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Solving Simultaneous Equations: Worksheets with Answers

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Solving simultaneous equations

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Lesson details

Key learning points.

• In this lesson, we will solve a word simultaneous equation. We will then interpret the problem and create two equations from it and model a solution.

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Starter quiz

5 questions, lesson appears in, unit maths / solving linear simultaneous equations algebraically.

Solving Simultaneous Equations

An introduction to simultaneous equations using the bar model, and with no scaling up. The questions are from Minimally Different Questions , which is definitely worth exploring.

Linear simultaneous equations starts with visual questions involving burgers and chips, and moves towards the algebraic method. Main task is differentiated and answers are included.

Solving linear simultaneous equations graphically. Make sure students can sketch linear graphs first.

Nonlinear simultaneous equations includes visual examples of solving, as well as algebraic. Main task is differentiated and answers are included.

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Unit 1: Linear simultaneous equations

Geometrical representation.

• Solutions to systems of equations: dependent vs. independent (Opens a modal)
• Number of solutions to a system of equations (Opens a modal)

Solution of simultaneous equations by use of graphs

• Systems of equations with graphing: 5x+3y=7 & 3x-2y=8 (Opens a modal)
• Number of solutions to a system of equations (graphically) Get 3 of 4 questions to level up!

Conditions of solvability of two linear simultaneous equations

• No videos or articles available in this lesson
• Number of solutions of system of equations Get 3 of 4 questions to level up!
• Number of solutions to systems of equations (intermediate) Get 3 of 4 questions to level up!

Method of substitution

• Systems of equations with substitution: -3x-4y=-2 & y=2x-5 (Opens a modal)
• Systems of equations with substitution Get 3 of 4 questions to level up!

Method of elimination

• Systems of equations with elimination (and manipulation) (Opens a modal)
• Solving systems of equations by elimination (old) (Opens a modal)
• Systems of equations with elimination Get 3 of 4 questions to level up!
• Systems of equations with elimination challenge Get 3 of 4 questions to level up!

Cross multiplication

• Solving system of equations through cross multiplication Get 3 of 4 questions to level up!

Equations reducible to linear form

• Identifying substitutions: Equations reducible to linear form Get 3 of 4 questions to level up!
• Solving equations reducible to linear form Get 3 of 4 questions to level up!
• Word problems: Writing equations reducible to linear form Get 3 of 4 questions to level up!

Linear equations word problems

• Age word problem: Ben & William (Opens a modal)
• Forming equations with two variables Get 3 of 4 questions to level up!
• Age word problems Get 3 of 4 questions to level up!
• Word problems involving pair of linear equations (advanced) Get 3 of 4 questions to level up!
• Systems of equations word problems Get 3 of 4 questions to level up!

Simultaneous Equations True or False ( Editable Word | PDF | Answers )

Solving Simultaneous Equations (Same y Coefficients) Fill in the Blanks ( Editable Word | PDF | Answers )

Solving Simultaneous Equations (Same y Coefficients) Practice Strips ( Editable Word | PDF |  Answers )

Solving Simultaneous Equations (Same x Coefficients) Fill in the Blanks ( Editable Word | PDF | Answers )

Solving Simultaneous Equations (Different y Coefficients) Fill in the Blanks ( Editable Word | PDF | Answers )

Solving Simultaneous Equations (Different y Coefficients) Practice Strips ( Editable Word | PDF | Answers )

Solving Simultaneous Equations (Different x Coefficients) Fill in the Blanks ( Editable Word | PDF | Answers )

Solving Simultaneous Equations Sort It Out ( Editable Word | PDF | Answers )

Linear Simultaneous Equations Crack the Code ( Editable Word | PDF | Answers )

Linear Simultaneous Equations Worded Problems Practice Strips ( Editable Word | PDF | Answers )

Worded Simultaneous Equations Name the Film ( Editable Word | PDF | Answers ​ )

Linear Simultaneous Equations Revision Practice Grid ( Editable Word | PDF | Answers )

Investigating Linear Simultaneous Equations and Graphs Activity ( Editable Word | PDF | Answers )

