You can find all my 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
Most math students are taught to solve simultaneous equations with two unknowns in middle school. The method that is taught is usually a system whereby one of the unknowns is eliminated by multiplying one of the equations by an appropriate constant, and then adding it to or subtracting it from the other equation. The resulting equation with just one unknown is solved for this unknown. The solution is then substituted in one of the given equations to produce an equation in just the other unknown, which can then be solved.
The method above can be quite laborious especially when the coefficients of the unknowns are such that both equations have to be multiplied by large numbers to get them to become equal so that addition or subtraction will eliminate the coefficients, allowing one to deduce the value of the other unknown. This becomes the case when the coefficients of the unknowns in the two equations are large prime numbers, for instance. Also notice that this method is almost always unsuitable for mental work, requiring the student to keep track of the equations, and the operations performed on each of them on paper.
In this lesson we will deal with the simplest form of simultaneous equations: 2 simultaneous equations with 2 unknowns. We will formalize the above method of solving simultaneous equations and reduce it down to a set of computations that will allow us to solve such simple simultaneous equations mentally, without the need to go through the laborious process of making the coefficients the same, subtracting or adding, etc.
Consider the set of simultaneous equations below:
2x + y = 7
x + 3y = 11
The solution values of x and y, in the most general case, are fractions. As such, they have numerators and denominators. To find the numerator of the value of x in the above case, follow the procedure below:
- Cross-multilply the coefficient of y in the first equation by the constant term (RHS) of the second equation
- Subtract from it the cross-product of the y coefficient in the second equation and the constant term (RHS) of the first equation
Following the procedure above, we get the numerator as: 1 x 11 - 3 x 7 = -10.
To find the denominator of the value of x, follow the procedure below:
- Cross-multiply the coefficient of y in the first equation by the coefficient of x in the second equation
- Subtract from it the cross-product of the y coefficient in the second equation and x coefficient in the first equation
Following the above procedure, we get the denominator as: 1 x 1 - 3 x 2 = -5.
Using the numerator and denominator together, we can then get the value of x as -10/-5 = 2.
We can either derive the value of y simply by substituting the above value of x in one of the equations and solving for y, or we can follow the same cross-multiplication procedure to get the value of y. If it makes it more convenient to remember, one can rewrite the equations so that the y terms are ahead of the x terms in both equations when we solve for y.
Note that the denominator of the value of y is exactly the same as the denominator of the value of x. Thus, even though it looks like 8 cross-products and 4 subtractions are required to find all the numerators and denominators, we only need 6 cross-products and 3 subtractions.
What is the algebraic basis for why this method works? Consider the equations below:
ax + by = c
dx + ey = f
To solve for x from this system of equations, one would make the coefficients of y equal zero in a third equation derived by multiplying one or both of the equations above by appropriate constants and then adding or subtracting the resulting equations. In this case, let us multiply the first equation by e and the second equation by b. We would then get:
aex + bey = ce
bdx + bey = bf
Now, we can subtract the first of these equations from the second to obtain:
(bd - ae)x = bf - ce
This directly leads to the value of x being (bf - ce)/(bd - ae).
We can immediately recognize that the numerator and denominator in the above solution are the cross-products we calculated earlier in the step-by-step procedure explained earlier.
Instead of solving for x first, we could solve for y first by cross-multiplying the given equations by d and a respectively to get the following:
adx + bdy = cd
adx + aey = af
Subtracting the second equation from the first, we get:
(bd - ae)y = cd - af
This then gives us the solution y = (cd - af)/(bd - ae).
As mentioned earlier, the denominator for the value of y is exactly the same as the denominator for the value of x, thus eliminating the need for us to calculate it twice.
Thus, we make the following observations regarding this method:
- The numerator of any unknown is the difference in cross-products between the constant terms and the coefficients of the other unknown
- The denominator of any unknown is the difference in cross-products between the coefficients of the unknowns
- Remember to keep the signs in mind when calculating the differences.
To remember the direction in which to find the differences, use these rules:
- For the numerator, always make sure you start with coefficient next to the coefficient of the unknown you are trying to solve for, in the first equation. Thus, you will find that the numerator of x has b in the first cross-product (b is the coefficient next to the coefficient of x in the first equation), and the numerator of y has c in the first cross-product (c is the coefficient next to the coefficient of y in the first equation). Thus, when solving for x, always start with the y coefficient of the first equation. When solving for y, always start with the constant term in the first equation.
- For the denominator, always start with the the coefficient of y in the first equation. Thus the denominator of both x and y has b (the coefficient of y in the first equation) in the first cross-product.
