Algebraic fractions & formulas

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.

Assumed knowledge recap

This chapter builds on expanding and factorising polynomials. If you have not studied those topics yet, use the callout below as a minimal toolkit.

Simplifying algebraic fractions

To simplify, factorise the numerator and denominator completely, then divide out any common factors.

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 \((x+1)\): \[ \begin{aligned} \frac{(x-1)(x+1)}{x(x+1)} &= \frac{x-1}{x}, \\ x &\neq 0,\; x \neq -1 \end{aligned} \] The restrictions matter: at \(x = -1\) the original denominator is zero.

Example. Simplify \(\dfrac{2x^2 - 8}{x^2 - 4}\).

\[ \begin{aligned} \frac{2(x^2-4)}{x^2-4} &= \frac{2(x-2)(x+2)}{(x-2)(x+2)} \\ &= 2, \\ x &\neq 2,\; x \neq -2 \end{aligned} \]

Adding and subtracting

Use a common denominator (the LCM of the denominators), as with numeric fractions.

Example. \(\dfrac{2}{x} + \dfrac{3}{x+1}\).

LCD is \(x(x+1)\): \[ \begin{aligned} \frac{2}{x} + \frac{3}{x+1} &= \frac{2(x+1) + 3x}{x(x+1)} \\ &= \frac{5x+2}{x(x+1)}, \\ x &\neq 0,\; x \neq -1 \end{aligned} \]

Example. \(\dfrac{1}{x-2} - \dfrac{3}{x^2 - 4}\).

Note \(x^2 - 4 = (x-2)(x+2)\). With LCD \((x-2)(x+2)\): \[ \begin{aligned} \frac{x+2}{(x-2)(x+2)} - \frac{3}{(x-2)(x+2)} &= \frac{x-1}{(x-2)(x+2)}, \\ x &\neq 2,\; x \neq -2 \end{aligned} \]

Multiplying and dividing

Multiply numerators and denominators; factorise before cancelling. To divide, multiply by the reciprocal of the divisor.

Example. \(\dfrac{x^2 - 9}{2x} \times \dfrac{4}{x+3}\).

\[ \begin{aligned} \frac{(x-3)(x+3)}{2x} \times \frac{4}{x+3} &= \frac{4(x-3)}{2x} \\ &= \frac{2(x-3)}{x}, \\ x &\neq 0,\; x \neq -3 \end{aligned} \]

Example. \(\dfrac{x^2 - 1}{x+1} \div \dfrac{x-1}{2}\).

\[ \begin{aligned} \frac{(x-1)(x+1)}{x+1} \times \frac{2}{x-1} &= 2, \\ x &\neq -1,\; x \neq 1 \end{aligned} \]

Harder examples

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 \(x^2 - 5x + 6 = (x-2)(x-3)\): \[ \frac{x}{(x-2)(x-3)}, \quad x \neq 2,\; x \neq 3. \]

Changing the subject of a formula

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 \(I \neq 0\)): \(R = \dfrac{V}{I}\).

History

For centuries, algebra was written in words. In the 16th and 17th centuries, mathematicians such as François Viète began to use letters systematically for both unknowns and known parameters. That made formulas — general relationships such as \(v = d/t\) — easy to rearrange without repeating long verbal rules.

Rational expressions (fractions built from polynomials) appear whenever you divide one polynomial by another. Scientists from Galileo onward used ratios and proportionalities in motion; later, Newton’s laws and electrical laws (e.g. \(V = IR\)) relied on changing the subject of a formula to solve for the quantity you can measure.

Today, algebraic fractions show up in simplifying models, in calculus limits, and whenever a relationship is written as one quantity divided by another.

Derivation

Why can we cancel a common factor?

If \(c \neq 0\), then \(\dfrac{ac}{bc} = \dfrac{a}{b}\) because multiplying numerator and denominator by the same non-zero number does not change the value of a numeric fraction; the same idea holds when \(c\) is a polynomial factor shared by top and bottom.

You cannot cancel in \(\dfrac{a+c}{b+c}\) in general — that would mean, for example, \(\dfrac{1+2}{3+2} = \dfrac{1}{3}\), which is false. Only factors, not terms, may be cancelled.

Why reverse the order of operations?

A formula like \(v = d/t\) builds \(v\) by first dividing \(d\) by \(t\). To recover \(t\), undo division by multiplying: \(vt = d\), then undo multiplication by \(v\) (when \(v \neq 0\)) to get \(t = d/v\). Each step reverses one operation that was applied to the subject you want.

Checkpoints

Pause and reason before continuing. Discuss with a partner or write your work in a notebook.

Applications

Uniform motion

For constant speed, distance \(d\), speed \(v\), and time \(t\) are related by \(v = d/t\). If a train travels \(240\) km in \(3\) h, then:

\[ \begin{aligned} v &= \frac{240}{3} \\ &= 80\ \text{km/h} \end{aligned} \]

If you know \(v\) and \(t\) but need distance, rearrange to \(d = vt\).

