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Wednesday, November 11, 2009

Vedic Mathematics Lesson 26: Solving Equations 1

Solving equations is an essential skill that is an integral part of Algebra. In this lesson, we will learn about some patterns of equations and formulae for solving them, so that instead of deriving the solution from first principles, we can apply these simple formulae to get the answer right away.

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.
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
Divisibility Rules
Simple Osculation
Multiplex Osculation

Why is the ability to identify patterns of equations, and solve them by applying formulae important? After all, algebra is about deriving the answer by isolating the unknown variable and equating it to the solution. Consider the case of a generic quadratic equations of the form ax^2 + bx + c = 0. Students are initially taught how to solve such equations by separating b into two parts such that the parts b1 and b2 add up to b, and the ratios a/b1 and b2/c are equal.

That procedure will then lead to the formation of an equation of the form below:

dx(ex + f) + g(ex + f) = 0

which in turn is then simplified to the form:

(dx + g)(ex + f) = 0.

This then gives us the solutions -g/d and -f/e.

Afterwards, we find that this process only works when it is easy to split b up according to the rules. When the required split of b involves fractions and decimals, it is not easy to derive b1 and b2 by trial and error without a lot of effort.

At that point, one may have been taught the derivation of the standard quadratic formula which involves the process of "completing the square". The end result of this derivation is a formula that is etched deep in most high-schoolers' brains.

Going forward, when encountering a quadratic equation that needs to be solved, students are expected to use the formula directly rather than going through the process of justifying the use of it by deriving the formula from first principles. In fact, the formula takes the place of going through the process of trying to isolate the unknown variable in order to solve the equation. The formula is a time-saving device that tells us that if and when we go through the process of isolating x (if we actually want to do it that way), we will end up with solutions that are predicted by this formula. The use of the formula is now so widespread that most people who are perfectly capable of solving quadratic equations can not derive the quadratic formula from scratch!

In this lesson, we will deal with four simple types of equations for which one can derive formulae similar to the quadratic formula. Going forward, one can then use these formulae directly instead of either deriving them or going through the steps involved in isolating the unknown variable following first principles. This will not only save us lots of time and effort, but also reduce the probability of errors during the process of isolation itself. Just as the quadratic formula is used widely to solve quadratic equations, the formulae in this lesson should be taught and used widely to solve equations instead of expecting students and others to go through the process of deriving them from first principles.

Note that all the simple forms of equations are provided in their most general form. Many equations may be missing some elements, which makes their identification and classification into one of these general forms a little tricky, but, with practice, it should become easier. As always, the application of the formulae itself will become much easier (and can easily be done mentally) with practice.

All these formulae are derived primarily by using a Vedic Sutra that reads "Paravartya Yojayet". The literal translation of this sutra is "transpose and adjust". This is a fairly general phrase that covers pretty much all that we do to solve simple equations.

For instance, given an equation such as 2x - 4 = 0, we "transpose" the -4 to the other side and get 2x = 4. We then "adjust" the coefficient of x to isolate it and get x = 4/2, which is the same as x = 2.

In fact, the example above is a simple form of a more general type of simple linear equation for which a formula is readily available to derive the solution. The general form is: ax + b = cx + d. This is the first type of simple equations we will solve using the Paravartya Yojayet sutra. The solution after the necessary transpositions and adjustments is x = (d-b)/(a-c).

So, for instance, if one were to encounter an equation such as 13x + 11 = 3x - 6, one should be able to solve it instantly using the above formula: x = (-6 - 11)/(13 - 3) = -1.7. No actual transposing or adjusting needs to be done from first principles to derive the solution. That is the power of using a formula, just the same way as use of the quadratic formula allows us to solve a quadratic equation without completing the square, transposing or adjusting anything.

Similarly, the solution of the equation 4x - 1 = 3x + 5 is simply x = (5 + 1)/(4 -3) = 6. What about 2x - 4 = 0? In this case, we simply have to recognize that c and d are both zero. We are left with a = 2 and b = -4. Substituting these values in the formula readily gives us the solution x = -(-4)/2 = 2.

How about an equation such as x + 2 = x + 5? Obviously, it is possible to identify on sight that this is not a valid equation with a valid solution since 2 is not equal to 5. Applying the formula to this equation validates this by giving us a solution of x = (5 - 2)/(1 - 1). Since (1 - 1) = 0, the solution involves a division by zero, which means that the solution is undefined, as it should be!

The second type of simple equation is of the form (ax + b)/(cx + d) = p/q. After one goes through the first principles process of cross-multiplying, transposing and isolating x, we would get the solution as x = (pd - bq)/(aq - cp). It is now possible to use this formula directly without having to go through the labor involved in deriving it whenever we encounter an equation of the above variety.

Thus, for instance, if we need to solve (3x + 2)/(x - 5) = 1/3, we would simply use the formula and say that x = (1*(-5) - 2*3)/(3*3 - 1*1) = -11/8.