Obviously, the two equations have to be written out in the right order for these rules to make sense. Make sure the unknowns are lined up in both equations, with the constant terms on the right hand side of the equal to sign. Label the first unknown as x and the second as y. Then apply the rules above. Or you may choose to simply the commit the two formulae to memory, and after writing the equations in the standard form, just substitute the values of a through f in the formulae to get the solution.
To illustrate the application of these formulae, let us apply them to a few sets of simultaneous equations. First, take the set of equations below:
2x + 4y = 11
x + y = 4
We recognize that a = 2, b = 4, c = 11, d = 1, e = 1 and f = 4. We could substitute the values in the formulae to calculate the values. Rather than do that, let us apply the cross-multiplication rules so that we become more familiar with them.
For the numerator of x, the cross-products involve the coefficients of y and the constant terms in both equations. The difference is calculated with the coefficient of y in the first equation (this is the one next to the coefficient of x in the first equation) first. Thus, the numerator is calculated as 4 x 4 - 1 x 11 = 5. The denominator is always calculated with the y coefficient of the first equation first. Thus, the denominator is 4 x 1 - 2 x 1 = 2. Thus the value of x = 2.5.
One can immediately recognize that the value of y = 1.5 by substituting the above value of x in the second equation. But, assuming we want to proceed with the cross-multiplication method to become more familiar with it, we recognize first of all that the denominator of y is the same as that of x. Thus the denominator of y = 2. The numerator of y is calculated with the constant coefficient in the first equation as the first term in the difference (the constant term is the one next to the coefficient of y in the first equation). Thus, the numerator = 11 x 1 - 4 x 2 = 3. Thus, the value of y, once again, comes out to be y = 1.5.
Now, let us consider the following set:
3 + 4x = 3y
x + y = 8
Notice that the equations are not written out in the right order. Unless this is corrected, we are likely to run into problems when calculating the differences of cross-products. Thus, the first step in the solution is creating the correctly ordered set of equations as below:
4x - 3y = -3
x + y = 8
Now, notice that at least one (in fact, two) of the coefficients is negative. Make sure you take these into account during the cross-multiplications, otherwise, the cross-products are going to be calculated wrong.
Now, we can quickly calculate that the numerator of x = -3 x 8 - (-3 x 1) = -21. The denominator of x is -3 x 1 - 4 x 1 = -7. Thus x = 3. The numerator of y = -3 x 1 - 8 x 4 = -35. This gives us a value of 5 for y (we use the same denominator for y as we derived for x).
Now, consider the set of equations:
2x + y = 0
5x - y = 7
In this case, from the first equation, it is easy to see that y = -2x. Thus one can substitute this value of y in the second equation and quickly derive a third equation (7x = 7) from which it is trivially easy to calculate the value of x to be 1. This would then enable us to solve for the value of y as being equal to -2.
However, assuming one wanted to tackle this using the cross-multiplication method explained here, we first note that one of the coefficients is zero. Make sure this is taken into account during the cross-multiplications. We find the numerator of x to be 1 x 7 - (-1 x 0) = 7. The denominator is 1 x 5 - (2 x -1) = 7. Thus x = 7/7 = 1. The numerator of y is 0 x 5 - 7 x 2 = -14. Thus, y = -14/7 = -2.
What happens when the simultaneous equations we set out to solve are not really independent equations at all? Do the formulae we derived in this lesson alert us to this fact? Consider the set of equations below:
x + y = 2
3x + 3y = 6
In the case of 2 simultaneous equations with 2 variables, non-independence is pretty easy to spot. However, this can be quite difficult when we are dealing with multiple simultaneous equations in several variables. Thus, the ability of any formula we derive to spot non-independence is quite important.
In this case, we get the numerator of x as 1 x 6 - 3 x 2 = 0. We get the denominator as 1 x 3 - 3 x 1 = 0. Thus, our formula gives us an undefined value (0/0) for the value of x. One can verify that the value of y by our formula is undefined too. This shows that if the set of equations is not independent, the values of x and y are undefined according to our formulae also, which is as it should be.
Notice that any time we are provided two equations in which the values of the coefficients of x and y are increased in the same proportion from one equation to the other (in this case, they were multiplied by 3 from the first equation to the second), our denominators will become zero, thus alerting us to our inability to solve for the unknowns using the given equations. This happens whether the set of equations is non-independent as above or inconsistent as below:
x + y = 2
3x + 3y = 3
The method provided in this lesson for the solution of simple simultaneous equations with two unknowns is very easy to work out. Obviously, it takes practice to be able to do it fast and without mistakes (especially when there is a mix of negative and positive coefficients in each equation). I hope you will practice by working with lots of examples so that you can solve such sets of equations mentally and on sight without any delays or hesitation. Good luck, and happy computing!