E2.5 - Solving fractional equations (Extended) | Cambridge IGCSE International Mathematics

Can a value solve a rearranged equation but fail the original one? Clearing fractions can reveal a familiar linear or quadratic equation, but a zero denominator is never allowed. This Extended E2.5 lesson solves fractional equations with numerical and linear algebraic denominators, then checks and rejects invalid candidate roots.

Clearing denominators

The equation $\frac{x+1}{3}=2$ says that one third of $x+1$ is 2. Multiplying both sides by 3 gives $x+1=6$ . You have performed the same reversible operation on both sides, so their equality is preserved.

Clearing denominators

To clear denominators, multiply every term on both sides of an equation by a common denominator. First exclude any value that makes an original denominator zero.

If the denominator contains the unknown, the multiplier is non-zero only on the permitted inputs. Keep brackets around a complete numerator or denominator when multiplying.

Worked example 1: a numerical denominator

Question. Solve $\frac{x+1}{3}=2$ .

Plan. Multiply both sides by 3, then isolate $x$ .

\begin{aligned} x+1&=6 &&\text{multiply both sides by 3}\\ x&=5 &&\text{subtract 1} \end{aligned}

Answer. $x=5$ .

Check. In the original equation, $\frac{5+1}{3}=\frac63=2$ . Its denominator is the non-zero number 3.

Replace the numerical denominator by $x-3$ . The numerator and right side stay the same, but the denominator now adds a restriction and produces an $x$ -term on the right.

Worked example 2: a linear denominator

Question. Solve $\frac{x+1}{x-3}=2$ .

Plan. Exclude $x=3$ , multiply by $x-3$ , then solve the linear equation.

\begin{aligned} x+1&=2(x-3) &&\text{multiply both sides by }x-3\\ x+1&=2x-6 &&\text{expand the whole product}\\ 7&=x &&\text{collect and isolate} \end{aligned}

Answer. $x=7$ .

Check. The denominator is $7-3=4\ne0$ , and $\frac{7+1}{7-3}=\frac84=2$ . The candidate is valid.

With several numerical denominators, use their lowest common multiple. For example, multiplying $\frac{x-1}{2}+\frac{x+2}{3}=4$ throughout by 6 gives $3(x-1)+2(x+2)=24$ . Every term is multiplied, including the 4 on the right.

Watch out

$\frac{x+1}{x-3}=2$ gives $x+1=2(x-3)$ , not $x+1=2x-3$ . The multiplier applies to the complete denominator expression. Expand the bracket after clearing it.

State the restriction before solving.

Two denominators

Different linear denominators can lead to a quadratic when you clear them. The method is still to multiply every term by a common denominator, simplify, and solve the resulting equation.

For an equation $\frac AB=\frac CD$ , with $B,D\ne0$ , multiplying both sides by $BD$ gives $AD=BC$ . This explains cross-multiplication. If either side is a sum, multiplying every term is safer than trying to cross-multiply selected pieces.

Worked example 3: a linear equation after clearing

Question. Solve $\frac1{x+2}=\frac3{x-2}$ .

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

\begin{aligned} x-2&=3(x+2) &&\text{clear both denominators}\\ x-2&=3x+6\\ -8&=2x\\ x&=-4 \end{aligned}

Answer. $x=-4$ .

Check. Neither denominator is zero. The original sides are $\frac1{-2}=-\frac12$ and $\frac3{-6}=-\frac12$ , so they agree.

Change only the first numerator from 1 to $x$ . Multiplication now produces $x(x-2)$ , so a quadratic replaces the linear equation.

Worked example 4: a quadratic equation after clearing

Question. Solve $\frac{x}{x+2}=\frac3{x-2}$ .

Plan. Keep $x\ne-2,2$ , clear both denominators, then factorise the quadratic.

\begin{aligned} x(x-2)&=3(x+2)\\ x^2-2x&=3x+6\\ x^2-5x-6&=0 &&\text{put all terms on one side}\\ (x-6)(x+1)&=0 \end{aligned}

Answer. $x=6$ or $x=-1$ ; neither is excluded.

Check. For 6, both sides are $\frac34$ . For $-1$ , both sides are $-1$ . Both checks use the original fractions, whose denominators are non-zero.

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E2.5 - Solving fractional equations (Extended)

Clearing denominators

Two denominators

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