Solving Linear Simultaneous Equations Graphically Practice Grid ( Editable Word | PDF | Answers )

Solving Linear Simultaneous Equations by Substitution Practice Strips ( Editable Word | PDF | Answers )

Solving Non-Linear Simultaneous Equations Fill in the Blanks ( Editable Word | PDF | Answers )

Non-Linear Simultaneous Equations Practice Strips ( Editable Word | PDF | Answers )

Harder Simultaneous Equations Practice Grid ( Editable Word | PDF | Answers )

A Level Maths

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Simultaneous Equations

The aim of this section is to understand what are simultaneous equations and how we can solve them? After reading this section we will be able to write down a word problem in the form of simultaneous equations and be able to find out the solution.

Introduction

Types of simultaneous equations, methods for solving simultaneous equations..

Who to verify the solution?

• Problems on Simultaneous Equations

Mathematics plays a vital role in our life; without mathematics many situations go wrong. There is no problem in physical science that can be solved without converting it into mathematics. Whenever a problem is converted to mathematics it gives an equation in some variable form. Like this situation if a person buys two cupcakes in £3, then what will be the price of one cupcake?

Here the problem in mathematical form is

Where x represents the price of cupcake. We can find the value of x by dividing 2 on both sides, but sometimes problems give the two or more equations. These equations involve two or more unknown variables, as x is an unknown value in above equation, which we have to determine.  These types of equations are called simultaneous equations .

Word simultaneous represents “at a same time”. So simultaneous equations are those equations which are correct for the certain values of unknown variables at a same time.

For examples

Above equations are simultaneous equations in unknown variables ‘x’   and ‘y’ . Both equations are true for x = 1 and y = 4 . Let’s check that these equations are true or not?

Put x = 1 and y = 4

Now the other equation

Now put x = 1 and y = 4

See both the equations are true at a same time for single value of x and y .

There are two types of simultaneous equations which we will see in this section.

1) Linear simultaneous equations

Linear simultaneous equations are called those equations in which power of each unknown variable is one. i.e.

x = 2y        5x – y = 15

2) Nonlinear simultaneous equations

Nonlinear simultaneous equations are those equations in which power of at least one unknown variable must be greater than one. i.e.

In above set first equation comes with second degree so this set will be called nonlinear simultaneous equations.

Note: An equation involves a variable with second degree is known as quadratic equation. In addition, above nonlinear equation is also quadratic simultaneous equations.

There are well known three methods we use to solve simultaneous equations, as are listed below.

1) Elimination Method : In this method we eliminate one variable to find the value of other variable.

In this method first we multiply both equations with different numbers to make coefficient same of any one variable and then subtract these equations, after subtraction one variable vanishes out so that we can find the value of another unknown variable easily. After finding out the value of one unknown variable we put this in any one equation and find out the other equations. We will see this method in examples.

Example 1: Solve the simultaneous equations 2x + 3y = 8   and 3x + 2y = 7

First give the name to both equations.

2x + 3y = 8       (1)

3x + 2y = 7       (2)

We will solve these equations by elimination method. To eliminate the one unknown variable, we make the coefficient same of one variable, here we are going to eliminate the unknown variable ‘x’ first multiply equation (1) by 3 and equation (2) by 2. (multiplying by coefficient of ‘x’ in equation (1) with equation (2) and coefficient of ‘x’ in equation (2) by equation (1) is an easy way to make coefficient same

3×1⇒  6x+9y=24

2×2⇒  6x+4y=14

Now subtract equation (2) from equation (1)

In this step ‘x’ eliminates, we get the equation in term of ‘y’ only. Now divide by ‘5’ on both side.

Now put the value of ‘y’ in any equation, we get the same results, let’s put the value of ‘y’ in equation (1)

2x + 3(2) = 8

⇒  2x + 6 = 8

Subtracting ‘6’ on both sides

2x + 6 – 6 = 8 – 6

Dividing by ‘2’ on both sides of the above equation, we get:

The solution of simultaneous equation is x = 1   and y = 2 .

