# Simultaneous Linear Equations

> Edexcel International GCSE Mathematics A · 4MA1 S2
> Source: https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-simultaneous-linear-equations/

This guide covers solving 2-variable linear simultaneous equations for Edexcel IGCSE Maths A (4MA1), including elimination and substitution methods, verification steps, and Higher tier graphical interpretation of solutions.

**Prerequisites:** [Rearranging linear equations](https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-rearranging-equations/); [Solving linear equations in one variable](https://www.owlsprep.com/study/edexcel-igcse-math-a-s2-linear-equations/)

## Learning objectives

- Solve 2-variable linear simultaneous equations using the elimination method accurately
- Solve 2-variable linear simultaneous equations using the substitution method correctly
- Verify solutions by substituting calculated values back into original equations
- (Higher only) Interpret solutions as the intersection point of two straight line graphs

## Introduction to Simultaneous Linear Equations

**Simultaneous Linear Equations** — Two linear equations with two unknown variables (e.g. x and y) that hold true for the same pair of values for the unknowns.

*Example:* x + y = 10, 2x - 3y = 5

All problems in this topic use exactly two linear equations (no quadratic, cubic or other non-linear terms) with two unknown variables. The solution is the unique pair of values that makes both equations true at the same time.

## Solving by Elimination Method

**Elimination Method** — A method that removes (eliminates) one unknown variable by adding or subtracting scaled copies of the two equations, leaving a linear equation in one variable to solve.

1. If necessary, multiply one or both equations by a constant so the coefficient of one variable is equal in size in both equations
2. Add equations if coefficients have opposite signs, or subtract if they have the same sign, to eliminate that variable
3. Solve the resulting one-variable linear equation
4. Substitute your answer back into one of the original equations to find the second unknown
5. Check your solution in the other original equation

**Worked example:** Solve the simultaneous equations: $2a + 5b = 12$, $3a + b = 5$

1. First, eliminate b by multiplying the second equation by 5 to match the coefficient of b in the first equation:
2. $$5 \times (3a + b = 5) \implies 15a + 5b = 25$$
3. Subtract the first original equation from the new equation to eliminate b:
4. $$(15a + 5b) - (2a + 5b) = 25 - 12 \implies 13a = 13$$
5. Solve for a:
6. $$a = 1$$
7. Substitute $a=1$ back into the original second equation $3a + b = 5$:
8. $$3(1) + b = 5 \implies b = 2$$
9. Check by substituting $a=1, b=2$ into the first original equation: $2(1) + 5(2) = 12$, which matches the right-hand side. The solution is $a=1, b=2$.

> **Exam tip:** Label equations (e.g. Equation 1, Equation 2) when solving to avoid mixing them up during scaling or subtraction. Always explicitly state both unknown values in your final answer.

*Calculator:* allowed

## Solving by Substitution Method

**Substitution Method** — A method that rearranges one equation to express one unknown in terms of the other, then substitutes this expression into the second equation to leave a one-variable linear equation to solve.

This method is especially efficient when one of the equations already has a variable with a coefficient of 1 or -1, making rearrangement very straightforward.

**Worked example:** Solve the simultaneous equations: $x + y = 14$, $x - y = 2$

1. Rearrange the first equation to make x the subject:
2. $$x = 14 - y$$
3. Substitute this expression for x into the second equation:
4. $$(14 - y) - y = 2 \implies 14 - 2y = 2$$
5. Solve for y:
6. $$-2y = -12 \implies y = 6$$
7. Substitute $y=6$ back into $x = 14 - y$:
8. $$x = 14 - 6 = 8$$
9. Check by substituting $x=8, y=6$ into the second original equation: $8 - 6 = 2$, which is correct. The solution is $x=8, y=6$.

*Calculator:* allowed

## Higher Tier: Graphical Interpretation

For Higher tier candidates, you need to connect simultaneous linear equations to straight line graphs. Each linear equation in x and y corresponds to a straight line when plotted on a coordinate grid.

