Name: Elimination Method Questions: Worksheet And Solutions
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Introduction

Solving two equations at once can often feel like a puzzle with too many moving parts. In algebra, this is known as a system of linear equations. While there are several ways to find where two lines meet, the elimination method is frequently the most direct and efficient strategy.

By strategically adding or subtracting equations, we can "cancel out" one variable, allowing us to solve for the other with ease. This guide will walk you through the logic behind the method, provide a clear roadmap for solving any problem, and offer plenty of original practice to sharpen your skills for your GCSE or A-Level exams.

Theory: Understanding Elimination

The goal of the elimination method is to create a single equation with only one variable. This is achieved by ensuring the coefficients (the numbers multiplying the variables) of either x or y are identical or exact opposites.

When to Choose Elimination

The elimination method is most effective when:

Both equations are presented in the form ax + by = c.
The coefficients of one variable are already the same.
One coefficient is a simple multiple of the corresponding coefficient in the other equation.

The Standard Procedure

Step 1: Align the Equations - Ensure both equations are in the same standard form so that the variables and constants are lined up vertically.
Step 2: Balance the Coefficients - If the coefficients do not match, multiply one or both equations by a constant value.
Step 3: Eliminate a Variable -
- If the matching coefficients have different signs, add the equations.
- If the matching coefficients have the same sign, subtract one equation from the other.
Step 4: Solve and Substitute - Solve the resulting equation for the remaining variable, then plug that value back into one of the original equations to find the second variable.
Step 5: Verify - Always check your final answers in the equation you didn't use for substitution to ensure accuracy.

Method Comparison

The following data highlights which method to prioritize based on the equation structure:

Equation Structure	Recommended Method	Efficiency
Both in ax + by = c form	Elimination	High
One variable is isolated (e.g. y = 2x + 3)	Substitution	High
Large or fractional coefficients	Elimination	Medium
Visualising the intersection point	Graphing	Low

Worked Example: A Step-by-Step Walkthrough

Solve the following simultaneous equations:

Equation 1:

4x + 3y = 25

Equation 2:

2x - y = 5

1. Match the coefficients: To eliminate y, we look at the coefficients.

Equation 1 has +3. Equation 2 has -1.

We can multiply Equation 2 by 3 to give:

6x - 3y = 15

2. Add the equations: Now we add Equation 1 and the New Equation 2 to eliminate the y variable:

(4x + 3y) + (6x - 3y) = 40

10x = 40

3. Solve for x: Divide both sides by 10:

x=4

4. Substitute to find y: Substitute x = 4 into the original Equation 2:

2(4) - y = 5

8 - y = 5

y = 3

Final Solution: x = 4, y = 3

Common Errors & Tips

The Distributive Property: When multiplying an equation by a constant, ensure you multiply every single term, including the constant on the right side of the equals sign.

Subtraction Hazards: When subtracting equations, it is helpful to put the second equation in parentheses to remind yourself to distribute the negative sign across all terms.

Check Your Work: A quick substitution of both values into the original equations can save you from simple arithmetic mistakes.

Practice Questions & Solutions

Score:

Solve by matching x:

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Solution

Subtract the second from the first:

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Substitute into the second:

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Solve by multiplying one equation:

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Solution

Multiply the second by 2:

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Subtract from the first:

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Substitute into the second:

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Solve the following:

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Solution

Multiply the first by 2:

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Subtract the second from this:

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Substitute into the first:

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Solve by multiplying both:

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Solution

Multiply first by 2 and second by 3:

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Subtract first from second:

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Substitute into the first:

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Solve:

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Solution

Add the equations:

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Substitute into the second:

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Solve:

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Solution

Multiply the second by 3:

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Add to the first:

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Substitute into the first:

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Solve:

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Solution

Multiply the second by 2:

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Subtract the first from this:

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Substitute into the first:

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Solve:

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Solution

Multiply the second by 2:

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Subtract the first from this:

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Substitute into the first:

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Solve:

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Solution

Multiply the first by 2:

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Add the second:

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Substitute into the first:

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Solve:

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Solution

Multiply the second by 2:

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Subtract the first from this:

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Substitute into the first:

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Solve:

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Solution

Multiply first by 3 and second by 5:

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Add them:

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Substitute into the second:

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Solve:

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Solution

Multiply the first by 3:

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Subtract the second:

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Substitute into the second:

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Solve:

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Solution

Multiply the first by 2:

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Subtract the second:

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Substitute into the first:

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3 Pens and 2 Pencils cost £2.10. 5 Pens and 4 Pencils cost £3.70. Find the cost of one pen (x) and one pencil (y).

