WORD PROBLEMS ON SIMULTANEOUS EQUATIONS

Problem 1 :

In a school, there are 880 students in total. If there is 20% more boys than girls, find the number of boys and girls in the school.

Let "x" and "y" be the no. of boys and girls respectively.

Then, we have

x + y  =  880 ------(1)

Given :  There is 20% more boys than girls.

So, we have

x  =  120% of y

x  =  1.2y  ------(2)

Substitute x  =  1.2y  in (1)

1.2y  +  y  =  880

2.2y  =  880

Divide by 2.2

y  =  400

Substitute y  =  400  in (2).

x  =  1.2  ⋅  400

x  =  480

So, the number of boys in the school is 480 and girls is 400. 

Problem 2 :

Mr. Lenin invests some amount in deposit A and some amount in deposit B. The total money invested is $2500. He gets 10% income on deposit A and 20% income on deposit B. If the total income earned be $380, find the amount invested in A and B separately    

Let "x" and "y" be the amounts invested in A and B respectively. 

x + y  =  2500

Subtract y from both sides.

x  =  2500 - y ------(1)

Given :  Mr. Lenin gets 10% income on deposit A, 20% income on deposit B and the total income earned be $380.

10% of x  +  20% of y  =  380

0.1x + 0.2y  =  380

Multiply both sides by 10.

x + 2y  =  3800 -------(2)

From (1), we can plug x  =  2500 - y in (2).

(2)------> 2500 - y + 2y  =  3800

2500 + y  =  3800

Subtract 2500 from both sides.

y  =  1300

Substitute y  =  1300 in (1).

(1)------> x  =  2500 - 1300

x  =  1200

So, the amount invested in the deposit A is $1200 and B is $1300.

Problem 3 :

The sum of two numbers is 209. If one number is 7 less than two times of the other, then find the two numbers.   

Let "x" and "y" be the two numbers. 

Given :  The sum of two numbers is 209.

x + y  =  209 ------(1)

Given :  One number is 7 less than two times of the other.

x  =  2y - 7 -------(2)

From (2), we can plug x  =  2y - 7 in (1). 

(1)------> 2y - 7 + y  =  209

3y - 7  =  209

Add 7 to both sides.

3y  =  216

Divide both sides by 3. 

y  =  72

Substitute y  =  72 in (2). 

x  =  2  ⋅ 72 - 7

x  =  144 - 7

x  =  137

So, the two numbers are 137 and 72.  

Problem 4 :

The perimeter of the rectangle is 158 cm. If the length is 7 more than 3 times the width, find the area of the rectangle. 

Let "x" and "y" be the length and width of the rectangle respectively.  

Given :  Perimeter of the rectangle is 158 cm.

2x + 2y  =  158

Divide both sides by 2.

x + y  =  79 ------(1) 

Given :  T he length is 7 more than 3 times the width.

So, we have 

x  =  3y + 7 --------(2)

From (2), we can plug x  =  3y + 7 in (1).

(1)------> 3y + 7 + y  =  79

4y + 7  =  79

Subtract 7 from both sides.

4y  =  72

Divide both sides by 4.

y  =  18

Substitute   y  =  18  in (2).

x  =  3  ⋅  18 + 7

x  =  61

Therefore, the length of the rectangle is 61 cm and width is 18 cm.

Then, area of the rectangle is

=  length  ⋅  width 

=  61  ⋅  18 

=  1098  

So, the area of the rectangle is  1098 sq.cm. 

Problem 5 :

A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. Total profit earned on these two products is $43.  The sum of the cost prices of two products is $150. Find the cost price of each product. 

Let "x" and "y" be the cost prices of the two products.

Then, we have  

x + y  =  150

Subtract y from both sides. 

x  =  150 - y -------(1)

Let the trader gain one third of the cost price as profit on the product whose cost price is x.

Then, profit on the product whose cost price is x :  

=  1/3  ⋅  x  

=  x / 3

Let the trader gain one fourth of the cost price as profit on the product whose cost price is y.

Then, profit on the product whose cost price is y :  

=  1/4  ⋅ y   

=  y / 4

Given :  Total profit earned on these two products is $43.

x/3  +  y/4  =  43

L.C.M of (3, 4) is 12.

(4x / 12)  +  (3y / 12)  =  43

(4x + 3y) / 12  =  516

Multiply both sides by 12.

