# Simultaneous Equations

Why do we have simultaneous equations? Well, there are two ways of looking at it.

1. The first is that it solves a problem that seems insoluble: how do you work out two variables at once? For example, if x + y = 10, what are x and y? That’s an impossible question because x and y could literally be anything. If x was 2, then y would be 8, but if x was 100, then y would be -90, but if x was 0.5, then y would be 9.5 and so on.Simultaneous equations help us solve that problem by providing more data. Yes, we still can’t solve each equation individually, but having both of them allows us to solve for one variable and then the other.
2. The second way of looking at simultaneous equations is to imagine that they describe two lines that meet. The x and y values are obviously different as you move along both lines, but they are identical at the point where they meet, and that is the answer to the question.

The next question is obviously ‘How do we solve simultaneous equations?’ The answer is simple in theory: you just have to add both equations together to eliminate one of the variables, at which point you can work out the second one and then put it back into one of the original equations to work out the first variable.

However, it gets more and more complicated as the numbers get less and less ‘convenient’, so let’s take three examples to illustrate the three different techniques you need to know.

The first step in solving simultaneous equations is to try and eliminate one of the variables by adding or subtracting them, but you can only do that if the number of the variable is the same in both. In theory, you could choose the first or the second term, but I find the one in the middle is the easiest, eg

4x + 2y = 10

16x – 2y = 10

Here, the number of the variables in the middle of the equations is the same, so adding them together will make them disappear:

20x = 20

It’s then simple to divide both sides by 20 to work out x:

x = 1

Once you have one variable, you can simply plug it back into one of the original equations to work out the other one, eg

4x + 2y = 10

4 x 1 + 2y = 10

4 + 2y = 10

2y = 6

y = 3

Answer: x = 1, y = 3

## Multiplying One Equation

If the number of variables in the middle is not the same, but one is a factor of the other, try multiplying one equation by whatever number is needed to make the number of the variables match, eg

4x + 2y = 10

7x + y = 10

Multiplying the second equation by 2 means the number of the y’s is the same:

4x + 2y = 10

14x + 2y = 20

The rest of the procedure is exactly the same, only this time we have to subtract rather than add the equations to begin with:

10x = 10

x = 1

The next part is exactly the same as the first example as we simply plug in x to find y:

4x + 2y = 10

4 x 1 + 2y = 10

4 + 2y = 10

2y = 6

y = 3

Answer: x = 1, y = 3

## Multiplying Both Equations

If the number of variables in the middle is not the same, but neither is a factor of the other, find the lowest common multiple and multiply the two equations by whatever numbers are needed to reach it, eg

4x + 2y = 10

x + 3y = 10

The lowest common multiple of 2 and 3 is 6, which means we need to multiply the first equation by 3:

12x + 6y = 30

…and the second by 2:

2x + 6y = 20

As the number of variables in the middle is now the same, we can carry on as before by subtracting one from the other in order to find x:

10x = 10

x = 1

Again, the final part of the technique is exactly the same as we plug x into the first of the original equations:

4x + 2y = 10

4 x 1 + 2y = 10

4 + 2y = 10

2y = 6

y = 3

Answer: x = 1, y = 3

## Practice Questions

Job done! Now, here are a few practice questions to help you learn the rules. Find x and y in the following pairs of simultaneous equations:

1. 2x + 4y = 16
4x – 4y = 8
2. 3x + 2y = 12
5x + 2y = 16
3. 12x – 4y = 28
3x – 2y = 5
4. 2x – y = 12
3x – 2y = 17
5. 4x + 3y = 24
5x – 2y = 7
6. 4x + 3y = 31
5x + 4y = 40
7. x + 4y = 23
5x – 2y = 5
8. 4x + 3y = 37
2x – 3y = -13
9. 2x + 4y = 16
3x – 5y = -9
10. 2x + 4y = 20
3x + 3y = 21

# Algebra

Nothing makes the heart of a reluctant mathematician sink like an algebra question.

Algebra is supposed to make life easier. By learning a formula or an equation, you can solve any similar type of problem whatever the numbers involved. However, an awful lot of students find it difficult, because letters just don’t seem to ‘mean’ as much as numbers. Here, we’ll try to make life a bit easier…

## Gathering Terms

X’s and y’s look a bit meaningless, but that’s the point. They can stand for anything. The simplest form of question you’ll have to answer is one that involves gathering your terms. That just means counting how many variables or unknowns you have (like x and y). I like to think of them as pieces of fruit, so an expression like…

2x + 3y – x + y

…just means ‘take away one apple from two apples and add one banana to three more bananas’. That leaves you with one apple and four bananas, or x + 4y.

If it helps, you can arrange the expression with the first kind of variables (in alphabetical order) on the left and the second kind on the right like this:

2x – x + 3y + y

x + 4y

Just make sure you bring the operators with the variables that come after them so that you keep exactly the same operators, eg two plus signs and a minus sign in this case.

Here are a few practice questions:

1. 3x + 4y – 2x + y
2. 2m + 3n – m + 3n
3. p + 2q + 3p – 3q
4. 2a – 4b + a + 4b
5. x + y – 2x + 2y

## Multiplying out Brackets

This is one of the commonest types of question. All you need to do is write down the same expression without the brackets. To take a simple example:

2(x + 3)

In this case, all you need to do is multiply everything inside the brackets by the number outside, which is 2, but what do we do about the ‘+’ sign? We could just multiply 2 by x, write down ‘+’ and then multiply 2 by 3:

2x + 6

However, that gets us into trouble if we have to subtract one expression in brackets from another (see below for explanation) – so it’s better to think of the ‘+’ sign as belonging to the 3. In other words, you multiply 2 by x and then 2 by +3. Once you’ve done that, you just convert the ‘+’ sign back to an operator. It gives exactly the same result, but it will work ALL the time rather than just with simple sums!

