Solving Equations with Algebraic Fractions

Equations with algebraic fractions all give in to the same move: multiply every term by the lowest common denominator, and the fractions disappear. In the Edexcel IGCSE course the twist is that clearing an x from a denominator usually creates a quadratic, so you factorise, find two solutions, and discard any value that would make a denominator zero. This page walks through the whole method with worked examples, from quick linear warm-ups to worded speed problems, then lets you master it on free auto-marked practice questions below.

Prior Knowledge This page builds on equations with fractions, solving quadratics by factorising, and simplifying and combining algebraic fractions.
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How to solve equations with algebraic fractions

Whatever sits in the denominators, the plan is the same: clear the fractions first. Multiply every term on both sides by the lowest common denominator (LCD). With numbers in the denominator that leaves a linear equation; with x in the denominator it usually leaves a quadratic, which you solve by factorising.

First: what the LCD is

LCD is short for lowest common denominator. It is the smallest expression that every denominator in the equation divides into exactly. That is what makes it useful: multiply every term by it and all the fractions cancel at once, in a single step.

To find it, look at the denominators together:

Finding the LCD
denominators \(2\) and \(3\) LCD \(=6\) plain numbers: the lowest number both divide into
denominators \(x\) and \(x+6\) LCD \(=x(x+6)\) nothing in common, so multiply them together
denominators \(x-3\) and \(x-2\) LCD \(=(x-3)(x-2)\) two different brackets, so again just the product

The one case where it is smaller than the product: if a denominator factorises and already contains the other one, do not multiply by that bracket twice. With \(x+1\) and \(x^2+4x+3\), the second is \((x+1)(x+3)\), so the LCD is just \((x+1)(x+3)\).

The method, step by step

  1. Find the LCD of all the denominators. With x in the bottom the LCD is an expression, such as \(x\) or \((x-3)(x-2)\).
  2. Multiply every term on both sides by the LCD, including lone x terms and plain numbers, then cancel.
  3. Expand carefully. Watch the signs when you subtract a bracketed numerator.
  4. Rearrange into standard form, with zero on one side, if an \(x^2\) has appeared.
  5. Factorise and solve. A quadratic usually gives two solutions.
  6. Check each answer in the original equation and discard any value that makes a denominator zero.

🎯 The LCD lands on every term, not just the fractions

6 (x − 3) − 4 (x − 2) = 1 this term counts as well × (x − 3)(x − 2) on every term

Miss the lone term on the right and the equation stops balancing, which is the most common way marks are lost here.

Clear by the LCD
\(\times\) every term
Multiply both sides by the LCD so every fraction cancels to a whole expression.
x in the denominator
\(x + a = \dfrac{b}{x}\)
Clearing an x from the bottom usually creates an \(x^2\), so expect a quadratic with two solutions.
Check and discard
denominator \(\ne 0\)
A value that makes any original denominator zero is not a solution, even if the algebra produced it.

Worked examples

In each example the orange factor shows the LCD multiplying every term, not just the fractions.

💡 Example 1: numerical denominators

Solve \(\dfrac{x+5}{2} - \dfrac{x-4}{3} = 3\).

\[ \begin{array}{rcl} \dfrac{x+5}{2} - \dfrac{x-4}{3} &=& 3 \\ \textcolor{#ea580c}{6\times}\dfrac{x+5}{2} - \textcolor{#ea580c}{6\times}\dfrac{x-4}{3} &=& \textcolor{#ea580c}{6\times}3 \\ 3(x+5) - 2(x-4) &=& 18 \\ 3x + 15 - 2x + 8 &=& 18 \\ x + 23 &=& 18 \\ x &=& -5 \end{array} \]
What is happening?

The LCD of \(2\) and \(3\) is \(6\), so multiply every term by \(6\). Subtracting the second bracket changes both signs: \(-2(x-4)\) becomes \(-2x+8\). Substituting \(x=-5\) back in gives \(0 + 3 = 3\), which checks out.

💡 Example 2: a fraction on each side

Solve \(\dfrac{3x-2}{5} = \dfrac{2x+1}{4}\).

\[ \begin{array}{rcl} \dfrac{3x-2}{5} &=& \dfrac{2x+1}{4} \\ \textcolor{#ea580c}{20\times}\dfrac{3x-2}{5} &=& \textcolor{#ea580c}{20\times}\dfrac{2x+1}{4} \\ 4(3x-2) &=& 5(2x+1) \\ 12x - 8 &=& 10x + 5 \\ 2x &=& 13 \\ x &=& \dfrac{13}{2} \end{array} \]
What is happening?

The LCD of \(5\) and \(4\) is \(20\); multiplying both sides by it is the same as cross-multiplying. Keep each numerator in a bracket while you expand. A fraction answer like \(\dfrac{13}{2}\) is perfectly fine.

