6*(2x + 3) + 2*(6x + 5) = 24x + 28. On the right hand side, we get 3*(4x + 8) + 4*(3x + 1) = 24x + 28. Since the two quantities are equal, we can tell that the test is passed. We then equate 24x + 28 to zero, getting us the solution x = -7/6.

Let us try it out on a different example now. Consider the equation 2/(2x + 3) - 1/(x + 1) = 6/(6x + 7) - 3/(3x + 2). We calculate the sum of the cross-products on the left hand side of the equation as 2*(x + 1) - 1*(2x + 3) = -1. Similarly, the sum of the cross-products on the right hand side of the equation becomes -2.

Now, both of them are constants, and we can reduce them to the same lowest term, 1. That tells us that the equation can probably be solved using the sutra, but the sums of the cross-products themselves do not lead to the solution since these sums can not be equated to zero.

The first thing we need to recognize about this test is that it is applicable regardless of the order of terms in the equation. As long as there are two terms on either side of the equation, each with a numerator (that is a constant) and a denominator, we can use the sum of cross-products test to verify applicability of the sutra, and then solve the equation if the sutra is applicable.

Therefore, if as above, the sum of the cross-products does not help us given a particular order of terms in the equation, we can try shuffling the terms around and see if things improve. In this case, we notice that all four terms don't have the same sign. So, one obvious shuffling we can think of is to transpose the negative terms to the opposite sides of the equation, so that we obtain the equation below:

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

Crossmultiplication and addition gives us 12x + 13 on both sides of the equation. Thus, the test is satisfied once again, and the sum of cross-products came out in a form that is usable for solving the equation. Setting the sum of cross-products to zero, we obtain the solution to the equation as x = -13/12.

Now consider the equation 5/(2x - 9) + 3/(3x - 3) = 10/(4x - 3) + 2/(2x - 5). Crossmultiplying and adding on the left hand side of the equation gives us 5*(3x - 3) + 3*(2x - 9) = 21x - 42. On the right hand side though, we get 10*(2x - 5) + 2*(4x - 3) = 28x - 56. We notice that these sums are not the same on either side of the equation.

However, we notice that we can reduce 21x - 42 to 21*(x - 2), so, reducing the left hand side to lowest terms gives us (x - 2). On the right-hand side, we notice that 28x - 56 can also be written as 28*(x - 2). Thus, the right-hand side can also be reduced to the same lowest terms as the left-hand side, thus satisfying the test. This gives us x = 2 as the solution to the equation.

The crossmultiplication test in this lesson is very powerful. Its power comes from its ability to identify equations that are amenable to solution using this sutra that don't have the same numerator in all the terms. In the previous lesson, two of the tests we conducted to see if a given equation could be transformed to the standard form (so that we could attempt to solve it using this sutra) dealt with making sure that the numerators were all the same in all the terms of the equation. That step is actually not needed at all, as we will discover in this lesson.

As an example of the kind of problem we dealt with in the previous lesson, consider the following problem:

(4x + 8)/(2x + 3) + (3x + 5)/(3x + 2) = (12x + 20)/(6x + 7) + (x + 2)/(x + 1)

Let us now apply the three tests we devised in that lesson to this problem. For the first test, we take the coefficients of the unknown term and check whether 4/2 + 3/3 = 12/6 + 1/1. We immediately find that the equation does indeed pass the first test.

Therefore, we proceed to the second test. Here we find that |8 - (3)*4/2| = 2. |5 - 2*(3/3)| = 3. Since 3 is obviously not equal to 2, the equation does not pass the second test. At this point, we may be tempted to give up on using the Samuccaya sutra and simply proceed to solve the equation by brute-force.

What we need to do at this point is to recognize that this might be a combination of two different transformations. So, when it fails the second or third test of the previous lesson, we must not give up. Instead, let us ignore the failed tests of the previous lesson's transformation and proceed to its final step where we get the following equation:

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

At this point, we switch over to the test we have devised in this lesson, and find that this equation does indeed satisfy this test. Thus, by setting the sum of crossproducts on one side of the equation to zero, we get x = -13/12. One can indeed verify that this is the solution of the given equation.

Thus, the tests from the previous lesson and this lesson can be combined as below:

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 need the equation to pass this test so that we can safely transform it as below, and have the constant terms that fall out, cancel out on either side of the equation.