Here are a few practice questions:

1. 2(a + 5)
2. 3(y + 2)
3. 6(3 + b)
4. 3(a – 3)
5. 4(3 – p)

## Solving for x

Another common type of question involves finding out what x stands for (or y or z or any other letter). The easiest way to look at this kind of equation is using fruit again. In the old days, scales in a grocery shop sometimes had a bowl on one side and a place to put weights on the other.

To weigh fruit, you just needed to make sure that the weights and the fruit balanced and then add up all the weights. The point is that every equation always has to balance – the very word ‘equation’ comes from ‘equal’ – so you have to make sure that anything you do to one side you also have to do to the other. Just remember the magic words: BOTH SIDES!

There are three main types of operation you need to do in the following order:

1. Multiplying out any brackets
2. Adding or subtracting from BOTH SIDES
3. Multiplying or dividing BOTH SIDES

Once you’ve multiplied out any brackets (see above), what you want to do is to simplify the equation by removing one expression at a time until you end up with something that says x = The Answer. It’s easier to start with adding and subtracting and then multiply or divide afterwards (followed by any square roots). To take the same example as before:

2(x + 3) = 8

Multiplying out the brackets gives us:

2x + 6 = 8

Subtracting 6 from BOTH SIDES gives us:

2x = 2

Dividing BOTH SIDES by 2 gives us the final answer:

x = 1

Simple!

Here are a few practice questions:

1. b + 5 = 9
2. 3y = 9
3. 6(4 + c) = 36
4. 3(a – 2) = 24
5. 4(3 – p) = -8

## Multiplying Two Expressions in Brackets (‘FOIL’ Method)

When you have to multiply something in brackets by something else in brackets, you should use what’s called the ‘FOIL’ method. FOIL is an acronym that stands for:

First
Outside
Inside
Last

This is simply a good way to remember the order in which to multiply the terms, so we start with the first terms in each bracket, then move on to the outside terms in the whole expression, then the terms in the middle and finally the last terms in each bracket.

Just make sure that you use the same trick we saw earlier, combining the operators with the numbers and letters before multiplying them together. For example:

(a + 1)(a + 2)

First we multiply the first terms in each bracket:

a x a

…then the outside terms:

a x +2

…then the inside terms:

+1 x a

…and finally the last terms in each bracket:

+1 x +2

Put it all together and simplify:

(a + 1)(a + 2)

= a² + 2a + a + 2

=a² + 3a + 2

Here are a few practice questions:

1. (a + 1)(b + 2)
2. (a – 1)(a + 2)
3. (b + 1)(a – 2)
4. (p – 1)(q + 2)
5. (y + 1)(y – 3)

## Factorising Quadratics (‘Product and Sum’ Method)

This is just the opposite of multiplying two expressions in brackets. Normally, factorisation involves finding the Highest Common Factor (or HCF) and putting that outside a set of brackets containing the rest of the terms, but some expressions can’t be solved that way, eg a² + 3a + 2 (from the previous example).

There is no combination of numbers and/or letters that goes evenly into a², 3a and 2, so we have to factorise using two sets of brackets. To do this, we use the ‘product and sum’ method.

This simply means that we need to find a pair of numbers whose product equals the last number and whose sum equals the multiple of a. In this case, it’s 1 and 2 as +1 x +2 = +2 and +1 + +2 = +3.

The first term in each bracket is just going to be a as a x a = a². Hence, factorising a² + 3a + 2 gives (a + 1)(a + 2). You can check it by using the FOIL method (see above) to multiply out the brackets:

(a + 1)(a + 2)

= a² + 2a + a + 2

=a² + 3a + 2

## Subtracting One Expression from Another*

Here’s the reason why we don’t just write down operators as we come across them. Here’s a simple expression we need to simplify:

20 – 4(x – 3) = 16

If we use the ‘wrong’ method, then we get the following answer:

20 – 4(x – 3) = 16

20 – 4x – 12 = 16

8 – 4x = 16

4x = -8

x = -2

Now, if we plug our answer for x back into the original equation, it doesn’t balance:

20 – 4(-2 – 3) = 16

20 – 4 x -5 = 16

20 – -20 = 16

40 = 16!!

That’s why we have to use the other method, treating the operator as a negative or positive sign to be added to the number before we multiply it by whatever’s outside the brackets:

20 – 4(x – 3) = 16

20 – 4x + 12 = 16

32 – 4x = 16

4x = 16

x = 4

That makes much more sense, as we can see:

20 – 4(4 – 3) = 16

20 – 4 x 1 = 16

20 – 4 = 16

16 = 16

Thank Goodness for that!

Here are a few practice questions:

1. 30 – 3(p – 1) = 0
2. 20 – 3(a – 3) = 5
3. 12 – 4(x – 2) = 4
4. 24 – 6(x – 3) = 6
5. 0 – 6(x – 2) = -12

## Other Tips to Remember

• If you have just one variable, leave out the number 1, eg 1x is just written as x.
• When you have to multiply a number by a letter, leave out the times sign, eg 2 x p is written as 2p.
• The squared symbol only relates to the number or letter immediately before it, eg 3m² means 3 x m x m, NOT (3 x m) x (3 x m).