Similarly, if (3x + 5)/(2x + 1) = 2, notice that this is actually an equation of the second type, even though the right hand side is not in the form of a fraction. In this case, p/q = 2, which can be rewritten as 2/1. This simply means that p = 2 and q = 1. Using these values, we should be able to solve the equation mentally using our formula, and get x = (2*1 - 5*1)/(3*1 - 2*2) = 3.

What about an equation such as (2x + 1)/x = 3/4? We have to recognize, once again, that this is an equation of the second type which simply has d = 0. Thus we have a = 2, b = 1, c = 1, d = 0, p = 3 and q = 4. Plugging these values into the formula x = (pd - bq)/(aq - cp), we readily get x = (3*0 - 1*4)/(2*4 - 1*3) = -4/5.

Now consider the equation 2(4x + 2) = 4(x - 5). One can either convert it into an equation of the first form by expanding out the multiplications to get 8x + 4 = 4x - 20. We can then apply the formula for the first type and get x = -6. One can also manipulate the equation into one of the second form and get (4x + 2)/(x - 5) = 4/2. Now applying the formula for the second form of equations gives us the same solution, x = -6. In fact, equations of the first and second forms can easily be transformed back and forth between the two types, and depending on the state they are in at any given moment, one can use the appropriate formula to solve them.

What happens when we try to solve (3x + 2)/(x + 3) = 3? Once again, if we cross-multiply the terms, we get a simpler-to-analyze linear equation that reads 3x + 2 = 3x + 9. This is obviously a meaningless equation with no well-defined solution. Applying the formula to this situation (after setting q = 1), gives us x = (3*3 - 2*1)/(3*1 - 1*3). The denominator becomes zero, alerting us immediately that the equation is not a valid, meaningful one with any valid solutions.

The third type of equations looks a lot scarier, but is equally easy to deal with once the formula has been derived and is ready for use. These are equations of the form (x + a)(x + b) = (x + c)(x + d). The reason this may look scary at first glance is because it looks as if this should lead to a quadratic equation once the terms have been expanded out. But in reality, the x^2 terms on either side of the equation cancel out, leaving us with a simple linear equation.

After expanding out the terms and transposing and adjusting, we can derive the solution to this equation as x = (cd - ab)/(a + b - c - d). And, in the future, we can use this formula to solve all such equations instead of going through the laborious process of expanding out the terms and solving the equation from first principles.

Thus, if (x + 2)(x + 6) = (x + 1)(x+3), then x = (1*3 - 2*6)/(2 + 6 - 1 - 3) = -9/4. Isn't it amazing that this is all there is to solving an equation which would take at least a few minutes just to expand out to its linear form?

What about (x + 1)(x + 4) = (x + 2)^2? Here, we need to understand that (x + 2)^2 is just a shorthand for writing (x + 2)(x + 2), which should then enable us to correctly identify and classify this equation as belonging to the third type. After that, it is easy to solve it and get the solution as x = (2*2 - 1*4) / (1 + 4 - 2 - 2) = 0.

Now, let us see what happens when we try to solve an equation such as (x + 1)(x + 3) = (x + 2)(x + 2). It is tricky to identify at a glance that this equation is actually one of those meaningless, invalid equations with no solution. But our formula can tell us this right away because (a + b - c - d), which is the denominator in our formula, becomes zero. At this point, out of curiosity, we may choose to expand out the equation to its linear form. We then find that it actually reads 4x + 3 = 4x + 4. This is obviously meaningless, but the fact could be well-concealed by the way the original equation is written. The formula, though, alerts us to the problem right away even if we don't choose to expand out the equation to its linear form.

This third form can be generalized to an equation of the form (ax + b)(cx + d) = (ex + f)(gx + h). This equation simplifies to a linear form ONLY WHEN a*c = e*g. When that condition is true, we can then do the necessary expansion of the terms, transpositions and adjustments to derive the formula for the solution as: x = (fh - bd)/(ad + bc - eh - fg).

For instance, (2x + 3)(3x + 2) = (6x + 1)(x + 8) can be solved easily using this formula to give us x = -1/18. However, the formula is not applicable to (x + 5)(2x + 3) = (3x + 5)(2x -3) because a*c = 2, and e*g = 6, and therefore, a*c is not equal to e*g.

This generalized form can be used to solve certain equations that may not look like they fall into the pattern covered by this type of equation. Consider the equation (3x + 4)/(x + 5) = (6x + 1)/(2x + 2). This looks complicated and does not seem to fall under this category of equations, but once you cross-multiply the terms, you get (3x + 4)(2x + 2) = (6x + 1)(x + 5). Since a*c = e*g, we can actually solve this equation using the general formula to get x = 3/17.

The fourth type of equation we will deal with in this lesson is of the form (p)/(x + a) + (q)/(x + b) = 0. This also looks like a pretty messy type of equation that would take some time to unwind to a simple form that is readily amenable to solution.