Instead of now performing the two additional tests from the previous lesson, simply transform the equation, and rewrite it 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)

Now, transpose the terms in the transformed equation (if necessary) to get an equation in the standard form (all four terms being positive). At this point, if all the numerators are the same, check whether cx + d + gx + h = mx + n + rx + s. If the two quantities are the same, then the sutra is applicable. Set either of these quantities to zero to find x.

However, one can skip the previous step entirely if one chooses to. Instead, directly apply the cross-multiplication test we have devised in this lesson as the second test. It does not matter what the values of the numerator are. And it does not matter whether the equation has all four terms positive or not either.

Calculate (b - d(a/c))*(gx + h) + (f - h(e/g))*(cx + d). Verify whether it is equal to (k - n(j/m))*(rx + s) + (q - s(p/r))*(mx + n). If they are equal, set them equal to zero to solve the equation by the samuccaya sutra. Notice that this is the sum of cross-products of numerators and denominators on each side of the equation, and therefore, is the test we devised for the kind of transformation we have been focusing on in this lesson.

If the two quantities above are not equal to each other, verify whether we can reduce both of them to the same lowest terms by taking out constant factors on both sides. If we can, then the test is satisfied. Set the lowest terms to zero and obtain the solution to the equation by the samuccaya sutra. If the sum of cross-products becomes a constant on both sides of the equation, simply shuffle two of the terms to the opposite sides of the equation and try once again.

Thus, the above method combines both transformations into one set of two tests that can be applied sequentially to see if the equation can be solved using the samuccaya sutra. The algebra looks complicated because of the use of letters instead of actual numbers. We can actually do all the calculations outlined in the steps above mentally if we work with actual numbers. Let us now solve a couple of problems using these tests to make sure we understand how the tests are to be applied, and also to illustrate that the actual arithmetic involved is not as complicated as it looks.

First, let us consider the equation (3x + 11)/(x + 4) + (2x + 8)/(2x + 7) = (4x - 9)/(2x - 5) + (2x - 17)/(x - 8). First we apply the first test and find that 3/1 + 2/2 = 4/2 + 2/1. So, the equation passes the first test.

At this point, we have two options: we could proceed with the tests outlined in the previous test, and see if we are able to solve the equation that way. Let us try that just for illustration:

Let us now apply the second test: |11 - (3/1)*4| = 1. |8 - (2/2)*7| = 1. |-9 - (4/2)*(-5)| = 1. |17 - (2/1)*(-8)| = 1. Thus, the equation passes the second test also. Therefore, we proceed to the third test from the previous lesson.

We see that 11 - (3/1)*4 = -1. 8 - (2/2)*7 = 1. -9 - (4/2)*(-5) = 1. -17 - (2/1)*(-8) = -1. Thus there is one positive and one negative term on each side of the equation, thus satisfying the third test of the first set. We now proceed with the transformation mentioned in the procedure above. We get -1/(x + 4) + 1/(2x + 7) = 1/(2x - 5) - 1/(x - 8). Transposing the negative terms to opposite sides of the equation, we get 1/(2x + 7) + 1/(x - 8) = 1/(2x - 5) + 1/(x + 4).

Now, we can verify that the sum of the denominators is 3x - 1 on each side of the equation, so by applying the samuccaya sutra, we get x = 1/3 as the solution to the equation. One can substitute x = 1/3 in the given equation and verify that this is indeed the solution of the equation.

The other option is to forego the last two tests from the previous lesson. Instead, we directly apply the transformation after the first test is satisfied and obtain:

-1/(x + 4) + 1/(2x + 7) = 1/(2x - 5) - 1/(x - 8)

Transposing the negative terms to the opposite sides of the equation we obtain:

1/(2x + 7) + 1/(x - 8) = 1/(2x - 5) + 1/(x + 4)

Now, we can verify mentally that the numerators are all the same, and the sum of the denominators is the same on both sides of the equation. Thus, we can set the sum of the denominators to zero, giving us x = 1/3.

