For some reason, the concept of simultaneous equations strikes dread into many hearts. Perhaps it’s the rather lengthy words – or perhaps it’s because of poor teaching in schools. But I was fortunate enough to have excellent maths teachers in my secondary school in the UK, and found I really started to enjoy maths about the time we started doing simultaneous equations, around Year 9.

If you aren’t familiar with equations in general, there’s a basic introduction on my page Algebra for 6-year-olds which I’d recommend reading first. Around six or seven is a good age to introduce basic algebraic concepts in fairly concrete ways, using objects.

#### Simultaneous equations and abstract thinking

Simultaneous equations require more abstract thought, and a measure of logic, which for most children won’t be well-developed until at least age 11 or 12. Of course a child can learn anything at almost any age, so don’t be put off by that if your 8-year-old is a logic whizz and wants to know how these work. But don’t worry, either, if your 12-year-old struggles with the concept. Sooner or later, they will be ready and may even find it enjoyable so long as there’s no pressure. This is a great benefit of home education!

For the purposes of this page, you do need the basic understanding of what an equation is, and how letters of the alphabet can stand for unknown variables, as described in the above article.

Let’s say that your local stationery shop sells pencils and rulers, but you’ve forgotten how much they charge for each. And let’s also say – unlikely though it may seem – that you do have a couple of reference points, in the form of two handwritten receipts from the shop.

One of them says: 10 pencils and 5 rulers – total £2.20 The other says: 8 pencils and 10 rulers – total £2.96

As a final assumption, let’s say that you know the price of each hasn’t changed, and you bought the same type of pencil and ruler each time. (Why might you want so many rulers? I’ve no idea. But that could be the subject of a creative writing exercise, if your child doesn’t grasp the idea of simultaneous equations, or gets distracted by the story).

If you’re in the USA or any other country that doesn’t use pounds sterling, just use your own currency so long as it divides into 100 units. For pounds read dollars, or euros, or whatever… for pence read cents, etc.

The first step in discovering how much each item costs individually is to write simple algebraic equations. Let’s call the cost of a pencil x and the cost of a ruler y. Let’s also count our money in pence, to get rid of that decimal point.

So:

10x + 5y = 220

8x + 10y = 296

These two equations are simultaneously true… hence they are a pair of simultaneous equations. If you only had one of them, there could be all kinds of possibilities. But with two of them, and two unknowns (ie the cost of pencils and the cost of rulers) you can find just one answer.

HOW do we find the answer? That, of course, is the difficulty, and it takes a bit of practise to spot ways of solving each one. The easiest method is usually to do a bit of multiplying or dividing until one of the variables has the same number in front of it in each equation. A glance at the pair here shows that there’s 5y in the first, 10y in the second. So we multiply the first equation by 2, giving:

20x + 10y = 440

(If you have a hard time imagining how you can multiply an equation, think back to the original problem. 10 pencils and 5 rulers cost you �2.20. If you bought another 10 pencils and 5 rulers, you’d pay another £2.20. Hence 20 pencils and 10 rulers would cost £4.40)

The next step also sounds a bit odd, but keep thinking in terms of pencils and rulers. We have 10y in both the equations, and need that to be on its own, on one side. So we subtract – or take away – the part that involves x, on both sides of each equation. This gives:

10y = 440 – 20x

10y = 296 – 8x

(Confused? Here you’ll have to stretch your imagination slightly further as you ponder on that stationery shop. Your 8 pencils and 10 rulers of the second visit cost £2.96. So let’s say you pay the bill, and your child suddenly reminds you that you already bought 10 pencils last time, and you really don’t need any more. So you apologetically ask the shop assistant if you can return them. Fortunately they’re not too busy, and they’re friendly and want to encourage your custom, so they say that’s fine.

What happens? You return the 8 pencils to the shop (thus ‘subtracting’ 8 pencils and their cost from your side of the agreement) and the shop assistant returns to you the amount you paid for them, which we’re calling 8x here. So, the second equation now tells us that the cost of 10 rulers is £2.96 minus the cost of 8 pencils. You should be able to think through a similar scenario for the first equation.)

So now we have two equations which are simultaneously true, each of which tells us the cost of 10 rulers (10y). The cost is £2.96 – 8x and it’s also £4.40 – 20x. Thus we can make a new equation which misses out the rulers altogether:

440 – 20x = 296 – 8x

Now is the part where the storyline rather breaks down, and where it’s easy to go wrong, but think it through slowly: you’ve temporarily put the rulers aside (we’ll get back to them later) and have one equation which is just in terms of the cost of pencils. Except that they’re all negative pencils at the moment. But with equations, we can always add the same thing onto each side, and it will still be an equation. So let’s start by adding 20x onto each side. Why? Because that way there will be no x on the left, and we want all the x to be on one side. This gives:

440 = 296 – 8x + 20x

which is the same as

440 = 296 + 12x

The penultimate step is then to subtract 296 from each side, so all we have left on the right is the 12x:

440 – 296 =12x

and as 440 – 296 = 144, we can say that

144 = 12x

In other words, 12 pencils cost £1.44

The last stage is to divide both sides by 12, to find the cost of just one pencil. If you know the multiplication tables well, this may be easy but if not you can always use a calculator. 144 divided by 12 is 12, so our last equation is:

x=12

so the cost of one pencil is 12p (or 12c in other countries).

Of course we still have to find the cost of a ruler, and the way to do that is to substitute 12 for x in one of the original equations that involved both x and y. Any of them will do, but let’s take the first one: 10x + 5y = 220. Now we know that x is 12, that means that 10x is 120.

So:

120 + 5y = 220

Subtracting 120 from both sides gives

5y = 220 – 120

and as 220 – 120 = 100

5y = 100

So:

y = 20

and the cost of a ruler is 20p.

(If you’re not certain this is the correct answer, you can check with the other equation too).

This isn’t the only method to solve simultaneous equations, but it’s the easiest one for many of them, and will always work even though you may find yourself having to do complex arithmetic for some of the stages. But try it with a few simple ones of this sort, taking everyday scenarios, and it can even become a game you play with your children.

Note that if you have three unknowns, you need three equations, and so on. It’s best to stick with two unknowns and two equations to start with!

More articles about teaching maths at home:

Prime numbers and factors

Multiplying on fingers

Maths and the home educated teen