E2.3 Algebraic fractions
- Syllabus
- 0580–2028–2029
- Topic
- E2.3
- Level
- Extended
Algebraic fractions follow the same operation rules as numerical fractions. Denominators must be non-zero, and the final expression should be simplified without changing its value.
| Operation | Reliable move |
|---|---|
| add or subtract | use a lowest common denominator, rewrite every numerator, then combine |
| multiply | factor first, multiply numerators and denominators, then cancel common factors |
| divide | multiply by the reciprocal of the second fraction, then simplify |
rac{2}{x-1}+rac{3}{x+2}=rac{2(x+2)+3(x-1)}{(x-1)(x+2)}=rac{5x+1}{(x-1)(x+2)}
The common denominator must contain every required factor. In the example, $x
e1,-2$ because those values make an original denominator zero.
For multiplication, rac{4a}{5} imesrac{15}{8a}=rac32 for $a
e0.Fordivision,rac{3p}{7}\divrac{9p}{14q}=rac{3p}{7} imesrac{14q}{9p}=rac{2q}{3},withp
e0andq
e0$.
Cancel only common factors in a product. Terms joined by + or − cannot be cancelled: in racx+3x, the x is not a factor of the whole numerator.
A rational expression simplifies when its numerator and denominator are written as products and a factor common to both is cancelled. The cancelled factor must be non-zero.
Factorise the numerator fully; factorise the denominator fully; identify identical factors; cancel only those factors; state every value excluded by the original denominator; expand the remaining factors only if a different final form is required.
rac{2x^2-5x-12}{3x^2-12x}=rac{(2x+3)(x-4)}{3x(x-4)}=rac{2x+3}{3x}
The original denominator is 3x(x−4), so $x
e0,4.Although(x-4)disappearsfromthesimplifiedexpression,x=4$ is still excluded because it made the original expression undefined.
Check by multiplying the simplified numerator and denominator by the cancelled factor: rac{2x+3}{3x} imesrac{x-4}{x-4} reconstructs the factorised original expression whenever $x
e4$.
Cancellation removes factors, not matching-looking terms. For example, racx+5x cannot be reduced, while racx(x+5)x=x+5 is valid only for $x
e0$.