4x + 3y  =  516 ------(2)

From (1), we can plug x  =  150 - y in (2). 

4(150 - y) + 3y  =  516 

600 - y  =  516

Subtract 600 from both sides.

- y  =  - 84

y  =  84

Substitute y  =  84 in (1).

(1)------> x  =  150 - 84

x  =  66

So, the cost prices of two products are $66 and $84. 

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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

linear equations

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,

problems on simultaneous equations

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

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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.

x^{2}+y = 5

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)

simultaneous equations word problems examples

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

\frac{5y}{5}=\frac{10}{5}

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:

\frac{2x}{2}=\frac{2}{2}

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.

x^{2}+y^{2}=10

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)

(4-y)^{2}+y^{2}=10

Adding same terms and rearranging above equation

2y^{2}-8y+16=10

Subtracting ’10’ on both side

2y^{2}-8y+16-10=10-10

Dividing by two ‘2’ on both sides

\frac{2y^{2}}{2}-\frac{8y}{2}+\frac{6}{2}=0

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

y^{2}-3y-y+3=0

Taking common similar terms

y(y-3)-1(y-3)=0

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.

simultaneous equations word problems examples

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.

simultaneous equations word problems examples

Dividing on both sides by ’19’

\frac{19y}{19}=\frac{133}{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

x^{2}+y^{2}=34

Rearranging equation (1), we get

Put this value of y in equation (2)

x^{2}+(8-x)^{2}=34

Collecting same terms

2x^{2}-16x+64-34=0

Dividing by 2 on both side

x^{2}-8x+15=0

Factorize the above quadratic equation

x^{2}-5x-3x+15=0

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)

6x^{2}+3y^{2}=12

Rearrange the equation (1), we will get

Put this value in equation (2)

6x^{2}+3(2-x)^{2}=12

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

x=\frac{4}{3}

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

simultaneous equations word problems examples

GCSE Maths Takeaway – Simultaneous Equation Word Problems

If I had to choose one thing you should have as a takeaway from GCSE maths i.e. one ‘poster’ topic to represent GCSE maths, I think it should be simultaneous equations – in particular word problems leading to simultaneous equations 1 .

This is because I think if you can solve a sufficiently varied set of simultaneous equation word problems, you must have a good understanding, for a sufficiently involved type of problem, of the complete ‘mathematical process’:

  • Translating a ‘real world’ problem into a mathematical model
  • Solving a mathematical model
  • Interpreting the solution(s) in the context of the ‘real world’.

Most maths questions just cover stage (2) processes of formally computing things by applying rules, often according to a set procedure, and while I think such skills are important to have and understand, it is stages (1) and (3) I’m most interested in as these are the parts that can’t easily be done by a computer algorithm, and they are also what makes maths useful.

Simultaneous equation word problems are great because they can be genuine problems that exist outside the world of formal mathematics, and yet a sophisticated formal mathematical process is required to solve them, so they give one of the first glimpses into the power of formal mathematics.

To understand something well it is important to have a good stock of examples, that is ‘complete’ in some sense. So I have tried to curate a list of my favourite simultaneous equation word problems. I have tried to choose questions such that:

  • Ideally, it is possible to imagine being faced with the problem or a similar problem
  • Ideally, the question is memorable and/or short
  • Ideally, the solution is not obvious without formally solving the equations
  • Any equations to be solved are linear
  • The calculations involved should be fairly straightforward (at least possible to do by hand)
  • Any relationship the problem requires use of should be knowable to the intuition or well known
  • Ideally, the question itself shouldn’t mention too many numbers
  • The set of questions as a whole is in some sense ‘complete’ i.e. covers ‘all kinds’ of linear simultaneous equation word problems 2
  • No two questions are of the same ‘formula’
  • There are no more than 15 questions
  • There are questions of varying difficulty

Here is the list of questions I’ve come up with. Some of them I have taken from books and the internet and others I have written myself 3 . The questions are roughly in order of increasing difficulty of forming the equations (difficulty of solving the equations is random), with some starred questions at the end. While the starred questions do not require any specific knowledge outside of GCSE maths, they may be more challenging.