2) Substitution Method: In this method first we write any one unknown variable in terms of second unknown variable from one equation. Then substitute the value of this variable in the other equation. Then one variable vanishes out and we find out the value of other one. After it, the procedure is same as discussed in elimination method. We will see this method in the following example.

Since it is nonlinear simultaneous equations, we first list them by numbering.

x + y = 4          (2)

We solve them by substitution method, firstly write down the equation two in term of ‘x’ only,

x = 4 – y     (3)

Now substitute the value of ‘x’ from equation (1) to equation (2)

Adding same terms and rearranging above equation

Subtracting ’10’ on both side

Dividing by two ‘2’ on both sides

Since it is a quadratic equation in term of ‘y’ , we can solve it by factorization.

Taking common similar terms

From above equation we can write as

y – 3 = 0   or   y-  1 = 0

Rearranging above equations we get

y = 3 or y = 1

So, here is the values of ‘y’ , now put these values in equation (3) one by one

x = 4 – 3 = 1

Now put y = 1

x = 4 – 1 = 3

So, there are two different answers, one is x = 1 , when y = 3 and the other is x = 3 , when y = 1 .

3) Graphical Method: In this method we draw the graph of each line and trace out the intersection of these lines. Basically, this intersection is the solution of these equation. Normally we use graphically method for linear. In nonlinear simultaneous equations graphically, method is not so effective because its solution out in surd form.

Example 3: Solve the simultaneous equations by graphical method. 6x + y = 40;   4x + 3y = 36

First of all, we draw the graph of both equation one by one and then trace out the intersection of lines, which will be the our required solution.

In this graph point A representing the point of intersection. At point A the value of x-axis is 6 and y-axis is 4, so point of intersection is 6,4, which is the required solution.

Many times, we find the solution but forget to check that even it is true or false. In mathematics it is very important to find out the collect solution for carrying good grades. So, for checking the solution that it is true of false we put the answer or values of unknown variables in both equation and see that either both sides are same or not, if same then our solution is correct, if that then we have to check our calculation again.

For example, in the last section we find out the solution of 6x + y = 40 ;   4x + 3y = 36 , which is x = 6 and y = 4 . Put the values in both equation one by one and see that either it is correct or not.

Put in first equation.

66 + 4 = 36 + 4 = 40

Now in second equation

46 + 34 = 24 + 12 = 36

Since values of x and y satisfy both equations, so our solution is correct.

Word Problems for Simultaneous Equations

Problem 1: If Jon bought three packets of chips and 2 packets of biscuits in £29, and Charlie bought one packet of chips and seven packets of biscuits in £54. Then what is the price of each packet of chips and biscuits?

Here we first give the name to the objects for which we have to determine the price.

Let ‘x’ is the price of one packets of chips and ‘y’ is the price of one packet of biscuit.

Now write down the statement in mathematical form step by step.

Mathematical expression for Jon:

Jon bought three packets of chips so, amount for chips will be ‘3x’ and he bought two packets of biscuits, amount for biscuits will be ‘2y’ . According to statement he spends £29 on chips and biscuits. Mathematical expression for Jon is written below.

3x + 2y = 29       (1)

Mathematical expression for Charlie:

Jon bought one packets of chips so, amount for chips will be ‘x’ and he bought seven packets of biscuits, amount for biscuits will be ‘7y’ . According to statement he spends £54 on chips and biscuits. Mathematical expression for Charlie is written below.

x + 7y = 54       (2)

From equation (1) and equation (2) we will determine the value of x and y . Since these are the simultaneous equations. So, we can solve them elimination method.

According to method first make the coefficient same of a one variable, here we make the same coefficient of x .

3×(2)⇒   3x + 21y = 162

Subtract the equation (1) from the above equation.

Dividing on both sides by ’19’

Put the value of y = 7 , in equation (2)

x + 7(7) = 54

x + 49 = 54

x = 54 – 49

So, the price of one packet of chips is £5 and price of one packet of biscuit is £7.

Problem 2: If the sum money in the pocket of two person A and B is $8 and sum of square their amount is $34, then how much amount each person has?

Let consider Person A have x and Person B have y in their pockets.