💡 Example 3: x in the denominator, quadratic

Solve \(x - 3 = \dfrac{10}{x}\).

\[ \begin{array}{rcl} x - 3 &=& \dfrac{10}{x} \\ \textcolor{#ea580c}{x\times}x - \textcolor{#ea580c}{x\times}3 &=& \textcolor{#ea580c}{x\times}\dfrac{10}{x} \\ x^2 - 3x &=& 10 \\ x^2 - 3x - 10 &=& 0 \\ (x-5)(x+2) &=& 0 \end{array} \]

So \(x = 5\) or \(x = -2\).

What is happening?

The LCD is \(x\), and it must multiply the \(x\) and the \(-3\) as well as the fraction. Clearing it creates an \(x^2\), so rearrange to standard form and factorise. Both answers check out, and neither makes the denominator zero, so keep both.

💡 Example 4: two x denominators

Solve \(\dfrac{6}{x-3} - \dfrac{4}{x-2} = 1\).

\[ \begin{array}{rcl} \dfrac{6}{x-3} - \dfrac{4}{x-2} &=& 1 \\ \textcolor{#ea580c}{(x-3)(x-2)\times}\dfrac{6}{x-3} - \textcolor{#ea580c}{(x-3)(x-2)\times}\dfrac{4}{x-2} &=& \textcolor{#ea580c}{(x-3)(x-2)\times}1 \\ 6(x-2) - 4(x-3) &=& (x-3)(x-2) \\ 6x - 12 - 4x + 12 &=& x^2 - 5x + 6 \\ 2x &=& x^2 - 5x + 6 \\ 0 &=& x^2 - 7x + 6 \\ 0 &=& (x-1)(x-6) \end{array} \]

So \(x = 1\) or \(x = 6\).

What is happening?

The LCD is the product of the two denominators, \((x-3)(x-2)\). Each fraction cancels one bracket and keeps the other. The denominators are zero at \(x=3\) and \(x=2\); neither solution matches those values, so both survive the check.

💡 Example 5: a worded speed problem

A courier cycles \(24\) km at \(x\) km/h, then a further \(18\) km at \((x+6)\) km/h. The whole journey takes \(3\) hours. Find \(x\).

\[ \begin{array}{rcl} \dfrac{24}{x} + \dfrac{18}{x+6} &=& 3 \\ \textcolor{#ea580c}{x(x+6)\times}\dfrac{24}{x} + \textcolor{#ea580c}{x(x+6)\times}\dfrac{18}{x+6} &=& \textcolor{#ea580c}{x(x+6)\times}3 \\ 24(x+6) + 18x &=& 3x(x+6) \\ 42x + 144 &=& 3x^2 + 18x \\ 0 &=& 3x^2 - 24x - 144 \\ 0 &=& x^2 - 8x - 48 \\ 0 &=& (x-12)(x+4) \end{array} \]

So \(x = 12\) or \(x = -4\). A speed cannot be negative, so reject \(x=-4\): the courier rides at \(12\) km/h.

What is happening?

Time equals distance over speed, so each leg contributes a fraction of hours and the two legs add to \(3\). The LCD is \(x(x+6)\). Dividing the whole equation by \(3\) before factorising keeps the numbers small. The algebra offers two roots, but only the positive one is a speed, so state the rejection: this final line is worth a mark in the exam.

🔑 Key Points

  • Multiply every term on both sides by the LCD, including lone x terms and plain numbers.
  • Keep each numerator in a bracket when it is multiplied, then expand carefully.
  • Once an \(x^2\) appears, rearrange to standard form before factorising.
  • A quadratic usually gives two solutions; give both unless the context rules one out.
  • Substitute each answer back and discard any value that makes a denominator zero.

⚠️ Common Pitfalls

  • Multiplying only the fraction terms by the LCD and missing the lone x or the constant.
  • Sign errors when subtracting a bracketed numerator: \(-2(x-4)\) is \(-2x+8\), not \(-2x-8\).
  • Keeping an answer that makes a denominator zero; it must be discarded.
  • Dividing both sides by x instead of factorising, which silently loses a solution.
  • Stopping after one root when the quadratic has two.
⇩ Practise now ⇩

When the cleared quadratic will not factorise, the same set-up still works: you just finish it with the quadratic formula instead.

Next: The Quadratic Formula →

Equations with Algebraic Fractions: Practice Room

Free auto-marked practice solving equations with algebraic fractions, built for Edexcel IGCSE Maths. Room 1 warms up at Starter to Builder pace with equations that clear to linear; Room 2 clears x from the denominator to leave a quadratic you factorise for two solutions; Room 3 brings harder denominators, three-fraction equations and worded speed problems at Challenger to Master level; Room 4 mixes every type. Type the value of x in each box: use / for a fraction (for example 3/2) or an exact decimal such as 1.5; unsimplified fractions like 6/4 are accepted. When an equation has two solutions you get two boxes, and the order does not matter. Press Enter or click away to mark; every correct card adds to your streak.

✓ Correct 0
✗ Re-attempts 0
🔥 Streak 0
🏆 Best 0

Always check a solution does not make a denominator zero: the generators here never produce one, but in an exam you must show the check. If a card has two boxes, the equation has two valid solutions; find both.