  • A bat and a ball costs £1.10. The bat costs one pound more than the ball. How much does the ball cost?
  • At a community swimming pool, the sign saying the price of children’s admission and adult’s admission has disappeared and no-one remembers what the prices were. It is recorded that on Monday the pool had 12 children and 33 adults, which brought in £177, and that same week on Tuesday 42 children and 33 adults came to the pool, which brought in £207. What are the admission prices for children and adults?
  • It takes 11 hrs to knit 2 hats and 3 jumpers. It takes 4 hours to knit 1 hat and 1 jumper. How long would it take to knit 5 hats and 6 jumpers?
  • Warmsnug calculate the prices of their windows according to the area of glass used and the length of frame needed. These are two example windows and their prices: 3 x 5 costs £470, 2 x 4 costs £320. What would be the quote to do a square window of width 3?
  • A sports pitch has length 30m and width 15m. It is required that the dimensions be changed so that the length is at least three times the width and the perimeter is increased by at least 70m. What are the minimum new dimensions of the pitch?
  • A taxi company has a cost structure which is an initial fee plus a fee proportional to the distance travelled. I did one journey that costs £9.50; then I remembered I had forgotten something important and had to get another taxi home and back again (same company). The second journey cost £17.50. How much is the initial fee?
  • In a certain family, each girl has just as many sisters as brothers but each boy has twice as many sisters as brothers. How many children are there?
  • When I was 14 my mother was 41, and she is now twice as old as I am; how old am I?
  • In a two digit number, the units digit is thrice the tens digit and if 36 is added to the number the digits are interchanged. Find the number.
  • Is there a two digit number such that the units and tens digit add up to 9, and when 36 is added to the number the digits are interchanged?
  • Is there a number such that when you add it to any two digit number where the units digit is 2 higher than the tens digit, the digits are interchanged?
  • It takes a motor boat 2 minutes to travel 2 km when going with the current. When the boat is going against the current it takes 4 minutes. The current is always constant. How long would it take the boat to do the same journey in slack water, when there is no current at all?
  • There are two neighbouring countries Foo and Bar. Fooish is the primary language of Foo and Barish is the primary language of Bar, but both languages are spoken across both countries (no other language is spoken in these countries and no other country speaks these languages). Half of people who speak Barish can also speak Fooish. Only 5 percent of people who speak Fooish can also speak Barish. The total population of Foo and Bar is 21 million. How many of these people only speak Fooish?
  • * In a food-eating competition it took me 5 minutes to eat 7 burgers and 2 hotdogs. I know it takes me at least 20 seconds to eat a hotdog. Is it possible for me to eat 12 burgers and 8 hotdogs within 10 minutes (without any more practice)?
  • * Bill and Ben’s combined age is 91. Bill is now twice as old as Ben was when Bill was as old as Ben is now. How old are they?
  • * Two horsemen started out at daybreak. They travelled the same distance and arrived at their destination at the same time. A rode twice as long as B rested and B rode for three times long as A rested for. Who rode the fastest?
  • Other runner up GCSE ’poster’ topics – Pythagoras’ theorem, Coordinate geometry, Quadratic equations.  ↩
  • Don’t know if this list is ‘complete’ enough yet – might add to this list if I find other interesting examples that are sufficiently different.  ↩
  • Questions 7, 8, 10, 13, 14 are from the book Mind-bending Classic Logic Puzzles . Question 1 is from the book Thinking Fast and Slow . Question 4 is inspired by an Nrich question . Question 2 and 9a are taken from somewhere on the internet. Questions 3, 5, 6, 9b, 9c, 11, 12 were written by me.  ↩

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

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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 exam questions! The focus will be on word problems and problem solving questions.

Solving simultaneous equations graphically

Solving simultaneous equations by elimination, solving simultaneous equations by substitution, solving simultaneous equations questions graphically, linear simultaneous equations, non-linear simultaneous equations questions, hard gcse simultaneous equations questions, looking for more simultaneous equations questions and resources, looking for more ks3 and ks4 maths questions.

Free GCSE maths revision resources for schools As part of the Third Space Learning offer to schools, the personalised online GCSE maths tuition can be supplemented by hundreds of free GCSE maths revision resources from the secondary maths resources library including: – GCSE maths past papers – GCSE maths worksheets – GCSE maths questions – GCSE maths topic list

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.

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

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

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.

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.

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.

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.