Then mathematical form for first condition, which is the sum of their amount is 8

x + y = 8      (1)

Then mathematical form for first condition, which is the sum of square their amount is 34

Rearranging equation (1), we get

Put this value of y in equation (2)

Collecting same terms

Dividing by 2 on both side

Factorize the above quadratic equation

x – 5 = 0    or     x – 3 = 0

⇒   x=5   or    x=3

Put the value of x , in y

y = 8 – 5 = 3

y = 8 – 3 = 5

Here is the solution x = 5 , when y = 3 and x = 3 , when y = 5 .

Its mean one of them have $3 and one of them have $5.

Problem 3: Solve the following Nonlinear Systems of Equations.

x + y = 2         (1)

Rearrange the equation (1), we will get

Put this value in equation (2)

3x = 0     or       3x – 4 = 0

x = 0        or      x = 43

Now put the values of x into y = 2 – x , one by one

x = 0   ⇒ y = 2 – 0 = 2

• Pure mathematics 1 by Sophie Goldie
• Advance Level Mathematics (Pure Mathematics 1) by Hugh Neill and Douglas Quadling

Word Problems on Simultaneous Linear Equations

Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations.

Problems of different problems with the help of linear simultaneous equations:

We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations.

In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols.

Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem.

Worked-out examples for the word problems on simultaneous linear equations: 1. The sum of two number is 14 and their difference is 2. Find the numbers. Solution: Let the two numbers be x and y.

x + y = 14 ………. (i)

x - y = 2 ………. (ii)

Adding equation (i) and (ii), we get 2x = 16

or, 2x/2 = 16/2 or, x = 16/2

or, x = 8 Substituting the value x in equation (i), we get

or, 8 – 8 + y = 14 - 8

or, y = 14 - 8

or, y = 6 Therefore, x = 8 and y = 6

Hence, the two numbers are 6 and 8.

2. In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number. Solution:

Let the digit in the units place is x

And the digit in the tens place be y.

Then x = 3y and the number = 10y + x

The number obtained by reversing the digits is 10x + y. If 36 is added to the number, digits interchange their places,

Therefore, we have 10y + x + 36 = 10x + y

or, 10y – y + x + 36 = 10x + y - y

or, 9y + x – 10x + 36 = 10x - 10x

or, 9y - 9x + 36 = 0 or, 9x - 9y = 36

or, 9(x - y) = 36

or, 9(x - y)/9 = 36/9

or, x - y = 4 ………. (i) Substituting the value of x = 3y in equation (i), we get

or, y = 4/2

or, y = 2 Substituting the value of y = 2 in equation (i),we get

or, x = 4 + 2

Therefore, the number becomes 26. 

3.  If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions. 

Solution:   Let the fraction be x/y. 

If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have

(x + 2)/(y + 2) = 9/10

or, 10(x + 2) = 9(y + 2) 

or, 10x + 20 = 9y + 18

or, 10x – 9y + 20 = 9y – 9y + 18

or, 10x – 9x + 20 – 20 = 18 – 20 

or, 10x – 9y = -2 ………. (i)  If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have 

(x – 3)/(y – 3) = 4/5

or, 5(x – 3) = 4(y – 3) 

or, 5x – 15 = 4y – 12

or, 5x – 4y – 15 = 4y – 4y – 12 

or, 5x – 4y – 15 + 15 = – 12 + 15

or, 5x – 4y = 3 ………. (ii) 

So, we have 10x – 9y = – 2 ………. (iii) 

and 5x – 4y = 3 ………. (iv)  Multiplying both sided of equation (iv) by 2, we get

10x – 8y = 6 ………. (v) 

Now, solving equation (iii) and (v) , we get

10x – 9y = -2

10x – 8y =  6         - y = - 8

          y = 8 

Substituting the value of y in equation (iv) 

5x – 4 × (8) = 3

5x – 32 = 3

5x – 32 + 32 = 3 + 32

Therefore, fraction becomes 7/8. 4.  If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son.  Solution:  Let father’s age be x years

Son’s ages = y years

Then 2y + x = 56 …………… (i) 

