# Linear equations

> Edexcel International GCSE Mathematics A · 4MA1
> Source: https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-linear-equations/

This guide covers single-variable linear equation solving for Edexcel IGCSE Maths A (4MA1) Foundation and Higher tiers, including fractional coefficients, unknowns on both sides, and context setup.

**Prerequisites:** [Basic algebraic manipulation and bracket expansion](https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-algebraic-manipulation/); [Fraction operations and LCM calculation](https://www.owlsprep.com/study/edexcel-igcse-math-a-s1-fractions/)

## Learning objectives

- Solve single-variable linear equations with integer and fractional coefficients, including unknowns on both sides
- Set up linear equations from geometric and real-world context problems
- Apply required Edexcel steps: eliminate fractions via LCM, expand brackets before collecting terms
- Verify solutions via substitution and present exact simplified fractional answers

## Solving Basic Linear Equations (Foundation Tier)

**Linear equation in one unknown** — An equation where the highest power of the single variable is 1, with no products or powers of the variable.

*Example:* $5x + 8 = 12$ is linear; $x^2 + 3 = 7$ is not

The golden rule for solving all linear equations is: perform the identical operation to both sides of the equals sign to isolate the unknown. Follow these standard steps: 1. Expand any brackets 2. Collect terms with the unknown on one side, constants on the other 3. Divide by the coefficient of the unknown to find the solution.

**Worked example:** Solve $7(x + 3) = 5x - 8$

1. Expand the left-hand bracket

   $$7x + 21 = 5x - 8$$
2. Subtract 5x from both sides to collect x terms on the left

   $$2x + 21 = -8$$
3. Subtract 21 from both sides to collect constants on the right

   $$2x = -29$$
4. Divide both sides by 2 to isolate x

   $$x = -\frac{29}{2}$$

> **tip**
>
> Always verify your solution by substituting it back into the original equation. For this example, substituting $x=-\frac{29}{2}$ gives equal values on both sides, confirming the solution is correct.

> **Exam tip:** Double check bracket expansion: multiply every term inside the bracket by the coefficient outside, not just the first term.

*Calculator:* allowed

## Linear Equations with Fractional Coefficients

For equations containing fractions, eliminate denominators first by multiplying every term in the equation by the lowest common multiple (LCM) of all denominators. This simplifies the equation to an integer form that you can solve using the basic steps above.

**Worked example:** (Foundation) Solve $\frac{4x + 5}{2} = 3$

1. Multiply all terms by the only denominator, 2, to eliminate the fraction

   $$4x + 5 = 6$$
2. Subtract 5 from both sides

   $$4x = 1$$
3. Divide by 4 to isolate x

   $$x = \frac{1}{4}$$

**Worked example:** (Higher) Solve $\frac{2x - 3}{6} + \frac{x + 2}{3} = \frac{5}{2}$

1. Find LCM of denominators 6, 3, 2 = 6. Multiply every term by 6

   $$2x - 3 + 2(x + 2) = 15$$
2. Expand the remaining bracket

   $$2x - 3 + 2x + 4 = 15$$
3. Collect like terms

   $$4x + 1 = 15$$
4. Isolate x and simplify the solution

   $$4x = 14 \implies x = \frac{7}{2}$$

> **warning**
>
> Never multiply only the fractional terms by the LCM: you must multiply every term on both sides, including constant terms, to keep the equation balanced.

> **Exam tip:** For Higher tier questions with multiple denominators, always use the LCM of all denominators to eliminate fractions in one step, rather than multiplying by each denominator individually.

*Calculator:* allowed

## Setting Up Linear Equations from Context

Many exam questions present a geometric or real-world scenario and ask you to find an unknown value. Your first task is to translate the written information into a linear equation, then solve it using the steps you have learned.

**Worked example:** The three interior angles of a triangle are $a^\circ$, $(a+10)^\circ$, and $(a+20)^\circ$. Find the value of $a$.

1. Recall the sum of interior angles of a triangle is 180°, so write the equation

   $$a + (a + 10) + (a + 20) = 180$$
2. Collect like terms

   $$3a + 30 = 180$$
3. Subtract 30 from both sides

   $$3a = 150$$
4. Divide by 3 to isolate a

   $$a = 50$$

> **info**
>
> Always check that your solution makes sense in context. For this example, the angles are 50°, 60°, 70°, which add up to 180° as expected.

