Sec 2 Algebraic Fractions & Formulae: Worked Examples

Secondary 2 G3 Mathematics · Study guide

Use factorisation to handle algebraic fractions, retain excluded inputs and rearrange formulae by collecting and isolating the required letter.

Concept → worked examples → practice13 min read · practice linked below

Fractions still follow the same rules

An algebraic fraction is a fraction whose numerator or denominator contains a letter. The familiar rules for number fractions still apply. Factorisation helps you see the whole factors that may cancel, and the factors that a common denominator needs.

Before you begin: be comfortable with numerical fractions, expanding brackets, factorising quadratics and solving equations. A denominator must never be zero. Record excluded values before simplifying, because cancellation can hide them.

Cancel factors, not terms: $\frac{3(x+2)}{3(x-1)}=\frac{x+2}{x-1}$ , for $x\ne1$ , because the complete numerator and denominator share a factor of 3. You cannot cancel the $x$ terms in $\frac{x+2}{x-1}$ : they are parts of sums, not factors of the whole expressions.

Simplify, multiply and divide using factors

To multiply, multiply the numerators and multiply the denominators. Factorise first to make cancellation easier. To divide by a fraction, multiply by its reciprocal; the fraction being inverted must be defined and non-zero.

Worked example 1 — keep the original restrictions

(a) Simplify $\dfrac{x^2-9}{x^2+x-6}$ . (b) Hence simplify $\dfrac{x^2-9}{x^2+x-6}\div\dfrac{x-3}{x+1}$ , stating every excluded value.

For (a), factorise the numerator and denominator: $\frac{x^2-9}{x^2+x-6}=\frac{(x-3)(x+3)}{(x+3)(x-2)}=\frac{x-3}{x-2},\qquad x\ne-3,2.$ The cancelled factor $x+3$ is non-zero only when $x\ne-3$ . That restriction remains even though the simplified fraction has a value at $x=-3$ .

The arrow divides the entire numerator and denominator by the same non-zero factor. It does not restore an input that was excluded in the original fraction.

For (b), the divisor also requires $x\ne-1$ , and it must not be zero, so $x\ne3$ . Multiply by its reciprocal:

\begin{aligned}\frac{x^2-9}{x^2+x-6}\div\frac{x-3}{x+1}&=\frac{(x-3)(x+3)}{(x+3)(x-2)}\times\frac{x+1}{x-3}\\&=\frac{x+1}{x-2},\qquad x\ne-3,-1,2,3.\end{aligned}

Check: at $x=0$ , the original calculation is $\frac{-9}{-6}\div\frac{-3}{1}=-\frac12$ . The final expression gives $\frac1{-2}=-\frac12$ . The permitted values must match as well as the values of the expressions.

Add and subtract with a common denominator

Factorise the denominators, then include each required factor the greatest number of times it occurs. Multiply each numerator by its missing factors. Put brackets around a numerator that is being subtracted.

Worked example 2 — a reversed bracket changes the sign

Write $\dfrac3{x^2-4}+\dfrac1{2-x}$ as one fraction in its simplest form.

The original denominators give $x\ne2,-2$ . Since $x^2-4=(x-2)(x+2)$ and $2-x=-(x-2)$ , rewrite the second fraction with a minus sign:

\begin{aligned}\frac3{x^2-4}+\frac1{2-x}&=\frac3{(x-2)(x+2)}-\frac1{x-2}\\&=\frac{3-(x+2)}{(x-2)(x+2)}\\&=\frac{1-x}{(x-2)(x+2)},\qquad x\ne-2,2.\end{aligned}

The second denominator needs one factor of $x+2$ , so its numerator also receives $x+2$ . The numerator becomes $3-x-2$ , not $3-x+2$ .

Check: at $x=0$ , the original sum is $-\frac34+\frac12=-\frac14$ . The answer gives $1/[(-2)(2)]=-\frac14$ . There is no common factor left to cancel.

