Today’s post is on a GCSE subject, but quite an advanced one… Solving Simultaneous equations by substitution

I say advanced; I think its not often covered, but I think it can be followed; There might just be some exciting algebra to do

Lets start with a refresher on how to solve simultaneous equations

To solve these equations

4x + y = 9

3x – y = 5

we know we need to add the equations together, to get rid of the y.

Sometimes we have to be a bit cleverer, and multiply one of the equations by something. For example

4x + y = 9

3x –2y = 4.

We have to multiply the first equation by 2 so the number in front of the y is the same. Otherwise the y won’t disappear when we add.

This is called ‘solving by addition/subtraction’

It only works sometimes. This new method words more often. It’s called ‘solving by substitution’

**Solving By Substitution **

Did you solve the first equations? You should have got x = 2 and y = 1

Now let’s change the first equation so its y = something. We can -4x from both sides. This gives us

y = 9 – 4x.

Now, this is where the ‘substitution’ comes in. We change the y in the other equation –

3x – (9-4x) = 5

We re-arrange this, remembering that – -4x = + 4x!

3x – 9 +4x = 5

7x = 14; x = 2.. and y = 1

Now that looks similar but more complicated. In fact I would recommend solving this equation the way we knew before.

So why am I showing you ‘substitution’?

To show the result is the same, and to get ready for when the other method doesn’t work.

For example

2x^{2 } + y^{2 } =11

y – x = 2

Here we have ‘squared’ in the mix, and this makes it hard to see how we can get rid of terms by adding or subtracting to two equations. This is where substitution comes into its own

Change the 2^{nd} equation to say y = x + 2 (I’ve just added the x to both sides)

Substitution y = x + 2 into the first

2x^{2} + (x + 2)^{2 }= 11

We have to a little work here, multiplying the brackets out and rearranging the terms

2x^{2 }+ x^{2} + 4x + 4 = 11

3x^{2} + 4x – 7 = 0 (I’ve combined the x^{2} terms and taken 11 from both sides)

Factorising this gives

(3x + 7 ) (x – 1)

That could be quite a hard factorisation to see straight off, though it helps that not many pairs of numbers multiply together to make -7. A little ‘trial and error’ may help.

So x = 1 or x = -7/3

Now remember y = x + 2 so

When x = 1, y = 3.. That’s quite easy to see

When x = -7/3, y = -1/3. Just add on 2;