Simultaneous Equations

Unfortunately, we can’t solve this equation. The reason is that it contains two letters (two unknown numbers). This means that we can choose any value we like for x and there will be a corresponding value for y. For instance, x = 1 is a possible value for x, which would mean that y would have to be 6. On the other hand, x could be 4, in which case y would have to be 4 also.

Now look at the following equation:

4x - 3y = 4

Again, this has cannot be solved on its own. You can choose any number you like for x, and it will give a value for y.

However, if we put the equations together and try to solve them at the same time, then we can get a single, unique solution. It will give us one single value for x and a single value for y, and they will be the only ones that work in both equations at the same time. The phrase "simultaneous" means "at the same time", so we call these equations Simultaneous Equations.

How do we go about solving them? There are several ways of solving simultaneous equations. They all have advantages and disadvantages. Here are some of the methods.

Variable elimination

Subtract 3y from both sides:	2x = 20 - 3y
Divide both sides by 2:	x = 10 - 1.5y

(Note in that last step, everything was divided by 2, i.e. all the terms on both sides)

Now, we take this equation over to the other one, and everywhere we see an x in that equation, we put in (10 - 1.5y) instead. It is always wise to put in the brackets, for reasons which will become apparent later:

4x - 3y = 4

becomes

4(10 - 1.5y) - 3y = 4

Now expand the brackets and simplify it:

4(10 - 1.5y) - 3y = 4	becomes	40 - 6y - 3y = 4
		40 - 9y = 4
Add 9y to both sides:		40 = 4 + 9y
Subtract 4 from both sides:		36 = 9y
Divide both sides by 9:		4 = y

Now we have done the hard work - we have found one of the unknowns. To get the other one, take our value for y (4) and go back to any of the equations that we have encountered in this problem. Put in y = 4 and crank the handle to get a value for x:

So x has the value 4, and y has the value 4.

Equation manipulation

You will notice that the only difference between the left sides of the two equations is that one of them contains 3x and the other contains 4x. Similarly, the only difference between the right sides is that one of them is 206 and the other 256. You can probably see that x has to be 50 in order to make this work.

In this method of solving simultaneous equations, we treat the equations like large addition or subtraction sums. In this case, we would subtract, as follows:

When we subtract 3x from 4x on the left we get just x. When we subtract 7y from 7y on the left, it disappears entirely. When we subtract 206 from 256 on the right, we just get 50. Now it becomes obvious that x must be 50. We have done the hard work and can go about finding out what y is. Again, take your value for x and substitute it in one of the equations:

We say that the coefficient of y in that equation was 7. The word "coefficient" means the number in front of a letter. Here's another example, this time with a slight twist:

This time, if we subtract one equation from the other, it doesn't work. In fact, the number in front of the n gets bigger, not smaller:

The reason that the number in front of the n is -14 is that we have done -7n - (+7n), which is equivalent to -7n - 7n i.e. -14n. So how can we solve the simultaneous equations? Well, in this case, we can do it by adding the equations. This sounds silly - how can adding them get rid of a number. The fact is that the coefficients of n are of different signs - one is positive and the other negative. This means that adding the two equations will tend to cancel them out, and since one coefficient is -7 and the other is +7, they will cancel out entirely!

In this case, we have done -7n + 7n = 0n, i.e. no n term in the final answer. We can now solve to get m (you can probably see that m = 10) and then substitute back into either one of the original equations to get n.

What happens if the terms with the same coefficient are both negative, as in this example?

In this case, if we add the equations, we make the problem worse! If we add 4p to 9p, then that becomes 13p - fair enough! We were expecting that. However, if we add -2p to -2p, then what we get is -2p + (-2p), which is equivalent to -2p - 2p, i.e. -4p. Instead of disappearing, the term with p in it has become larger (albeit in a negative direction!)

The secret is to subtract the equations in this case as well. If we do -2p - (-2p), then it is equivalent to -2p + 2p (as the minus signs on both sides of the opening bracket cancel out to produce a +), and this reduces to 0 (as -2p + 2p = 0). The whole problem then becomes the following:

Again, we now substitute the value -3 into either one of the equations to get a value for q.

The general rule is: Consider the terms in the equations which have the same coefficient in both equations (i.e. the same number in front of the same letter in both equations). If these coefficients are the same sign (i.e. both positive or both negative), then subtract the equations to make these terms disappear. If the coefficients are different signs (one positive, the other negative), then add the equations to make the terms disappear."

Click on the icon for a worked example	Click on the icon for a question

Harder Equation Manipulation

Adding the equations would give

and subtracting the equations gives

Instead, we need to multiply one or both equations by numbers until the coefficients of one of the letters do match. In this case, we can multiply the first equation by 2:

We can also multiply the second equation by 3:

Note that every term was multiplied by 2 (or 3 in the second case) throughout. It may seem that we have multiplied twice as much on the left side of the equation as on the right side, but it is mathematically correct. If we do the second equation again, this time on a "blow by blow" basis:

You see, it does work! Of course, the effect is less impressive with the first equation as the right side is 0, and that will stay at 0 regardless of what you multiply it by. Now we compare the two manipulated equations:

Suddenly, we can subtract the first equation from the second equation and e will disappear:

(Note that 21g - (- 10g) gives the value 31g as we are subtracting -10, which is equivalent to adding 10). Dividing 31g = 93 by 31 gives the value of g as 3. Substituting g = 3 in the first equation gives 3e - 15 = 0, i.e. 3e = 15 or e = 5.

So the key was to spot that, although 2e and 3e didn't match, we could multiply each equation by different numbers to make them match. Alternatively, we could have multiplied the first equation by 7 and the second equation by 5:

This time, the coefficient of g becomes identical (apart from the sign) in each equation, and we can add the equations to eliminate g. We end up with 31e = 155, which gives e = 5 (as we expect), and we substitute this value back in 2e + 7g = 31 to give 10 + 7g = 31. This means that 7g = 21, or g = 3.

Either of these two courses of action is perfectly adequate, although I think the first one is better as it involves smaller numbers. Usually, one particular set of multiplications will suggest itself, but whichever way you do it, you should get the same values for both of the letters. Sometimes you will only need to multiply one equation, for instance:

In this case, if you multiply the first equation by 2, you get the following:

Now we can add the equations to eliminate w entirely.

In this final exercise, the coefficients of the letters don't match, so you will need to multiply either one or both equations by numbers before eliminating one of the letters using equation manipulation.