Fractional equations: clear denominators without losing restrictions

An expression is simplified; an equation is solved. In an equation, multiply every term on both sides by a common denominator. This is reversible only where that denominator is non-zero. Check candidate solutions in the original equation.

Worked example 3 — a fractional equation becomes quadratic

Solve $\dfrac6{x-1}=x$ .

First record $x\ne1$ . For every permitted value, multiply both sides by $x-1$ :

\begin{aligned}6&=x(x-1)\\0&=x^2-x-6\\0&=(x-3)(x+2).\end{aligned}

A product is zero when at least one factor is zero. Thus $x=3$ or $x=-2$ . Neither is excluded.

Check both: when $x=3$ , $6/(3-1)=3$ ; when $x=-2$ , $6/(-2-1)=-2$ . Both satisfy the original equation, so both are solutions.

Do not reject a negative solution merely because it is negative. A stated context, such as a length or speed, can impose an additional positive-value restriction; this equation has no such context.

Make a different letter the subject

The subject is the letter standing alone on one side of a formula. Undo operations in reverse order. If the required letter appears twice, gather its terms and factorise it out before dividing.

Worked example 4 — the new subject appears twice

Given $y=\dfrac{2x+3}{x-1}$ , express $x$ in terms of $y$ , then find $x$ when $y=7$ .

The original formula requires $x\ne1$ . Clear its denominator, collect the terms containing $x$ and factorise:

\begin{aligned}y(x-1)&=2x+3\\xy-y&=2x+3\\xy-2x&=y+3\\x(y-2)&=y+3\\x&=\frac{y+3}{y-2},\qquad y\ne2.\end{aligned}

Why exclude $y=2$ ? Substituting it into $x(y-2)=y+3$ would give $0=5$ , so it cannot occur in the original formula. For permitted $y$ , the new expression never gives $x=1$ , since $y+3=y-2$ is impossible.

When $y=7$ , $x=(7+3)/(7-2)=2$ . Substitute into the original formula: $(2(2)+3)/(2-1)=7$ , as required.

When the subject is squared: from $v^2=u^2+2as$ , subtract $2as$ first to get $u^2=v^2-2as$ . Thus $u=\pm\sqrt{v^2-2as}$ , with $v^2-2as\ge0$ for real values. If $u$ is specified as a positive speed, choose the positive root. A length or time should not receive a negative answer.

Substitute signed values with brackets, keep full calculator precision during the calculation, and round only the final answer to the requested precision. Check that the units on both sides of a physical formula agree.

Explore why cancellation does not fill a missing value

Same values, with one excluded input

Compare $\dfrac{x^2-9}{x+3}$ with $x-3$ . For $x\ne-3$ , factorisation cancels $x+3$ and the values agree. Move the input to $-3$ to see why the original fraction still has no value there.

The diagram shows the original fraction at $x=1$ , where its value is $-2$ . The hollow point marks the excluded input $x=-3$ .

The hollow point belongs to the full straight line

y=x-3

, but not to the graph of the original fraction. At

x=-3

, its denominator is zero.

At x = 1, the numerator is −8 and the denominator is 4. The original fraction is −2, matching x − 3 = −2. This input is permitted.

Common mistakes and checking strategies

Cancelling across an addition: factorise the complete numerator and denominator before cancelling.
Forgetting a missing denominator factor: whatever multiplies the denominator must also multiply that fraction’s numerator.
Treating $2-x$ as $x-2$ : they are negatives, so a minus sign must accompany the change.
Inverting the wrong fraction: in $A\div B$ , keep $A$ and multiply by the reciprocal of $B$ .
Losing restrictions or a square-root branch: retain original exclusions and state why a context permits only one sign.

For an equation, test each answer in the original equation. For a rearranged formula, substitute the result back or test a permitted set of values. For a simplified expression, check both its value and its allowed inputs.

Practise fractions and formulae together

Continue with the Sec 2 algebraic fractions and formulae practice bank. Work from simplification to combined fractions, then changing the subject. If the factorisation step is difficult, revisit expansion and factorisation. The quadratic equations practice bank also includes equations involving fractions and contextual restrictions.