An algebraic fraction is a fraction whose numerator and denominator are polynomials (or monomials). The same ideas you use for numbers — simplify, find a common denominator, and combine — apply here, with extra care: you may only cancel common factors, and you must note values of the variable that would make a denominator zero.
This chapter builds on expanding and factorising polynomials. If you have not studied those topics yet, use the callout below as a minimal toolkit.
Before you start
To simplify, factorise the numerator and denominator completely, then divide out any common factors.
Rule
\[ \frac{ac}{bc} = \frac{a}{b} \quad (b \neq 0,\; c \neq 0) \]
Here \(c\) is a common factor of the top and bottom, not a term added to each.
Example. Simplify \(\dfrac{x^2 - 1}{x^2 + x}\).
Factorise: \(x^2 - 1 = (x-1)(x+1)\) and \(x^2 + x = x(x+1)\). Cancel the common factor \((x+1)\): \[ \frac{(x-1)(x+1)}{x(x+1)} = \frac{x-1}{x}, \quad x \neq 0,\; x \neq -1. \] The restrictions matter: at \(x = -1\) the original denominator is zero.
Example. Simplify \(\dfrac{2x^2 - 8}{x^2 - 4}\).
\[ \frac{2(x^2-4)}{x^2-4} = \frac{2(x-2)(x+2)}{(x-2)(x+2)} = 2, \quad x \neq 2,\; x \neq -2. \]
Use a common denominator (the LCM of the denominators), as with numeric fractions.
Example. \(\dfrac{2}{x} + \dfrac{3}{x+1}\).
LCD is \(x(x+1)\): \[ \frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1) + 3x}{x(x+1)} = \frac{5x+2}{x(x+1)}, \quad x \neq 0,\; x \neq -1. \]
Example. \(\dfrac{1}{x-2} - \dfrac{3}{x^2 - 4}\).
Note \(x^2 - 4 = (x-2)(x+2)\). With LCD \((x-2)(x+2)\): \[ \frac{x+2}{(x-2)(x+2)} - \frac{3}{(x-2)(x+2)} = \frac{x-1}{(x-2)(x+2)}, \quad x \neq 2,\; x \neq -2. \]
Multiply numerators and denominators; factorise before cancelling. To divide, multiply by the reciprocal of the divisor.
Divide
\[ \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} \]
Example. \(\dfrac{x^2 - 9}{2x} \times \dfrac{4}{x+3}\).
\[ \frac{(x-3)(x+3)}{2x} \times \frac{4}{x+3} = \frac{4(x-3)}{2x} = \frac{2(x-3)}{x}, \quad x \neq 0,\; x \neq -3. \]
Example. \(\dfrac{x^2 - 1}{x+1} \div \dfrac{x-1}{2}\).
\[ \frac{(x-1)(x+1)}{x+1} \times \frac{2}{x-1} = 2, \quad x \neq -1,\; x \neq 1. \]
These use the same rules; the algebra is longer. Work factorisation and the LCD carefully.
Example. \(\dfrac{1}{x} + \dfrac{1}{x+1}\) (already seen) gives \(\dfrac{5x+2}{x(x+1)}\).
Example. Simplify \(\dfrac{x}{x^2 - 5x + 6}\).
Factorise \(x^2 - 5x + 6 = (x-2)(x-3)\): \[ \frac{x}{(x-2)(x-3)}, \quad x \neq 2,\; x \neq 3. \]
A formula links several quantities. Changing the subject means rewriting the formula so that one chosen letter stands alone on one side — like solving an equation for that letter.
Undo operations in reverse order to how they build the right-hand side: if the target variable is multiplied, divide; if it is inside a bracket, expand or divide through; if it appears in a denominator, multiply both sides by that denominator.
Example. Make \(t\) the subject of \(v = \dfrac{d}{t}\) (\(t \neq 0\)).
Multiply both sides by \(t\): \(vt = d\). Divide by \(v\) (assuming \(v \neq 0\)): \[ t = \frac{d}{v}. \]
Example. Make \(A\) the subject of \(P = \dfrac{F}{A}\) (\(A \neq 0\)).
Multiply by \(A\): \(PA = F\), so \(A = \dfrac{F}{P}\) (\(P \neq 0\)).
Example. Make \(R\) the subject of \(V = IR\).
Divide by \(I\) (assuming \(I \neq 0\)): \(R = \dfrac{V}{I}\).
| Formula | Meaning (typical units) |
|---|---|
| \(v = d/t\) | speed = distance ÷ time |
| \(P = F/A\) | pressure = force ÷ area |
| \(V = IR\) | voltage = current × resistance |
| \(\rho = m/V\) | density = mass ÷ volume |