And 2x + y = 82 …………… (ii)  Multiplying equation (i) by 2, (2y + x = 56 …………… × 2)we get

or, 3y/3 = 30/3

or, y = 30/3

or, y = 10 (solution (ii) and (iii) by subtraction) Substituting the value of y in equation (i), we get;

2 × 10 + x = 56

or, 20 + x = 56

or, 20 – 20 + x = 56 – 20

or, x = 56 – 20

5. Two pens and one eraser cost Rs. 35 and 3 pencil and four erasers cost Rs. 65. Find the cost of pencil and eraser separately. Solution: Let the cost of pen = x and the cost of eraser = y

Then 2x + y = 35 ……………(i)

And 3x + 4y = 65 ……………(ii) Multiplying equation (i) by 4,

Subtracting (iii) and (ii), we get;

or, 5x/5 = 75/5

or, x = 75/5

or, x = 15 Substituting the value of x = 15 in equation (i) 2x + y = 35 we get;

or, 2 × 15 + y = 35

or, 30 + y = 35

or, y = 35 – 30

Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.

●   Simultaneous Linear Equations

Simultaneous Linear Equations

Comparison Method

Elimination Method

Substitution Method

Cross-Multiplication Method

Solvability of Linear Simultaneous Equations

Pairs of Equations

Practice Test on Word Problems Involving Simultaneous Linear Equations

●   Simultaneous Linear Equations - Worksheets

Worksheet on Simultaneous Linear Equations

Worksheet on Problems on Simultaneous Linear Equations

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15 Simultaneous Equations Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included

Beki Christian

Simultaneous equations questions involve systems of equations in which there are two or more unknowns. When solving simultaneous equations, we are finding solutions which work for all of the equations in the system. Simultaneous equations can be solved graphically or algebraically.

Here, you will find a selection of simultaneous equations questions of varying difficulty, from questions suitable for KS3 students through to some tricky GCSE-style exam questions! The focus will be on word problems and problem solving questions.

GCSE MATHS 2024: STAY UP TO DATE Join our email list to stay up to date with the latest news, revision lists and resources for GCSE maths 2024. We’re analysing each paper during the course of the 2024 GCSEs in order to identify the key topic areas to focus on for your revision. Thursday 16th May 2024: GCSE Maths Paper 1 2024 Analysis & Revision Topic List Monday 3rd June 2024: GCSE Maths Paper 2 2024 Analysis & Revision Topic List Monday 10th June 2024: GCSE Maths Paper 3 2024 Analysis GCSE 2024 dates GCSE 2024 results (when published) GCSE results 2023

How to solve simultaneous equations

In KS3 and KS4, we work with systems with two equations and two unknowns. There are three methods for solving simultaneous equations : graphically, by elimination or by substitution.

Solving simultaneous equations graphically

If we have two simultaneous equations, we can solve them by plotting their graphs. Once plotted, the solution or solutions will be the point or points of intersection of the two lines. 

15 Simultaneous Equations Questions And Practice Problems (KS3 & KS4) Worksheet

Download this free worksheet on simultaneous equations. This set of 15 simultaneous equations questions and answer key will help you prepare for GCSE Maths

Solving simultaneous equations by elimination

When solving simultaneous equations algebraically, elimination is a great method to use if we have two linear equations. Elimination works by making the coefficient of one of the variables the same in each equation and then subtracting one equation from the other to eliminate that variable.

For example, let’s solve the simultaneous equations

Step 1 : Use multiplication to make the coefficient of one of the variables the same in both equations.

Multiplying the first equation by 3, we get 9a+6b=57.

Multiplying the second equation by 2, we get 8a+6b=54.

The coefficient of b for both equations is now 6.

Step 2 : Subtract one equation from the other.

Step 3 : Substitute into one of the original equations to work out the other value.

Step 4 : Check your answer by substituting both values into the other original equation.

This works, so our solution is correct.

a=3 and b=5

Solving simultaneous equations by substitution

Substitution is a useful method to use when one or more of the equations is not linear. Substitution works by making one of the variables the subject of one of the equations and then substituting this into the other equation.

Step 1 : Make one of the variables the subject of one of the equations.