> **Exam tip:** If the question asks you to 'set up and solve' an equation, you will get separate marks for writing the correct equation and solving it, so never skip writing the equation step.

*Calculator:* allowed

## Practice and Verification

**Check your understanding**

1. Solve $3(2x - 5) = 9$

   - $x=4$
   - $x=2$
   - $x=-4$
   - $x=-2$

   *Why:* Expand to get $6x - 15 = 9$, so $6x = 24$ and $x=4$. Correct!

2. What is the first step to solve $\frac{x}{3} + \frac{x-1}{2} = 4$?

   - Multiply all terms by 6
   - Subtract $\frac{x-1}{2}$ from both sides
   - Add the fractions on the left
   - Divide both sides by 4

   *Why:* The LCM of 3 and 2 is 6, so multiplying all terms by 6 eliminates all fractions in one step, per Edexcel recommended method.

**Summary**

- Always perform identical operations to both sides of the equation to keep it balanced
- Eliminate fractions first using the LCM of all denominators, then expand any brackets
- Collect variable terms on one side, constants on the other, then divide by the coefficient of the variable
- Verify solutions via substitution, and present exact simplified fractions unless instructed otherwise

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to multiply constant terms by the LCM when eliminating fractions, e.g. for $\frac{x}{2} + 3 = 5$, only multiplying $\frac{x}{2}$ by 2 to get $x + 3 = 5$
  - Why it fails: You must apply the same operation to every term on both sides to maintain equation balance
  - Correct: Multiply every term by 2 to get $x + 6 = 10$, so $x = 4$
- **Wrong:** Incorrect bracket expansion, e.g. $7(x+3) = 7x + 3$
  - Why it fails: The coefficient outside the bracket must be distributed to every term inside the bracket
  - Correct: Multiply 7 by x and 7 by 3 to get $7x + 21$
- **Wrong:** Sign errors when moving terms across the equals sign, e.g. $2x + 21 = 5x - 8$ becomes $2x + 5x = -8 + 21$
  - Why it fails: To move a term across the equals sign you subtract it from both sides, so its sign flips
  - Correct: Rearrange to $2x - 5x = -8 - 21$, so $-3x = -29$ and $x = \frac{29}{3}$
- **Wrong:** Giving rounded decimal answers instead of exact fractions when not requested
  - Why it fails: Edexcel awards full marks only for exact answers unless rounding is explicitly specified
  - Correct: Simplify all fractions fully and present them as your final answer unless instructed otherwise
- **Wrong:** Skipping writing the equation for context questions, jumping straight to the final answer
  - Why it fails: Method marks are awarded for deriving the correct equation, even if your final solution is wrong
  - Correct: Always write the full linear equation you derived from the context before solving it

## Cheatsheet

| Step | Foundation Tier | Higher Tier |
| --- | --- | --- |
| 1 | Expand all brackets first | Find LCM of all denominators, multiply every term by LCM to eliminate fractions |
| 2 | Collect variable terms on one side, constants on the other | Expand any remaining brackets |
| 3 | Divide by coefficient of variable to isolate unknown | Collect like terms, isolate variable as per foundation steps |
| 4 | Check via substitution, present exact simplified fraction | Check via substitution, present exact simplified fraction |
| Context questions | Translate scenario to equation first, then solve | Same as foundation, expect multi-step scenarios |

## What's next

Mastering linear equations is the foundation for all higher algebra topics in Edexcel IGCSE Maths A, including simultaneous equations, quadratic equations, and algebraic graphs. You will encounter linear equations across both calculator and non-calculator papers, often combined with geometry, statistics, or real-world problem-solving contexts. Once you are confident solving linear equations and setting them up from context, you can move on to simultaneous linear equations, which require solving two equations with two unknowns, and rearranging formulae, which uses the same 'same operation to both sides' rule you learned here. Make sure to practice past paper questions regularly to build speed and accuracy, and always check your working for common mistakes like incorrect bracket expansion or sign errors.

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