Year 10 - Linear equations with algebraic fractions - Free practice

Year 10 Linear Equations with Algebraic Fractions Worksheets & Practice

Clear the fractions, solve with confidence. Worked solutions every time, free.

Quick answer

How do you solve linear equations with algebraic fractions?

Multiply both sides of the equation by the lowest common denominator (LCD) to eliminate all fractions, then solve the resulting linear equation. For example, to solve x/3 + x/4 = 7, multiply every term by 12 to get 4x + 3x = 84, giving x = 12.

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 10 linear equations with algebraic fractions: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Single algebraic fraction
Solve $\dfrac{x + 3}{4} = 7$ .
Show worked solution
Key idea: Multiply both sides by the denominator to eliminate the fraction, then solve.
1. Multiply both sides by $4$ :
  $\begin{gathered}4 \times \frac{x+3}{4} = 7 \times 4 \\ x + 3 = 28\end{gathered}$
2. Subtract $3$ from both sides:
  $x = 25$
3. Check: $\dfrac{25 + 3}{4} = \dfrac{28}{4} = 7$ ✓
Answer: $x = 25$
Example 2 · Unlike denominators on both sides
Solve $\dfrac{x + 1}{3} = \dfrac{2x - 1}{5}$ .
Show worked solution
Key idea: Find the LCD of both denominators, multiply every term by it to eliminate all fractions, then solve.
1. The LCD of $3$ and $5$ is $15$ . Multiply both sides by $15$ :
  $\begin{gathered}15 \times \frac{x+1}{3} = 15 \times \frac{2x-1}{5} \\ 5(x + 1) = 3(2x - 1)\end{gathered}$
2. Expand the brackets:
  $5x + 5 = 6x - 3$
3. Subtract $5x$ from both sides:
  $5 = x - 3$
4. Add $3$ to both sides:
  $x = 8$
5. Check: $\dfrac{8+1}{3} = \dfrac{9}{3} = 3$ and $\dfrac{2(8)-1}{5} = \dfrac{15}{5} = 3$ ✓
Answer: $x = 8$
Example 3 · Real-life problem with an algebraic fraction
Layla and her friend Marcus save up to buy a video game together. Marcus contributes $15 more than Layla. One quarter of Marcus's contribution covers the online activation fee, which costs $12. Let $x$ represent Layla's contribution in dollars. Write an equation and solve it to find how much each person contributes.
Show worked solution
Key idea: Translate the real-life information into an equation involving an algebraic fraction, then solve by multiplying both sides by the denominator.
1. Marcus contributes $x + 15$ dollars. One quarter of Marcus's contribution equals $12$ , so:
  $\frac{x + 15}{4} = 12$
2. Multiply both sides by $4$ :
  $x + 15 = 48$
3. Subtract $15$ from both sides:
  $x = 33$
4. Marcus's contribution: $x + 15 = 33 + 15 = 48$ dollars.
5. Check: $\dfrac{48}{4} = 12$ ✓
Answer: Layla contributes $33 and Marcus contributes $48.

Common mistakes

Where students usually lose marks

Multiplying only one term instead of the whole side

$\dfrac{x+1}{3} = \dfrac{2x-1}{5}$ becomes $x + 1 = \dfrac{2x-1}{5} \times 3$ (only the left side is multiplied by 3)

Whatever you do to one side, you must do to the entire other side. Multiply every term on both sides by the LCD. Using the LCD of 15 here gives $5(x+1) = 3(2x-1)$ .

Forgetting to expand brackets after multiplying

$5(x + 1) = 3(2x - 1)$ becomes $5x + 1 = 6x - 1$ (the 5 and 3 are not distributed)

After multiplying, expand fully: $5 \times x + 5 \times 1 = 3 \times 2x - 3 \times 1$ , giving $5x + 5 = 6x - 3$ . Always distribute the number outside the bracket to every term inside.

Using the wrong LCD

For $\dfrac{x}{4} + \dfrac{x}{6}$ , students use $24$ instead of $12$ as the LCD

The LCD is the lowest common multiple, not just any common multiple. List the multiples of each denominator and pick the smallest one they share. For 4 and 6, the multiples of 4 are 4, 8, 12, 16 and multiples of 6 are 6, 12, 18, so the LCD is 12.

Ignoring the unknown in the denominator

For $\dfrac{12}{x} = 4$ , students write $12 = 4$ (treating $x$ in the denominator as 1)

Multiply both sides by $x$ to move it out of the denominator: $x \times \dfrac{12}{x} = 4 \times x$ , which gives $12 = 4x$ , so $x = 3$ . Always check your answer by substituting back into the original equation.

Try one yourself

Your turn

Solve the equation $\frac{x+1}{3} = \frac{2x-1}{5}$

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About this worksheet

Why we built it

Algebraic fractions bring a new layer to linear equations in Year 10: students must identify a common denominator, multiply both sides to clear the fractions, and then solve what remains. These worksheets build that process step by step, from single-fraction equations through to problems with unlike denominators on both sides, with full worked solutions so students can pinpoint exactly where their method needs attention.

What's covered

Sub-skills your student will practise

✓Solving equations with a single algebraic fraction
✓Finding and using the lowest common denominator (LCD)
✓Multiplying both sides to eliminate all fractions
✓Solving equations with unlike denominators on both sides
✓Equations where the unknown appears in the denominator
✓Checking solutions by substituting back into the original equation

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Frequently asked

Questions parents ask about linear equations with algebraic fractions

What does 'clearing the fractions' mean and why do we do it?

Clearing the fractions means multiplying every term on both sides by the lowest common denominator so no fraction signs remain. It turns a harder-looking equation into a straightforward linear one that is much easier to solve.

What mistakes do Year 10 students commonly make with algebraic fraction equations?

The most common errors are forgetting to multiply every term (not just the fraction terms) by the LCD, and making sign errors when the numerator contains more than one term. Checking the answer by substituting it back in catches both mistakes quickly.

Are algebraic fraction equations in the Year 10 curriculum?

Yes, solving linear equations involving simple algebraic fractions is part of Year 10 in the Australian Curriculum, so it is fair game in school assessments and end-of-year exams. In WA it sits in the optional Year 10 content.

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