Rearranging this we get

Step 2 : Substitute this into the other equation.

Step 3 : Solve the equation.

Step 4 : Substitute into one of the original equations to find the other value.

The solutions are x=0 and y=-10 or x=6 and y=8 .

Step 5 : Check your answer by substituting into the other original equation.

These both equal 100 , so our solutions are correct.

KS3 simultaneous equations questions

Simultaneous equations will first be covered towards the end of KS3. To begin with, students are taught how to solve simultaneous equations graphically and are introduced to the method of elimination.

Solving simultaneous equations questions graphically

1. The following graph shows the straight lines x+y=4 and y=2x-5.

Use the graph to find the coordinates of the point which satisfies both of the equations

The equations have been plotted on the graph.

The solution to the simultaneous equations is the point at which the lines meet: x = 3, \ y = 1.

2. The following graph shows the line y=0.5x+3.

Use the graph to solve the simultaneous equations

We can plot the line y=4

Then the solution to the simultaneous equations is the point of intersection of the two lines: x=2, \ y=4.

Linear simultaneous equations

3. The difference between two numbers, m and n , is 6 . The sum of the two numbers is 22. For their difference, we can write the equation

Write an equation for their sum.

Solve the pair of simultaneous equations to find the values of m and n.

For the sum, the equation is m+n=22.

We can solve these equations by elimination.

The coefficient of m is 1 in both equations.

Subtracting them we get

\begin{aligned} m+n&=22\\ -~m-n&=6\\ \hline 2n&=16 \end{aligned}  

Therefore n=16 \div 2=8

Substituting back into one of the original equations

m + n = 22  

  We can check our answer by substituting into the other equation   \begin{aligned} &m-n=6 \\ &14-8=6 \end{aligned}   This works, so our answer is correct.   The two numbers are 14 and 8.

4. The cost of 2 apples and 3 bananas is 90p . The cost of 3 apples and 1 banana is 65p . Find the cost of 1 apple and 1 banana.

We can write two equations here

  \begin{aligned} &2a + 3b = 90\\ &3a + b = 65 \end{aligned}

We need to make the coefficients of either a or b the same.

Multiplying the second equation by 3 we get 9a + 3b = 195.

Subtracting one equation from the other

Therefore a=105\div7=15

Substituting into one of the original equations

We can check by substituting into the other original equation

The cost of an apple is 15p and the cost of a banana is 20p. Therefore the cost of an apple and a banana is 35p.

KS4 simultaneous equations questions

Simultaneous equations are covered more extensively in GCSE maths.

During KS4, the method of elimination is revisited and solving equations graphically is extended to include solving pairs of simultaneous equations where one equation may be a quadratic equation or another non-linear equation such as the equation of a circle. The method of substitution is also introduced as a way of solving quadratic simultaneous equations algebraically. 

Simultaneous equations is a part of the national curriculum and is examined by all exam boards including Edexcel, AQA and OCR. The questions included below are great practice for students working towards GCSEs and many are of a similar style to those found on past papers.

Simultaneous equations also feature at A Level, so it is important for those wanting to study maths at KS5 to be confident in each of the methods discussed here.

5. Lucy has £15 pocket money saved up. Lucy gets another £2 each week.

Plot a line on the axis below to show how Lucy’s money will increase over time.

Charlie has no money saved. Charlie gets £5 pocket money each week.

Plot a line showing Charlie’s money on the same set of axes.

Use your lines to determine how many weeks it will be until Lucy and Charlie have the same amount of money and the amount they will each have at that time.

8 weeks £40

3 weeks £21

5 weeks £20

5 weeks £25

The lines intersect at 5 weeks and £25.

Therefore, Lucy and Charlie will have the same amount of money in 5 weeks and they will both have £25.

6. The following diagram shows the line y=x^{2}-1.

On the same set of axes, draw the graph of y=3x-1.

