Algebraic Fractions

Mathematics revision notes · Cambridge (CIE) · 0580

1. Simplifying a single fraction

An algebraic fraction has an expression on the top, the bottom, or both
Every rule that works for ordinary fractions works here, because the letters simply stand for numbers
Simplifying means cancelling, and cancelling is only ever allowed between factors
So the first move is always to factorise the top and the bottom as far as they will go
A factor that appears above and below then divides out, whether it is a single term or a whole bracket
Marks are given for each correct factorisation, so a part-finished attempt still scores

A term that is added on is not a factor, so it cannot be cancelled
- In (4x + 8) ÷ (4x − 1) the 4x cannot be removed, because the top is 4(x + 2) and the bottom has no factor of 4
Test any cancelling by asking whether the thing being removed multiplies everything above and everything below
Where neither part factorises and no bracket is shared, there is nothing to cancel and the fraction stands as it is
When one of the two expressions is hard to factorise, factorise the easier one first: the same bracket almost always turns up in the other

Worked example

Factorising both parts before cancelling

Simplify (2x² − 14x) ÷ (x² − 9x + 14).

Solution:

Nothing can be cancelled while the top and bottom are sums, so factorise both first
The top has a common factor in each term: 2x² and 14x both contain 2x
Top = 2x(x − 7)
The bottom is a quadratic with no number in front of x², so find two numbers multiplying to +14 and adding to −9
Both must be negative to give a positive product and a negative sum: −7 and −2
Bottom = (x − 7)(x − 2)
The fraction now reads 2x(x − 7) ÷ [(x − 7)(x − 2)]
(x − 7) is a factor of both the top and the bottom, so it divides out
The answer is 2x ÷ (x − 2)
Only whole brackets cancel: the x in (x − 2) cannot be cancelled against the 2x on top, because it is a term rather than a factor
Each correct factorisation scores its own mark, so both lines are worth writing even if the cancelling is missed

Factorise every top and every bottom first, because that is what exposes the cancelling
Cancel across the multiplication sign as well as within a single fraction, since everything is being multiplied
Then multiply the tops together and the bottoms together
Leave the answer in factorised form and check once more for anything that still cancels

Worked example

Cancelling across a product

Express (x² − 36) ÷ (3x + 6) × (x + 2) ÷ (x − 6) as a single fraction in its simplest form.

Solution:

The top of the first fraction is a difference of two squares, giving (x − 6)(x + 6)
The bottom of the first fraction has a common factor of 3, giving 3(x + 2)
The product is now [(x − 6)(x + 6)] ÷ [3(x + 2)] × (x + 2) ÷ (x − 6)
The bracket (x − 6) sits above and below, so it divides out
The bracket (x + 2) also sits above and below, so it divides out too
What survives is (x + 6) ÷ 3

Turn the second fraction upside down and change the division into a multiplication
Only the fraction after the division sign is inverted
Once inverted, carry on exactly as for a product
So 5b ÷ 12 divided by 15b² ÷ 8 becomes 5b ÷ 12 × 8 ÷ 15b², which is 40b ÷ 180b², and that reduces to 2 ÷ 9b

Two fractions can only be combined once they share a denominator
Multiplying the two denominators together always produces a common denominator, but not always the lowest one
- For denominators 4x and 6x the lowest is 12x, not 24x²
- Where one denominator already contains the other, such as (x + 7) and (x + 7)(x − 3), the longer one is the lowest
Factorising each denominator first is what reveals the shared brackets
Writing the two fractions over that denominator earns a mark by itself, before any expanding happens

Whatever multiplies a denominator must multiply that fraction's top as well
Add or subtract the tops over the single denominator, then expand and collect
A subtraction applies to the whole of the second top, so bracket it before expanding
Finally check whether the new top factorises and shares a bracket with the bottom

Worked example

Two linear denominators

Write 5 ÷ (x + 3) + 2 ÷ (x − 4) as a single fraction in its simplest form.

Solution:

The two denominators share nothing, so the lowest they have in common is (x + 3)(x − 4)
The first fraction is multiplied above and below by (x − 4), giving 5(x − 4)
The second is multiplied above and below by (x + 3), giving 2(x + 3)
Over that denominator the top reads 5(x − 4) + 2(x + 3)
Expand: 5x − 20 + 2x + 6
Collect: 7x − 14, which factorises to 7(x − 2)
Neither bracket below is (x − 2), so nothing cancels
The answer is 7(x − 2) ÷ [(x + 3)(x − 4)]

Worked example

A subtraction with number denominators

Write 2x ÷ 5 − (x − 3) ÷ 4 as a single fraction in its simplest form.

Solution:

The lowest denominator 5 and 4 have in common is 20
Multiplying above and below by 4 turns the first fraction into 8x ÷ 20
Multiplying above and below by 5 turns the second into 5(x − 3) ÷ 20
Over 20 the top reads 8x − 5(x − 3), and the bracket keeps the subtraction acting on both terms
Expand: 8x − 5x + 15
Collect: 3x + 15, which factorises to 3(x + 5)
The 20 shares no factor with 3 or with (x + 5)
The answer is 3(x + 5) ÷ 20

The syllabus lists these under equations rather than under algebraic fractions, but the method rests on everything above
Two routes reach the same place: combine the fractions into one and then cross-multiply, or clear the denominators one at a time
Clearing them is usually faster, provided each denominator multiplies every term on both sides, including any term that was never a fraction
Bracket a denominator as you multiply by it, since it is a whole expression rather than a single number
Removing two different algebraic denominators normally leaves a quadratic, so gather everything on one side and solve it as usual
A value that would make a denominator zero cannot be a solution, so discard it if it appears

Worked example

Two algebraic denominators

Solve 4 ÷ (x + 2) + 3 ÷ (x − 3) = 1.

Solution:

Multiplying throughout by (x + 2) clears the first fraction: 4 + 3(x + 2) ÷ (x − 3) = (x + 2)
Multiplying throughout by (x − 3) clears what is left: 4(x − 3) + 3(x + 2) = (x + 2)(x − 3)
Expand the left: 4x − 12 + 3x + 6, which collects to 7x − 6
Expand the right: x² − x − 6
Bring everything to one side: 0 = x² − 8x
There is no constant term, so take out x: 0 = x(x − 8)
The solutions are x = 0 or x = 8
Check x = 0: 4 ÷ 2 + 3 ÷ (−3) = 2 − 1 = 1
Check x = 8: 4 ÷ 10 + 3 ÷ 5 = 0.4 + 0.6 = 1