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Simultaneous equations are a system of independent equations with multiple unknowns. Usually, the number of equations in the system is the same as the number of unknowns.

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.

The full lesson here builds on the material here by solving some examples of simultaneous equations. It also explains what happens when the set of given equations is not independent or they are inconsistent. The formulae presented in this lesson can help one spot such systems of equations right away so that no effort needs to be spent trying to solve them.

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!