E2.5 - Equations with algebraic denominators | Cambridge IGCSE Mathematics

An equation can appear to offer an answer that makes its original fractions impossible. You will identify excluded values, clear numerical and linear algebraic denominators, and check every candidate solution in the original equation.

Denominator restrictions

Can multiplying away the fractions create a false answer? Consider $\frac{x-2}{x-2}=\frac{3}{x-4}$ . The value $x=2$ must never become a solution, however neat a later line of algebra looks: its first fraction would be $\frac00$ .

A fraction means division. Division by zero is undefined, so identify the values that make any original denominator zero before simplifying or solving. These are the excluded values. The numerator may be zero; the denominator may not.

Denominator restriction

A denominator restriction states which values an original fraction cannot accept. For $\frac{p(x)}{ax+b}$ , where $a\ne0$ , require $ax+b\ne0$ , so $x\ne-\frac ba$ . For example, $\frac{0}{5}=0$ is defined, but $\frac50$ is not.

Restrictions survive cancellation. Cancelling a common factor gives an equivalent expression only at the values where the original expression exists.

Worked example 1: a linear denominator

Question. State the excluded value for $\frac{5}{x-3}$ .

Plan. Find where the denominator would be zero.

\begin{aligned} x-3&=0 && \text{locate the forbidden denominator}\\ x&=3 && \text{add three} \end{aligned}

Answer. $x\ne3$ .

Check. At $x=3$ , the fraction is $\frac50$ , which is undefined. At $x=4$ , it is $\frac51=5$ , which is allowed.

Change only the numerator to match the denominator. The restriction is unchanged, even though cancellation is now possible.

Worked example 2: cancellation retains a restriction

Question. Simplify $\frac{x-3}{x-3}$ and state its restriction.

Plan. Record the excluded value, then cancel the common nonzero factor.

\begin{aligned} x-3&\ne0 && \text{the original denominator must exist}\\ \frac{x-3}{x-3}&=1\quad(x\ne3) && \text{divide a nonzero number by itself} \end{aligned}

Answer. $1$ , with $x\ne3$ .

Check. At $x=4$ , $\frac{4-3}{4-3}=1$ . At $x=3$ , $\frac00$ is undefined, so the simplified value cannot fill that gap.

With several fractions, collect every restriction. For denominators $x+2$ and $2x-1$ , require both $x\ne-2$ and $x\ne\frac12$ . A nonzero numerical denominator, such as $6$ , excludes no value of $x$ .

Watch out

“The $x-3$ cancels, so $x=3$ is now allowed” is wrong. Cancellation changes the written form, not where the original fraction is defined.

Clearing denominators

To clear denominators, multiply every term on both sides by a common multiple of the denominators. This preserves equality, and division factors cancel. With algebraic denominators, the multiplier is nonzero at every allowed value because you have already recorded its restrictions.

For numerical denominators $3$ , $2$ and $6$ , multiplying by $6$ is enough. For $x+2$ , $2$ and $6$ , use $6(x+2)$ . Keep brackets until every term has been multiplied.

Worked example 3: numerical denominators

Question. Solve $\frac{x}{3}+\frac12=\frac56$ .

Plan. Multiply the whole equation by six, then solve the resulting linear equation.

\begin{aligned} 6\left(\frac{x}{3}\right)+6\left(\frac12\right)&=6\left(\frac56\right) && \text{multiply every term}\\ 2x+3&=5 && \text{cancel numerical factors}\\ 2x&=2 && \text{subtract three}\\ x&=1 && \text{divide by two} \end{aligned}

Answer. $x=1$ .

Check. In the original equation, $\frac13+\frac12=\frac26+\frac36=\frac56$ .

Replace only the first denominator by $x+2$ . Now the multiplier contains an expression, and a restriction is needed.

Worked example 4: a linear algebraic denominator

Question. Solve $\frac{x}{x+2}+\frac12=\frac56$ .

Plan. Exclude $x=-2$ , multiply every term by $6(x+2)$ , then solve.

\begin{aligned} 6x+3(x+2)&=5(x+2) && \text{clear all three denominators}\\ 6x+3x+6&=5x+10 && \text{expand the brackets}\\ 4x&=4 && \text{collect and subtract six}\\ x&=1 && \text{divide by four} \end{aligned}

Answer. $x=1$ , which is not excluded.

Check. $\frac{1}{1+2}+\frac12=\frac13+\frac12=\frac56$ . The original denominator is $3$ , not zero.

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E2.5 - Equations with algebraic denominators

Denominator restrictions

Clearing denominators

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