Algebraic Fractions
Algebraic fractions contain algebraic expressions in their numerator, denominator or both. The same fraction rules still apply, but factorising and denominator restrictions become especially important.
Higher: this lesson develops simplification, multiplication, division, addition, subtraction and equations involving algebraic fractions.
You should already be confident with simplifying expressions and factorising. Revisit Simplifying Algebraic Expressions and Factorising if needed.
Denominators and restrictions
In , the denominator is x − 3. A fraction cannot have a denominator of zero, so x ≠ 3. This is a restriction on the original expression.
Write restrictions before cancelling factors. A factor that later cancels does not make the original denominator restriction disappear.
Simplifying by factorising
You can cancel factors, not individual terms. Factorise first so common factors are visible.
Worked example
Simplify (x² − 9) / (x² + 5x + 6).
Factorise numerator and denominator: (x − 3)(x + 3) / [(x + 2)(x + 3)].
Cancel the common factor (x + 3): (x − 3) / (x + 2).
The original denominator was (x + 2)(x + 3), so the restrictions are x ≠ −2 and x ≠ −3. Both restrictions remain even though x + 3 cancels.
The cancellation trap
Cancellation works because a common factor divided by itself equals 1. It does not work across addition or subtraction.
For example, (x + 3)/(x + 5) cannot be simplified by cancelling the x terms.
But (x² + 3x)/x = x(x + 3)/x = x + 3, with restriction x ≠ 0, because x is a factor of the whole numerator.
Multiplying algebraic fractions
Multiply numerators together and denominators together, but factorising before multiplying often lets you cancel first.
Worked example
Simplify [(x² − 4)/(3x)] × [6x/(x + 2)].
Factorise x² − 4 = (x − 2)(x + 2).
The product is [(x − 2)(x + 2) × 6x] / [3x(x + 2)]. Cancel x, x + 2 and 6/3.
2(x − 2), with restrictions x ≠ 0 and x ≠ −2 from the original denominators.
Dividing algebraic fractions
To divide by a fraction, multiply by its reciprocal. Also remember that the divisor itself cannot equal zero.
Worked example
Simplify [x/(x − 1)] ÷ [2x/(x + 3)].
Multiply by the reciprocal: [x/(x − 1)] × [(x + 3)/(2x)].
Cancel x to get (x + 3) / [2(x − 1)].
Restrictions: x ≠ 1 and x ≠ −3 from denominators, and x ≠ 0 because the divisor 2x/(x + 3) must not be zero.
Adding and subtracting
Fractions can be added or subtracted only when they have a common denominator.
Same denominator
(3x)/(x + 2) + 5/(x + 2) = (3x + 5)/(x + 2), with x ≠ −2.
Different denominators
For 2/x + 3/(x + 1), use common denominator x(x + 1).
2/x = 2(x + 1)/[x(x + 1)] and 3/(x + 1) = 3x/[x(x + 1)].
Add the numerators: (5x + 2) / [x(x + 1)], with x ≠ 0 and x ≠ −1.
Watch subtraction signs
1/(x − 2) − 2/(x + 1) = [(x + 1) − 2(x − 2)] / [(x − 2)(x + 1)] = (5 − x) / [(x − 2)(x + 1)], with x ≠ 2 and x ≠ −1.
Solving equations with algebraic fractions
State restrictions first, then multiply every term by a common denominator to remove the fractions. Solve the resulting equation and check that no solution breaks a restriction.
Worked example — a linear result
Solve 2/(x + 1) = 3/x.
Restrictions: x ≠ −1 and x ≠ 0.
Multiply by x(x + 1): 2x = 3(x + 1).
2x = 3x + 3, so x = −3. This does not break either restriction.
Worked example — a quadratic result
Solve 1/x + 1/(x + 2) = 3/4.
Restrictions: x ≠ 0 and x ≠ −2.
Multiply by 4x(x + 2): 4(x + 2) + 4x = 3x(x + 2).
8x + 8 = 3x² + 6x, so 3x² − 2x − 8 = 0.
Factorise: (3x + 4)(x − 2) = 0.
x = −4/3 or x = 2. Both satisfy the original restrictions.
Higher — reject an excluded solution
Worked example — Clearing denominators can leave an invalid candidate
Solve (x² − 1)/(x − 1) = 2. The original restriction is x ≠ 1.
For allowed x, factorise and cancel: (x − 1)(x + 1)/(x − 1) = x + 1. Thus x + 1 = 2 suggests x = 1, but that makes the original denominator zero. There is no solution. Never accept a value just because it solves a transformed equation.