Use your graph to find the two solutions to the simultaneous equations y=x^{2}-1 and y=3x-1.

x = 2 and y = 3 or x = -2 and y = 3

x = 0 and y = -1 or x = 3 and y = 8

x = 3 and y = 8

3 = 0 and y = 0 or y = 3 and y = 9

The points of intersection are (0, \ -1) and (3, 8) , so the solutions to the simultaneous equations are x = 0 and y =-1 or x = 3 and y = 8.

7. Fiona has a box of chocolates containing 12 identical chocolates. The total weight of the box and the chocolates is 294g.

Fiona eats 7 of the chocolates and the total weight of the box and chocolates decreases to 210g . What is the weight of the box?

If we call the weight of the box B and the weight of each chocolate C, then we can write two equations:

We can solve these using the elimination method.

The coefficient of B is the same in both equations, so we can go ahead and subtract one from the other:

Therefore C=84\div 7=12  

Substituting this into one of the original equations   \begin{aligned} B+5C&=210\\ B+5\times 12&=210\\ B+60&=210\\ B&=150 \end{aligned}

  We can check this by substituting these values into the other original equation

  B + 12C = 294   150+12 \times 12=150+144=294   This works, so our answer is correct.   The box weighs 150g and the chocolates weigh 12g each.

8. 600 tickets to a village fair were sold. Adult tickets were sold for £5 and child tickets were sold for £3.

A total of £2500 was made from ticket sales. Work out the number of adult tickets and the number of child tickets sold.

300 adult tickets, 200 child tickets

450 adult tickets, 150 child tickets

200 adult tickets, 500 child tickets

350 adult tickets, 250 child tickets

We can write two equations.

If we call the number of adult tickets A and the number of child tickets C then

Multiplying the first equation by 3 we get 3A + 3C = 1800.

Therefore A=700 \div 2=350  

Substituting this into one of the original equations

  \begin{aligned} A+C&=600\\ 350+C&=600\\ C&=250 \end{aligned}

We can check our answer by substituting into the other original equation

This works, so our answer is correct.

350 adult tickets and 250 child tickets were sold.

9. Write and solve two simultaneous equations based on the puzzle below. Use your solution to find the value of each symbol in the puzzle and hence find the missing total.

There is one row and one column containing only lightning and sun symbols.

From these we can write the following equations

Multiplying the first equation by 2 gives us 6L + 2S = 44.

Subtracting one of these from the other

Therefore L=20\div 4=5  

  \begin{aligned} 3L+S&=22\\ 3 \times 5 + S&=22\\ 15+S&=22\\ S&=7 \end{aligned}

Using a different column: 2L+2M=28.

The lightning symbol is worth 5 , the sun symbol is worth 7 and the moon symbol is worth 9.

We can check this by trying different rows or columns.

The missing total is 5 + 7 + 9 + 9 = 30.

10. Find the perimeter of the following rectangle

Since it is a rectangle, we can write two equations

Firstly, rearranging each equation we get

Next, we multiply the first equation by 5 and the second equation by 3 to make the coefficients of y the same

  Subtracting one from the other

  \begin{aligned} &10x-15y&=5\\ -~&9x-15y&=0\\ \hline &x&=5 \end{aligned}

  \begin{aligned} 2x-3y&=1\\ 2\times 5-3y&=1\\ 10-3y&=1\\ 3y&=9\\ y&=3 \end{aligned}

We can check our answer by substituting these values into the other original equation

We can now calculate the length and width of the rectangle.

Non-linear simultaneous equations questions

11. The area of a rectangle is 48cm^2 . The perimeter of the rectangle is 32cm . Write two equations using this information and solve them simultaneously to find the length and width of the rectangle.

Length 12cm , width 4cm

Length 24cm , width 2cm

Length 6cm , width 10cm

Length 8cm , width 6cm

If we call the length of the rectangle a and the width b then

We can solve these using the substitution method.

Firstly, we rearrange one equation to make one of the variables the subject

Then we substitute this into the other equation

b = 4 or b = 12

Substituting these back into one of the original equations

If b = 4, \ 4a = 48 so a = 12

If b = 12, \ 12a = 48 so a = 4

The side lengths are 4 and 12.

Checking this for the perimeter

This works, so our answer is correct

12. The sum of two numbers is 16. The difference between the squares of the two numbers is 32. Find the difference between the two numbers.

