Algebraic Fractions

Algebra and Sequences

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