**Solution as Intersection Point** — The pair of values that solves both simultaneous linear equations is the (x,y) coordinate where the two lines representing the equations intersect on a graph.

**Worked example:** The equations $y = x + 1$ and $y = 3x - 1$ are plotted on a coordinate grid. What is the solution to these simultaneous equations?

1. At the intersection point, the y-values of both lines are equal, so set the expressions for y equal to each other:
2. $$x + 1 = 3x - 1$$
3. Solve for x:
4. $$2 = 2x \implies x = 1$$
5. Substitute $x=1$ into $y = x + 1$ to get $y=2$.
6. The lines intersect at (1,2), so the solution is $x=1, y=2$.

> **Exam tip:** If a question asks you to use a graph to find the solution, you do not need to solve algebraically: just read the (x,y) coordinate of the intersection point carefully.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to multiply every term in an equation when scaling for elimination, e.g. multiplying $3a + b = 5$ by 5 to get $15a + b = 25$
  - Why it fails: You only scale part of the equation, so the new equation is no longer equivalent to the original, leading to an incorrect solution
  - Correct: Multiply every single term on both sides of the equation by the scaling constant, so $3a + b = 5$ becomes $15a + 5b = 25$
- **Wrong:** Mixing up signs when adding or subtracting equations during elimination, e.g. subtracting $2a + 5b =12$ from $15a +5b=25$ to get $13a +10b=13$
  - Why it fails: You incorrectly combine the coefficients of the variable you are trying to eliminate, leading to a wrong one-variable equation
  - Correct: If coefficients have the same sign, subtract one full equation from the other; if opposite signs, add them, applying sign rules carefully for each term
- **Wrong:** Only stating one unknown value as the final answer, e.g. only writing $x=8$ for the equations $x+y=14$, $x-y=2$
  - Why it fails: The question asks for the solution to both equations, which requires both variable values to earn full marks
  - Correct: Always explicitly state both unknown values in your final answer, e.g. $x=8, y=6$
- **Wrong:** Substituting your calculated value back into the rearranged equation you used for substitution, instead of checking in a different original equation
  - Why it fails: Any mistake in your rearrangement will not be caught by this check, so you may not notice an incorrect solution
  - Correct: Always verify your solution by substituting both values into the original equation you did not use to find the second unknown
- **Wrong:** (Higher tier) Assuming parallel lines have a solution, e.g. trying to solve $y=2x+1$ and $y=2x+3$
  - Why it fails: Parallel lines have equal gradients and never intersect, so there is no valid solution to the simultaneous equations
  - Correct: Check if the gradients of the two lines are equal: if they are, and the y-intercepts are different, state that there is no solution

## Cheatsheet

| Method | Best For | Key Steps | Check Step |
| --- | --- | --- | --- |
| Elimination | Equations where coefficients of one variable can be easily matched by scaling | 1. Scale equations to match coefficient of one variable 2. Add/subtract to eliminate variable 3. Solve for remaining unknown 4. Substitute back to find second unknown | Substitute both values into the unused original equation |
| Substitution | Equations where one variable has coefficient 1 or -1 | 1. Rearrange one equation to make one variable the subject 2. Substitute into the second equation 3. Solve for remaining unknown 4. Substitute back to find second unknown | Substitute both values into the unused original equation |
| Higher Graphical | Questions that provide pre-plotted lines | 1. Locate the intersection point of the two lines 2. Read the x and y coordinates of the point | Confirm coordinates satisfy both original equations |

## What's next

Now that you have mastered solving 2-variable linear simultaneous equations, you are ready to move on to more complex equation types in the Edexcel IGCSE Maths A syllabus. The next key topic is simultaneous equations with one linear and one quadratic equation, which builds directly on the substitution method you learned here. You will also apply these solving skills to real-world problem-solving questions, such as calculating costs, speeds, and unknown lengths in geometry problems. For Higher tier students, this topic links directly to straight line graphs and coordinate geometry, so make sure you are confident plotting and identifying line properties to tackle graphical interpretation questions. Practice as many past paper questions as possible to build speed and accuracy, as this topic is tested in almost every exam series.

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