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

Simultaneous equations are two or more algebraic equations that share common variables and are solved at the same time (that is, simultaneously). For example, equations x + y = 5 and x - y = 6 are simultaneous equations as they have the same unknown variables x and y and are solved simultaneously to determine the value of the variables. We can solve simultaneous equations using different methods such as substitution method, elimination method, and graphically.

In this article, we will explore the concept of simultaneous equations and learn how to solve them using different methods of solving. We shall discuss the simultaneous equations rules and also solve a few examples based on the concept for a better understanding.

What are Simultaneous Equations?

Simultaneous equations are two or more algebraic equations with the same unknown variables and the same value of the variables satisfies all such equations. This implies that the simultaneous equations have a common solution. Some of the examples of simultaneous equations are:

  • 2x - 4y = 4, 5x + 8y = 3
  • 2a - 3b + c = 9, a + b + c = 2, a - b - c = 9
  • 3x - y = 5, x - y = 4
  • a 2 + b 2 = 9, a 2 - b 2 = 16

We can solve such a set of equations using different methods. Let us discuss different methods to solve simultaneous equations in the next section.

Solving Simultaneous Equations

We use different methods to solve simultaneous equations. Some of the common methods are:

  • Substitution Method
  • Elimination Method
  • Graphical Method

Simultaneous equations can have no solution, an infinite number of solutions, or unique solutions depending upon the coefficients of the variables. We can also use the method of cross multiplication and determinant method to solve linear simultaneous equations in two variables . We can add/subtract the equations depending upon the sign of the coefficients of the variables to solve them.

To solve simultaneous equations, we need the same number of equations as the number of unknown variables involved. We shall discuss each of these methods in detail in the upcoming sections with examples to understand their applications properly.

Simultaneous Equations Rules

To solve simultaneous equations, we follow certain rules first to simplify the equations. Some of the important rules are:

  • Simplify each side of the equation first by removing the parentheses, if any.
  • Combine the  like terms .
  • Isolate the variable terms on one side of the equation.
  • Then, use the appropriate method to solve for the variable.

Solving Simultaneous Equations Using Substitution Method

Now that we have discussed different methods to solve simultaneous equations. Let us solve a few examples using the substitution method to understand it better. Consider a system of equations x + y = 4 and 2x - 3y = 9. Now, we will find the value of one variable in terms of another variable using one of the equations and substitute it into the other equation. We have

x + y = 4 --- (1)

2x - 3y = 9 --- (2)

From (1), we have

x = 4 - y --- (3)

Substituting this in (2), we get

2(4 - y) - 3y = 9

⇒ 8 - 2y - 3y = 9

⇒ 8 - 5y = 9

Isolating the variable term to one side of the equation, we have

⇒ -5y = 9 - 8

⇒ y = 1/(-5)

Substituting the value of y in (3), we have

x = 4 - (-1/5)

= (20 + 1)/5

Answer: So, the solution of the simultaneous equations x + y = 4 and 2x - 3y = 9 is x = 21/5 and y = -1/5.

Solving Simultaneous Equations By Elimination Method

To solve simultaneous equations by the elimination method, we eliminate a variable from one equation using another to find the value of the other variable. Let us solve an example to understand find the solution of simultaneous equations using the elimination method. Consider equations 2x - 5y = 3 and 3x - 2y = 5. We have

2x - 5y = 3 --- (1)

and 3x - 2y = 5 --- (2)

Here, we will eliminate the variable y, so we find the LCM of the coefficients of y. LCM (5, 2) = 10. So, multiply equation (1) by 2 and equation (2) by 5. So, we have

[ 2x - 5y = 3 ] × 2

⇒ 4x - 10y = 6 --- (3)

[ 3x - 2y = 5 ] × 5

⇒ 15x - 10y = 25 --- (4)

Now, subtracting equation (3) from (4), we have

(15x - 10y) - (4x - 10y) = 25 - 6

⇒ 15x - 10y - 4x + 10y = 19

⇒ (15x - 4x) + (-10y + 10y) = 19

⇒ 11x + 0 = 19

⇒ x = 19/11

Now, substituting this value of x in (1), we have

2(19/11) - 5y = 3

⇒ 38/11 - 5y = 3

⇒ 5y = 38/11 - 3

⇒ 5y = (38 - 33) / 11

⇒ y = 5/(11×5)

So, the solution of the simultaneous equations 2x - 5y = 3 and 3x - 2y = 5 using the elimination method is x = 19/11 and y = 1/11.