If we do go through the process though, we will eventually find that the solution to this equation is simply x = (-pb - qa)/(p + q). Obviously, this is much easier to deal with than going through the laborious, time-consuming and error-prone process of deriving the solution from first principles!

Applying this method to 3/(x + 3) + 2/(x + 2) = 0, for instance, we instantly get the answer as (-3*2 - 2*3)/(3 + 2) = -12/5. Taking another example, we can solve 1/(x + 5) - 2/(x + 1) = 0 and get x = (-1*1 + 2*5)/(1 - 2) = -9.

Let us take the example now of 3/x - 2/(x - 2) = 0. This is actually an equation of the fourth type with a = 0 and b = -2. Substituting these values into the formula gives us the solution x = (-3*(-2) - 0*(-2))/(3 - 2) = 6.

With equations of this type, it can become harder to identify meaningless equations at first glance, but the formula can provide us instant confirmation as to whether a particular equation is valid or not. For instance take the case of 3/(x + 9) - 3/(x - 5) = 0. The formula instantly tells us that since p + q = 3 - 3 = 0, this equation has no valid solutions.

Similarly, take the equation 2/x - 3/x = 0. Here the solution actually works out to zero because both a and b are zero, and therefore -pb - qa = 0. However, at first glance this may seem counter-intuitive since 2/0 - 3/0 is actually undefined. However, if we expand out the equation to linear form first, we actually see that the equation becomes 2x = 3x. At this point, we can tell that zero is indeed the correct solution to the problem.

The fourth type of equation can also be expanded to a more general form. In this general form, p/(ax + b) + q/(cx + d) = 0. The solution to this general form of the equation can be derived as x = (-pd - qb)/(pc + aq). Using this more general form of the formula, one can solve the equation 1/(3x + 4) = 3/(5x + 1) to get x = -11/4 (make sure to note that the first step in applying the formula is to bring the right hand term of the equation to the left hand side to get the standard form of the equation on which the formula is based. This makes q = -3, not q = 3. The astute reader may have also noticed that it is easy to transform equations of the fourth form into ones of the first form and vice versa also).

Thus, not only do these formulae allow us to solve complex-looking equations on sight, mentally, they also allow us to identify when equations have no valid solutions, and sometimes they give us the right solution even when the solution is not intuitive. Just as the quadratic formula is used as soon as quadratic equation presents itself, without regard for first principles, these four types of linear equations should be identified on sight and the appropriate formula should be applied to solve them, without regard for first principles. We are not really ignoring first principles when we do this, it is just that we have taken the labor out of applying the first principles by deriving the end result directly.

The table below summarizes the different types of equations and their solutions into a handy cheat-sheet:

Equation Form


ax + b = cx + d

x = (d – b)/(a – c)

(ax + b)/(cx + d) = p/q

x = (pd – bq)/(aq – cp)

(x + a)(x + b) = (x + c)(x + d)

x = (cd – ab)/(a + b – c – d)

(ax + b)(cx + d) = (ex + f)(gx + h)

Valid only if a*c = e*g

x = (fh – bd)/(ad + bc – eh – fg)

p/(x + a) + q/(x + b) = 0

x = (–pb – qa)/(p + q)

p/(ax + b) + q/(cx + d) = 0

x = (–pd – qb)/(pc + aq)

Obviously, identifying the equations by type, and learning to apply the appropriate formula mentally and quickly, comes only with practice. Hope you will spend some time practicing with these different types of equations. As the examples illustrate, one sometimes has to manipulate the given equation in some simple fashion (cross-multiply, move a term from the left hand side to the right hand side or vice versa, etc.) to get it into a form covered by the types above, before the appropriate formula can be applied to the equation. It takes some practice to identify these manipulations also. Good luck, and happy computing!


Anonymous said...

Yes, why try to actually understand the process of solving linear equations when we can just memorize a huge list of forms? If the problem gets more complicated, not to worry, I'll just have to make an even bigger list of forms and memorize those.

What if I accidentally forget the exact form of one of the solutions? Well, I can't reconstruct it the solution for myself, since I have wisely devoted all my time to memorizing a list rather than understanding the process. No problem, I'll just have to memorize them better next time!

Math is easy when it's reduced to memorizing a massive (infinitely long) list! Why didn't anybody ever think of this before?

Now, if only I could find a list to help me solve equations of the type:

y''(t) + f(t)*y'(t) + g(t)*y(t) = h(t)

with appropriate boundary/initial conditions of course. I could waste time trying to understand what this equation means and therefore develop powerful methods for solving these equations, but it would be much better to just have a list and practice rote memorization until I go blind.

Blogannath said...

I am sure you have never used the quadratic formula, ever in your life (and would not consider using it in the future either), but instead go through the process of completing the squares or deriving the quadratic formula every time you need to solve a quadratic equation. Good for you!

I also hope that your "thorough understanding" of mathematics because of your unwillingness to use any formulae has lead to several great and exciting discoveries in mathematics, and has put you in line for a Fields Medal. Congratulations! Thanks for your gracious visit!!

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