From this information we can write the equations

We can solve these using substitution.

First, make one of the variables the subject of one of the equations

Next, substitute this into the other equation

We can check this by substituting into the other original equation

The difference between a and b is 9-7 = 2.

13. A circle has equation x^{2}+y^{2}=25 and a line has equation y=3x-5 . The circle and the line intersect at the points A and B. Find the coordinates of the points A and B.

(0, -5) and (3, 4)

(0, 5) and (3, 4)

(3, 4) and (10, 25)

(-3, -4) and (0, -5)

The coordinates of the points of intersection are the solutions to the simultaneous equations.

We know that y = 3x-5, so substituting this into the other equation

Substituting these values into one of the original equations

The points of intersection are (0, -5) and (3, 4).

We can check these values by substituting them into the equation for the circle

Both solutions work, so our answer is correct.

Hard GCSE simultaneous equations questions

14. The lines y=2x^{2}-13x+15 and x-y+3=0 intersect at the points P and Q. Find the length of the line PQ.

To find the points of intersection of the lines, we solve the equations simultaneously.

We can do this using substitution.

First, rearrange one equation so that one of the variables is the subject

Now, substitute into the other equation

Substituting these into one of the original equations

When x = 1, \ y = 4 When x = 6, \ y = 9

The points of intersection are (1, 4) and (6, 9).

To find the length of the line PQ, we use Pythagoras theorem

The length of the line is 5 \sqrt{2}

15. Solve the simultaneous equations

x = 2 and y = 7 or x = 5 and y = 1

\frac{3}{2} and y = 8 or x = 2 and x = 7

x = 1 and y = 9 or x = 5 and y = 1

\frac{16}{3} and \frac{1}{3} or x = 2 and y = 7

This can be solved using substitution.

x=\frac{16}{3} or x=2

When x=\frac{16}{3}:

The solutions are x=\frac{16}{3} and y=\frac{1}{3} or x=2 and y=7 .

We can check these solutions by substituting them back into the other original equation

Both solutions work, so our answers are correct.

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Simultaneous Equations Practice Questions – GCSE Maths

Try these simultaneous equations practice questions to make sure you’re ready for your GCSE maths exams. These questions get steadily more difficult and include both the elimination and substitution methods. Check out our simultaneous equations revision guide if you need to revise the methods before trying these questions. You can also contact us directly to ask any questions and book a one-to-one lesson with an expert maths tutors. Book your free GCSE maths consultation through the button below.

Elimination Method

Solve the following simultaneous equations using the elimination method: 

Do you need some more help before continuing? Try a lesson with our expert online tutors to get personalised support and guidance.

The next set of questions are a little trickier. Just keep using the same elimination method and you’ll be fine. Watch out for the different positive and negative terms in the questions.

Simultaneous Equations Practice Questions – Substitution Method

Now use the substitution method to solve these simultaneous equations. Again, remember to take a look at our revision guide on this topic if you get stuck with this method.

Take a look at the answers to these questions below. How did you get on? Simultaneous equations can appear quite difficult at first, but once you’ve practised the methods it should get much easier. You might find BBC bitesize quite useful again to keep practising this method.

You can revise many other GCSE maths topics online and for free with our revision resources . If you need any further help then remember to contact us directly with any questions. You can use the button below to book a free GCSE maths consultation. We’ll offer some free personalised advice for you and you can arrange a one-to-one lesson with one of our tutors.

• x = 2 and y = 3
• y=1 and x = 5
• x = 3 and y = 4
• a = 6 and b = 4
• a = 2 and b = 7
• l = -1 and m = 5
• y = 9 and x = -3
• a = 3 and b = 1
• x = 4 and y = 3
• x = 6 and y = 11
• y = 3 and x = 5
• x = 4 and y = 1
• x = 6 and y = 4
• a = 5 and b = 3
• y = 2 and x = 10

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Solve the Simultaneous Equations Algebra Problems Worksheets Math

Subject: Mathematics

Age range: 9 - 14

Resource type: Worksheet/Activity

Last updated

24 April 2024

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