Solving Simultaneous Equations Graphically

In this section, we will learn to solve the simultaneous equations using the graphical method. We will plot the lines on the coordinate plane and then find the point of intersection of the lines to find the solution. Consider simultaneous equations x + y = 10 and x - y = 4. Now, find two points (x, y) satisfying for each equation such that the equation holds.

For x + y = 10, we have

So, we have coordinates (0, 10) and (10, 0). Plot them and join the points and plot the line x + y = 10.

For equation x - y = 4, we have

So, we have coordinates (0, -4) and (4, 0). Plot them and join the points and plot the line x - y = 4.

Solving simultaneous equations

Now, as we have plotted the two lines, find their intersecting point. The two lines x + y = 10 and x - y = 4 intersect each other at (7, 3). So, we have found the solution of the simultaneous equations x + y = 10 and x - y = 4 graphically which is x = 7 and y = 3.

Important Notes on Simultaneous Equations

  • Simultaneous equations are two or more algebraic equations that share common variables and are solved at the same time.
  • Simultaneous equations can be solved using different methods such as substitution method, elimination method, and graphically.
  • We can also use the cross multiplication and determinant method to solve simultaneous linear equations in two variables.

☛ Related Articles:

  • Solutions of a Linear Equation
  • Simultaneous Linear Equations

Simultaneous Equations Examples

Example 1: Solve the simultaneous equations 2x - y = 5 and y - 4x = 1 using the appropriate method.

Solution: To solve 2x - y = 5 and y - 4x = 1, we will use the elimination method as it is easy to eliminate the variable y by adding the two equations. So, we have

2x - y = 5 --- (1)

y - 4x = 1 --- (2)

Adding (1) and (2), we get

(2x - y) + (y - 4x) = 5 + 1

⇒ 2x - y + y - 4x = 6

Substitute this value of x in (1)

2(-3) - y = 5

⇒ -6 - y = 5

⇒ y = -6 - 5

Answer: Solution of simultaneous equations 2x - y = 5 and y - 4x = 1 is x = -3 and y = -11.

Example 2: Find the solution of the simultaneous equations 2x - 4y + z = 2, x + 5y - 3z = 7, 3x + 2y - z = 10 using the substitution method.

Solution: We have

2x - 4y + z = 2 --- (1)

x + 5y - 3z = 7 --- (2)

3x + 2y - z = 10 --- (3)

z = 2 - 2x + 4y

Substituting this value of z in (2) and (3),

x + 5y - 3(2 - 2x + 4y) = 7

⇒ x + 5y - 6 + 6x - 12y = 7

⇒ 7x - 7y = 13 --- (4)

3x + 2y - (2 - 2x + 4y) = 10

3x + 2y - 2 + 2x - 4y = 10

⇒ 5x - 2y = 12 --- (5)

Now, solving the two-variable equations (4) and (5), multiply (4) by 2 and (5) by 7, we have

[7x - 7y = 13 ] × 2 and [5x - 2y = 12 ] × 7

⇒ 14x - 14y = 26 and 35x - 14y = 84

Now, subtracting the above two equations, we have

(14x - 14y) - (35x - 14y)= 26 - 84

⇒ 14x - 35x - 14y + 14y = -58

⇒ -21x = -58

⇒ x = 58/21 --- (A)

Substitute the value of x in (5)

5(58/21) - 2y = 12

⇒ 290/21 - 2y = 12

⇒ 2y = 290/21 - 12

= (290 - 252)/21

⇒ y = 19/21 --- (B)

Substituting the values of x and y in z = 2 - 2x + 4y, we have

z = 2 - 2(58/21) + 4(19/21)

= (42 - 116 + 76)/21

= 2/21 --- (C)

From (A), (B), (C), we have x = 58/21, y = 19/21, and z = 2/21

Answer: Solution is x = 58/21, y = 19/21, and z = 2/21.

Example 3: Find the solution of simultaneous equations x - y = 10 and 2x + y = 9.

Solution: We will solve the given equations using the elimination method.