Density

Density \(\rho\) (rho) is mass per unit volume: \(\rho = m/V\). A block of mass \(2.7\) kg with volume \(0.001\) m³ has:

\[ \begin{aligned} \rho &= \frac{2.7}{0.001} \\ &= 2700\ \text{kg/m}^3 \end{aligned} \]

To find mass from \(\rho\) and \(V\), use \(m = \rho V\).

Pressure and electric circuits

Pressure \(P = F/A\) links force \(F\) and area \(A\). Hydraulic and weather models often solve for \(F = PA\). In a resistor, \(V = IR\) gives voltage from current and resistance; rearrange to \(I = V/R\) when you know voltage and resistance.

Question bank

Work each question, then open the solution to check your reasoning. Difficulty increases through the sets.

Easy

Simplify \(\dfrac{6x}{3x}\).

Answer: \(2\), with \(x \neq 0\).

Steps: Cancel the common factor \(3x\).

Simplify \(\dfrac{x^2}{x}\) for \(x \neq 0\).

Answer: \(x\).

Explanation: Cancel one factor \(x\) from numerator and denominator.

Make \(d\) the subject of \(v = d/t\).

Answer: \(d = vt\) (\(t \neq 0\) when rearranging from the original).

Steps: Multiply both sides by \(t\).

Make \(R\) the subject of \(V = IR\).

Answer: \(R = V/I\) (\(I \neq 0\)).

Steps: Divide both sides by \(I\).

Intermediate

Simplify \(\dfrac{x^2 - 9}{x + 3}\).

Answer: \(x - 3\), with \(x \neq -3\).

Steps: \(x^2 - 9 = (x-3)(x+3)\); cancel \((x+3)\).

\(\dfrac{2}{x} + \dfrac{1}{x} = \) ______

Answer: \(\dfrac{3}{x}\), \(x \neq 0\).

Explanation: Same denominator already.

Simplify \(\dfrac{x^2 - 1}{x^2 + x}\).

Answer: \(\dfrac{x-1}{x}\), \(x \neq 0,\; x \neq -1\).

Steps: Factorise; cancel \((x+1)\).

Make \(m\) the subject of \(\rho = m/V\).

Answer: \(m = \rho V\) (\(V \neq 0\) in the original).

Steps: Multiply both sides by \(V\).

Difficult

\(\dfrac{1}{x} + \dfrac{2}{x+1} = \) ______

Answer: \(\dfrac{3x+1}{x(x+1)}\), \(x \neq 0,\; x \neq -1\).

Steps: LCD \(x(x+1)\); numerator \( (x+1) + 2x = 3x+1\).

\(\dfrac{x}{2} \times \dfrac{4}{x^2} = \) ______

Answer: \(\dfrac{2}{x}\), \(x \neq 0\).

Steps: Multiply; cancel a factor \(x\).

Simplify \(\dfrac{1}{x-2} - \dfrac{3}{x^2 - 4}\).

Answer: \(\dfrac{x-1}{(x-2)(x+2)}\), \(x \neq 2,\; x \neq -2\).

Steps: Write \(x^2-4 = (x-2)(x+2)\); common denominator \((x-2)(x+2)\).

Hardcore

\(\dfrac{x^2 - 1}{x+1} \div \dfrac{x-1}{2} = \) ______

Answer: \(2\), with \(x \neq -1,\; x \neq 1\).

Steps: Factorise \(x^2-1\); multiply by reciprocal; cancel \((x-1)\) and \((x+1)\).

Make \(t\) the subject of \(v = \dfrac{d}{t}\) and then find \(t\) when \(d = 150\) km and \(v = 75\) km/h.

Answer:

\[ \begin{aligned} t &= \frac{d}{v} \\ &= \frac{150}{75} \\ &= 2\ \text{h} \end{aligned} \]

Steps: \(vt = d \Rightarrow t = d/v\); substitute values.

Simplify fully: \(\dfrac{2x^2 - 8}{x^2 - 4}\).

Answer: \(2\), \(x \neq 2,\; x \neq -2\).

Steps: Factorise numerator \(2(x^2-4)\) and denominator \((x-2)(x+2)\); cancel \((x-2)(x+2)\).

A formula is \(P = \dfrac{F}{A}\). The force is \(120\) N and the area is \(0.05\) m². Find \(P\), then make \(F\) the subject and verify your force value.

Answer:

\[ \begin{aligned} P &= \frac{120}{0.05} \\ &= 2400\ \text{Pa} \end{aligned} \]

\[ \begin{aligned} F &= PA \\ &= 2400 \times 0.05 \\ &= 120\ \text{N} \end{aligned} \]

Explanation: Rearranging checks that the formula is consistent.

References

• Viète, In artem analyticem isagoge (1591) — early systematic use of letters in algebra.
• HKDSE Compulsory Part — Number and Algebra (algebraic fractions, formulas and identities).