But, as an alternative, we can even forego the step of transposing the negative terms to the opposite sides of the equation. Instead, we can apply the crossmultiplication test we have developed in this lesson and verify that 1*(x + 4) - 1*(2x + 7) = 1*(x - 8) - 1*(2x - 5). Setting either of them to zero also gives us x = -3 as another solution to the equation. One can actually verify that both 1/3 and -3 are solutions to the given equation!

By applying the sum of cross-products to the equation obtained after transposing the negative terms to the opposite sides of the equation, we actually do obtain x = 1/3 as another solution from the sum of cross-products method also. Thus, transposing terms to shuffle the equation around can not only find solutions where there were none before, but also help us find all the solutions of an equation if there are more than one. The methodology developed in the previous lesson does not enable us to do that. That is why the sum of cross-products method developed in this lesson is so much more powerful!

Now, let us consider a different equation. Let the given equation be:

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

We see that since 1/1 + 1/1 = 3/3 + 3/3, the equation passes the first test. So, now we transform the equation and try the second test. Transformation of the equation gives us:

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

Applying the second test to the equation as above results in a constant on either side of the equation, so we need to shuffle the terms around to see if a different configuration will allow us to solve the equation.

We transpose the two negative terms to the opposite sides of the equation to get:

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

Now applying the second test, we get:

For the left hand side of the equation, 3*(x - 3) + 1*(3x - 1) = 6x - 10. For the right hand side of the equation, we get 3*(x - 4) + 1*(3x + 2) = 6x - 10.

Since they are equal, the second test is also satisfied. We set the quantity calculated in the second test equal to zero, and find the solution of the equation as x = 5/3. We can substitute this value in the given equation and find that the solution is indeed correct.

As yet another example, consider the equation below:

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

We apply the first test and find that 2/2 + 3/3 = 2/1 + 0/2 (the coefficient of x in the numerator of the fourth term is 0). Thus, the first test is satisfied. Applying the second test we find the following results:

We then proceed to transform the equation as below for proceeding with the second test:

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

Now, we apply the second test. On the left hand side of the equation, we get:

5*(3x - 4) + 5*(2x - 6) = 25x - 50. Reducing it to lowest terms by removing the common multiple of 25 gives us (x - 2).

On the right hand side of the equation, we get:

-1*(2x + 1) + 1*(x + 3) = -x + 2. Removing the common multiple of -1 reduces it to the lowest terms of (x - 2).

Since we get the same sum of crossproducts on both sides of the equation, we can now say that the second test is also passed. We can then set x - 2 = 0 to obtain x = 2 as the solution. We can verify that this is indeed the correct solution by substituting x = 2 in the given equation.

Even though, we have solved the equation, we can proceed one step further, and randomly transpose two terms to the opposite sides of the equation above and reperform the second test. Let us get a new equation by transposing a couple of terms as below:

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

You can verify that this is indeed the same equation as above, with just the order of terms changed. Calculating the sum of cross-products on both sides gives us 7x + 9 when reduced to lowest terms. Thus, we discover that x = -9/7 is another solution of this equation. Thus, the method provided in this lesson is not only simple, but also very powerful in enabling us to ferret out all solutions to a given equation (in case there are more than one) simply by rearranging the terms of the equation!

Unfortunately, it is difficult to tell whether the tests covered in this lesson are both necessary and sufficient, or just sufficient. In other words, in spite of all these tests, we may not be guaranteed that every equation that can be solved using this sutra will be caught and transformed by the above procedure so that the sutra can actually be applied. Thus, the tests may be a sufficient, but not a necessary condition for being able to apply the sutra to solve it.

So, even though we can apply the tests and procedures in these lessons to try and solve equations using the samuccaya sutra (because of the tremendous savings in time and effort that would entail), the tests may not be able to infallibly spot every equation that can actually be transformed into a form that can be attempted to be solved by the samuccaya sutra. Sometimes, we might have to try our own transformations to see if something clicks.

The main guideline we have to keep in mind when we try these transformations is to see if we can make the numerator of all the terms the same. That is the first step in trying to check for the applicability of the sutra. Working with lots of equations and manipulating them in various ways so that we can perform manipulations quickly and easily is key to trying out different transformations to see if any of them can convert the given equation to the standard form. Once again, practice leads to perfection, so I hope you will take the time to practice manipulating equations. Good luck, and happy computing!

1 comment:

Vera Bradley said...

Amazing ! But so long for me to reading ^^

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