Adding x - y = 10 and 2x + y = 9, we have

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

⇒ x + 2x - y + y = 19

So, we have

19/3 - y = 10

⇒ y = 19/3 - 10

= (19 - 30)/3

Answer: The solution of x - y = 10 and 2x + y = 9 is x = 19/3 and y = -11/3.

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FAQs on Simultaneous Equations

Simultaneous equations are two or more algebraic equations that share common variables and are solved at the same time (that is, simultaneously).

How to Solve Simultaneous Equations?

What is the substitution method in simultaneous equations.

According to the substitution method, we obtain the value of one variable in terms of another and then substitute that into another equation to find the value of the other variable.

What is the Rule for Simultaneous Equations?

Some of the important rules of simultaneous equations are:

  • Combine the like terms.

What are Linear Simultaneous Equations?

Linear simultaneous equations refer to simultaneous equations where the degree of the variables is one.

How to Solve 3 Simultaneous Equations?

We can solve 3 simultaneous equations using various methods such as:

It also depends upon the number of variables involved.

What are the Three Methods to Solve Simultaneous Equations?

The three methods to solve simultaneous equations are:

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Systems of equations word problems

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  • System of equations word problem: infinite solutions
  • Systems of equations word problems (with zero and infinite solutions)
  • Systems of equations with elimination: TV & DVD
  • Systems of equations with elimination: apples and oranges
  • Systems of equations with substitution: coins
  • Systems of equations with elimination: coffee and croissants
  • Systems of equations: FAQ
  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
  • an exact decimal, like 0.75 ‍  
  • a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Solving Simultaneous Equations: Worksheets with Answers

Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you. And best of all they all (well, most!) come with answers.

Mathster keyboard_arrow_up Back to Top

Mathster is a fantastic resource for creating online and paper-based assessments and homeworks. They have kindly allowed me to create 3 editable versions of each worksheet, complete with answers.

Corbett Maths keyboard_arrow_up Back to Top

Corbett Maths offers outstanding, original exam style questions on any topic, as well as videos, past papers and 5-a-day. It really is one of the very best websites around.

Mathematics

Simultaneous equations - problem solving.

Equations that must be solved at the same time are simultaneous equations. In this tutorial, I am going to explain how a pair of simultaneous equations are formed from a given problem and then solve it. If you go through the worked examples in the order given, you will have enough experience to deal with any question involving simultaneous equations.

I assume you know how to solve simultaneous equations, either by elimination method or substitution method. If you want to brush up on your skill, please go through this tutorial first, and then use this one. If you think you need a bit more practise use the following question generator for the task.

Algebra Equation Generator

You can generate as many questions as you want with the following programme, along with the answers. First generate the question, then work them out and check with the answer.

Simultaneous Equations Generator

The sum of two numbers is 6 and the difference is 2. Find the numbers. Let the numbers be x and y. x + y = 6 1 x - y = 2 2 1 + 2 => 2x = 8 x = 4 Sub in 1 => 4 + y = 6 y = 2 The numbers are x = 4 and y = 2.

The sum of two books and a pencil is £6.00. The difference of cost between 3 books and 2 pencils is £2.00. Find the cost of a book and a pencil. Let the cost of a book be x and that of a pencil be y. 2x + y = 6 1 3x - 2y = 2 2 1 X 2 => 4x + 2y = 12 3 2 + 3 => 7x = 14 x = 2 Sub in 1 4 + y = 6 y = 2 The cost of a book and a pencil is £2.00 each.

If I double a number and add three times a second number, the answer is 1. If I multiply the first number by 3 and take away twice the second number, the answer is 8. Find the numbers. Let the numbers be x and y. 2x + 3y = 1 1 3x - 2y = 8 2 From 1 - 3y => 2x = (1 - 3y) x = (1 - 3y)/2 Sub in 2 => 3(1 - 3y)/2 - 2y = 8 X 2 => 3(1 - 3y) - 4y = 16 3 - 9y - 4y = 16 3 - 13y = 16 -3 => -13y = 13 y = -1 Sub in 1 => 2x - 3 = 1 + 3 => 2x = 4 x = 2 The numbers are x = 2 and y = -1.