Solution

Equations:

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Multiply the first by 2:

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Subtract the second:

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Substitute:

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Pen = £0.50, Pencil = £0.30.

At a local cinema, adult tickets cost £12 and child tickets cost £7. On a particular afternoon, a total of 50 tickets were sold, and the total revenue collected was £475. How many adult tickets (a) and how many child tickets (c) were sold?

Solution

First, we set up two equations based on the number of tickets and the total cost:

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To eliminate the variable c, multiply the first equation by 7:

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Now, subtract this new equation from the second equation:

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Divide by 5 to find the number of adult tickets:

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Substitute this value back into the first equation to find the number of child tickets:

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The cinema sold 25 adult tickets and 25 child tickets.

A rectangle has a perimeter of 40 cm. It is known that twice the length (l) plus three times the width (w) equals 52 cm. Find the dimensions of the rectangle.

Solution

The perimeter of a rectangle is calculated as 2l + 2w. We can form the following system of equations:

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Since the coefficient of l is the same in both equations, we can subtract the first equation from the second:

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The width of the rectangle is 12 cm. Now, substitute this value into the perimeter equation:

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Subtract 24 from both sides:

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The dimensions of the rectangle are 8 cm by 12 cm.

The sum of two numbers, x and y, is 25. The difference between twice the first number and the second number is 14. Determine the values of both numbers.

Solution

Based on the problem description, we can write the following equations:

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In this case, the y coefficients are opposites (+1 and -1). We can eliminate y by adding the two equations together:

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Divide by 3 to solve for x:

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Now, substitute x = 13 into the first equation to find y:

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The two numbers are 13 and 12.

Solve the following system which contains a mix of decimals and algebraic fractions:

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Solution

First, clear the decimals from the first equation by multiplying every term by 10:

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Next, clear the fractions from the second equation by multiplying every term by the lowest common multiple of the denominators, which is 12:

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Now we have a standard system. To eliminate y, multiply the first equation by 3 and the second by 5:

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Add the two equations together:

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Substitute x = 6 into the simplified version of the first equation:

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The solution is:

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Solve the system by first expanding the brackets and simplifying the expressions:

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Solution

Expand and simplify the first equation:

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Expand and simplify the second equation:

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To eliminate y, multiply the simplified second equation by 4:

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Now, subtract the simplified first equation from this new equation:

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Substitute x = 2 into the simplified first equation:

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The solution is:

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A student has 20 coins in a jar, consisting only of 50p coins (x) and 20p coins (y). The total value of the coins is £7.60. Determine the number of each type of coin held in the jar.

Solution

First, we create an equation for the total number of coins:

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Next, we create an equation for the total value in pounds. Note that 50p is £0.50 and 20p is £0.20:

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To make the second equation easier to work with, multiply every term by 10 to clear the decimals:

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Now, multiply the first equation by 2 so that the y-coefficients match:

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Subtract this equation from the simplified value equation:

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Divide by 3 to find the number of 50p coins:

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Substitute x = 12 back into the first equation to find the number of 20p coins:

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The student has 12 coins of 50p and 8 coins of 20p.

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UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.

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Abhishek Mittal

December 2025

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Vanessa - Editorial Manager Superprof UK

December 2025

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joshua muwanguzi

November 2025

nice work

Vanessa - Editorial Manager Superprof UK

November 2025

Thank you! 😊 We’re glad you found the worksheet helpful. Keep up the great work with your studies!

Summarise with AI:

Chapters

Introduction
Theory: Understanding Elimination
Worked Example: A Step-by-Step Walkthrough
Practice Questions & Solutions

Solving Simultaneous Equations Worksheet: Practice Made Clear

Introduction

Theory: Understanding Elimination

When to Choose Elimination

The Standard Procedure

Method Comparison

Worked Example: A Step-by-Step Walkthrough

Practice Questions & Solutions

System of Equations

Elimination Method

Graphing Systems of Equations

Substitution Method

Nonlinear Systems of Equations

Gaussian Elimination Method

System of Equations Word Problems

Solving Substitution Method Questions

Elimination Method Questions: Worksheet And Solutions

Cancel reply

Solving Simultaneous Equations Worksheet: Practice Made Clear

Introduction

Theory: Understanding Elimination

When to Choose Elimination

The Standard Procedure

Method Comparison

Worked Example: A Step-by-Step Walkthrough

Practice Questions & Solutions

Theory

System of Equations

Elimination Method

Graphing Systems of Equations

Substitution Method

Nonlinear Systems of Equations

Gaussian Elimination Method

Exercises

System of Equations Word Problems

Solving Substitution Method Questions

Elimination Method Questions: Worksheet And Solutions

Cancel reply