Algebraic Fractions

Algebra and Sequences

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

Write (x² − 25) ÷ (3x + 6) × (x + 2) ÷ (x − 5) as a single fraction in its simplest form.

Solution:

The top of the first fraction is a difference of two squares, giving (x − 5)(x + 5)
The bottom of the first fraction has a common factor of 3, giving 3(x + 2)
The product is now [(x − 5)(x + 5)] ÷ [3(x + 2)] × (x + 2) ÷ (x − 5)
The bracket (x − 5) 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 + 5) ÷ 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