Higher practice — Solve (x² − 4)/(x − 2) = 4.
The restriction is x ≠ 2. For allowed x, the fraction simplifies to x + 2.
x + 2 = 4 gives x = 2, which is excluded. Therefore there is no solution.
Think Like a Mathematician
Why does a cancelled factor still create a restriction?
Simplifying changes the appearance of an expression, not the values for which the original expression existed. If the original denominator was zero at a particular value, that input was never allowed.
This is why a simplified expression can look defined at a value that the original algebraic fraction excluded.
Common Mistakes
Cancelling terms instead of factors: you cannot cancel x in (x + 3)/(x + 5). Factorise first.
Forgetting restrictions: write values that make an original denominator zero before simplifying.
Flipping the wrong fraction: when dividing, keep the first fraction and multiply by the reciprocal of the second.
Adding denominators: 1/x + 1/(x + 1) is not 2/(2x + 1). Find a common denominator.
Not checking solutions: after solving an equation, reject any value excluded by the original denominators.
Practice — algebraic fractions
All questions are Higher. Keep restrictions with your answers and show factorising before cancellation where needed.
1. State the restriction on (x + 1)/(x − 4).
The denominator is zero when x − 4 = 0.
x ≠ 4.
2. Simplify 6x2 / 3x.
6x² / 3x = 2x, cancelling a factor of 3x.
2x, with x ≠ 0.
3. Simplify (x2 − 16)/(x + 4).
x² − 16 = (x − 4)(x + 4).
Cancel x + 4 to get x − 4, with x ≠ −4.
4. Simplify (x2 + 5x + 6)/(x + 2).
Factorise the numerator: (x + 2)(x + 3).
Cancel x + 2 to get x + 3, with x ≠ −2.
5. Simplify [x/(x + 1)] × [(x + 1)/4].
Cancel the common factor x + 1.
x/4, with x ≠ −1.
6. Simplify [3x/(x − 2)] ÷ [6x/(x + 1)].
Multiply by the reciprocal: [3x/(x − 2)] × [(x + 1)/(6x)].
Cancel 3x with 6x to get (x + 1)/[2(x − 2)].
Restrictions: x ≠ 2, x ≠ −1 and x ≠ 0.
7. Simplify (2x + 1)/(x − 3) + (x − 4)/(x − 3).
The denominators already match, so add the numerators: 2x + 1 + x − 4 = 3x − 3.
3(x − 1)/(x − 3), with x ≠ 3.
8. Simplify 1/x + 2/(x + 1).
Use denominator x(x + 1): (x + 1)/[x(x + 1)] + 2x/[x(x + 1)].
(3x + 1)/[x(x + 1)], with x ≠ 0, −1.
9. Simplify 3/x − 1/(x − 2).
Use denominator x(x − 2): [3(x − 2) − x]/[x(x − 2)].
The numerator is 3x − 6 − x = 2x − 6.
2(x − 3)/[x(x − 2)], with x ≠ 0, 2.
10. Solve 5/(x + 1) = 2.
Restriction: x ≠ −1.
Multiply by x + 1: 5 = 2x + 2, so 2x = 3.
x = 3/2, which is allowed.
11. Solve 1/x + 1/(x + 2) = 3/4.
Restrictions: x ≠ 0 and x ≠ −2.
Multiply by 4x(x + 2): 4(x + 2) + 4x = 3x(x + 2).
This gives 3x² − 2x − 8 = 0 = (3x + 4)(x − 2).
x = −4/3 or x = 2. Both are allowed.
12. Simplify (x2 − 1)/(x2 + 3x + 2) and state all restrictions.
Factorise: x² − 1 = (x − 1)(x + 1), and x² + 3x + 2 = (x + 1)(x + 2).
Cancel x + 1 to get (x − 1)/(x + 2).
From the original denominator, x ≠ −1 and x ≠ −2.
Frequently Asked Questions
Can I cancel the same letter from the top and bottom?
Only when it is a factor of the whole numerator and denominator. Factorise first; do not cancel across addition or subtraction.
Why do restrictions matter after simplification?
They record where the original fraction was undefined. Cancelling a factor does not make those original inputs valid.
Do I always need to factorise?
No. For adding fractions, a common denominator may be the main step. Factorising is most useful when it reveals common factors or simpler denominators.
What should I check after solving an equation?
Check the original restrictions and substitute every remaining candidate into the original equation. Reject any value that makes an original denominator zero or fails the original equation.