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What’s that, you cry? Algebra was bad enough, without throwing fractions into the mix? Well, unfortunately for you, it’s on the Edexcel Unit 3 spec., and you’re going to have to learn it. To find out about algebraic fractions, simplifying before multiplying, and calculations involved, read on…

Algebraic Fractions

It’s almost surprising how easy these are to begin with, when they aren’t even that different to normal fractions – the only difference is that there are letters in the fraction as well as numbers. You can get some more complicated ones, sure, but we’ll come onto that.

Numerical denominators

These are when you have a number for a denominator, and not a variable.

Firstly, you might have a standard equation. That’s all fine and dandy. But now, your equation might look something like this:

AlgebraicFractions_1

That’s not too bad. Obviously, you have to cancel the denominator out before you do anything else – so you multiply both sides by 3. This takes us to:

AlgebraicFractions_2

From here, it’s obvious that x = 5, by subtracting 7 from both sides of the equation.

Now, that was a really simple example. Next, we’ll go to 2 fractions – on either side of the equation, and with a variable on each side of the equation. It could be something like this:

AlgebraicFractions_3

So, again, we need to cancel the denominators first. So first we multiply by 4 (of course, you could always convert the denominators to the same denominator then multiply them both by the same value, although it isn’t all that much faster) – giving us:

AlgebraicFractions_4

Then, we need to multiply both sides by 6, to cancel the remaining denominator. This gives us:

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Now this is a fairly simple equation – we can expand the bracket and simplify down to find the value for x. When we expand the brackets we get:

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Then we collect the terms on one side and the numbers on the other:

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Then we find x by dividing by 2, giving us:

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Simple enough, eh?

Algebraic denominators

So this is when you have a variable as the denominator in your fraction. First, we’ll look at what it’s like when you only have 1 fraction. You could have something like this:

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Like normal, the first thing we want to do is cancel the denominator. To do this, we multiply by (x + 1) – this is what we get:

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From here, we can expand brackets, collect terms and simplify to find x. This goes as follows:

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And that’s it for a single denominator, with a neat answer. We’ll take a quick look at two denominators with variables, then we’re going to move onto some of the harder stuff – involving quadratics.

So here’s a question with two algebraic denominators:

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We need to cancel the denominators:

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Then we get to:

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From here we expand the brackets:

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And finally we collect terms and simplify in order to find the value for x:

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Once you get your head around how the algebraic denominators work, it’s all just as easy as the other types – you just have to cancel the denominators first, then work on other parts of the equation.

Harder algebraic fractions

This is where we get into having denominators that eventually culminate in a quadratic that we have to solve. Here’s an example question:

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Like before, we now need to cancel out the denominators – here’s how we do it in this example:

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Multiply both sides by (x + 1):

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Expand the brackets:
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Then, we have to collect terms on either side:

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As you can see, we now have a quadratic that we need to solve. Of course, we can do this in one of three ways – the quadratic formula, factorising, or completing the square. However, this quadratic cannot factorise nicely, so we’re going to go for the quadratic formula. After plugging the values into the formula and simplifying, we get:

AlgebraicFractions_25

Answers (to 3 significant figures) are: x = -4.24  or  x = 8.24

And that’s one type of quadratic you might see, that’s fairly difficult. However, that’s not all – we could get some with variables all over the place..

Variables… VARIABLES EVERYWHERE

Finally, we come to some of the most difficult of this type of question – where there are variables all over the question, and things get a bit more confusing again – because it’s a pair of fractions that equal a number, or even more variables. Ready? Here’s a particularly difficult example question:

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Now, your first instinct may be to try and get rid of the denominators, but before we do that, let’s look at our quadratics and how they can factorise to see if we can make it easier. By factorising, we can get to:

 AlgebraicFractions_27

Now comes the clever bit. Because that fraction on the right has a pair of brackets in both the numerator and the denominator (on the top and bottom), we can see if any will cancel – and a couple here do just that. Here’s the working:

 AlgebraicFractions_28

This means that our equation now looks like this:

 AlgebraicFractions_29

You know where we are now? In a position to cancel the denominators. But one thing you must remember in this type of question, and any algebra involving a denominator – you have to do the same to both sides, whatever you do. That means, if you’re multiplying to remove a denominator on one side, you have to do it on the other side as well. Here it is:

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Now multiply through by (x-2) as well, in order to cancel all the denominators out:

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See what I’ve done? BOTH sides have been multiplied completely by BOTH denominators – so we get 5(x + 4)(x – 2) on the side where we had 5 originally. After this we first multiply the 5 by the (x + 4), to make our next line of working a bit easier – here’s what this gives us:

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Now we expand brackets and eventually simplify down – here are the lines of working:

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Then, we simplify:

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Now we can collect terms, and eventually get to a single quadratic to find a value for x – but first, let’s collect terms:

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Now we have a quadratic that we can use in order to finally solve for a value of x – this time, because we have a value for a (as this is a quadratic in the form ax2 + bx + c) greater than 1, so we’ll use the formula. In other circumstances, you could use the other methods of solving a quadratic, but for the purposes of this question, using the formula is probably the easiest method. This eventually gives us:

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When we round to 3 significant figures, we get:

x = 2.41   or   x = -1.75

And that’s one of the hardest styles of algebraic fraction questions you are likely to get! It’s all simple skills, just manipulating them for the purpose you need – if you do anything to one side, you have to do it to the other.

Here’s the summary of working for that last question:

Starting question:

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Working:

(FACTORISE)

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(CANCEL IF POSSIBLE)

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(REMOVE THE DENOMINATORS)

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To:

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(EXPAND BRACKETS)

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(COLLECT LIKE TERMS – X, X2, ETC.)

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(SIMPLIFY TO A SINGLE EQUATION)

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(SOLVE THE QUADRATIC – USING THE FORMULA IN THIS CASE)

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(GIVE ANSWER TO CORRECT NUMBER OF SIGNIFICANT FIGURES – IN THIS CASE, 3)

x = 2.41   or   x = -1.75    

And that’s really all there is to algebraic fractions! If you have any questions or problems with any questions, post a comment. Hope this helped!

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