The sum of twice the cost of a box biscuits and the cost of a box chocolates is £8.00. The difference between the cost of box of chocolates and the box of biscuits is £1.00. Find the cost of each. Let the cost of the box chocolates and the box of biscuits be y and x respectively. 2x + y = 8 1 y -x = 1 2 1 => y = 8 - 2x Sub in 2 => y = 8 - 2x - x = 1 8 - 3x = 1 -8 => -3x = -9 :- -3 => x = 3 Sub in 1 => 6 + y = 8 -6 => y = 2 The cost of box of chocolates =£2.00 and that of biscuits = £3.00.

Simultaneous Quadratic Equations

These equations are simultaneous as there are two unknowns in them; since one of the unknown is in quadratic form, they are quadratic too. Therefore, these equations have two sets of solutions, one for each unknown.

The equation of a circle is x 2 + y 2 = 45. It intersects with, y = 2x, at two points. Find the coordinates of the points of intersection. x 2 + y 2 = 45 1 y = 2x 2 Sub y in 1 => x 2 + 4x 2 = 45 5x 2 = 45 :- 5 => x 2 = 9 x = ± 3 Sub in 2 => y = ±6 Solutions: (3,6); (-3,-6)

Solving Simultaneous Equations - fully interactive

The following applet help you solve simultaneous equations instantly. Just type in the two equations into the text boxes, exactly in the form shown, and then press enter. You may move the grid to see the point of intersection of the two lines.

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simultaneous equations word problems examples

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Questions for Practice

Now, in order to complement what you have just learnt, work out the following questions:

  • The sum of two numbers is 18. The difference is 4. Find the numbers.
  • A straight line passes through (3 , -4) and (5 , 8). Find the equation of the line.
  • The cost of 3 DVD's and 4 CD's is £62.00. The cost of 4 DVD's and 3 CD's is £64.00. Find the cost of each.
  • A diver swims downstream a distance of 40 miles in two hours. If he swims upstream, he can only move 16 miles during the same time. Find his swimming speed and the speed of the river.
  • Half the difference between two numbers is 8. The average of the numbers is 12. Find the numbers.
  • Nicole has 23 notes of £20 and £5 in her hand bag. The amount of money she has in the bag is £340.00. Find the number of notes of each type.
  • There are two angles on a straight line. One angle is 15 more than twice the other. Find the size of each angle.
  • The sum of ages of an uncle and his nephew two years ago was 40. In two years time from now, the age of the uncle will be three times that of his nephew by then. Find their ages in 7 years time.
  • The curve, px 2 + qx, passes through (3,6) and (1,-2). Find the values of p and q.
  • The numerator of a fraction is 3 smaller than its denominator. If both the numerator and denominator are increased by by 1, the fraction is 5/8. Find the original fraction.
  • A car covers a distance of 220 miles at 3o mph and 20 mph respectively. Find the time taken for each part of the journey.
  • The curve, px 2 + qx + r, passes through (0,5) and (2,5) and (3,11). Find the values of p, q and r.

Move the mouse over, just below this, to see the answers:

  • y = 6x - 22

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Word Problems on Simultaneous Linear Equations | Simultaneous Linear Equations Questions

By solving the system of linear equations in two variables, you will get an ordered pair having x coordinate and y coordinate values (x, y) that satisfies both equations. Here those simultaneous linear equations are in the form of word problems. So, you can solve different word problems with the help of linear equations.

We have already learned some steps and methods to solve the simultaneous linear equations in two variables. Assume the unknown quantities in the question as x, y variables and represent them in the form of a linear equation according to the condition mentioned in the question. And follow the methods to solve the formed system of linear equations to get the values of unknown quantities. We have also provided simultaneous equations problems with solutions that help you to grasp the concept.

Simultaneous Equations Word Problems

One number is greater than thrice the other number by 6. If 4 times the smaller number exceeds the greater by 7, find the numbers?

Let x, y be the two numbers

So that x > y

Given that, one number is greater than thrice the other number by 6.

Then, we can write it as

x = 3y + 6 ——— (i)

According to the question,

4 times the smaller number exceeds the greater by 7.

4y – x = 7 ——— (ii)

Substitute (3y + 6) in equation (ii).

4y – (3y + 6) = 7

4y – 3y – 6 = 7

y – 6 = 7

Substitute y = 13 in equation (i)

x = 3(13) + 6

So, the two numbers are 45, 13.

The sum of two numbers is 25 and their difference is 5. Find those two numbers?

Let the two numbers be x and y.

The sum of two numbers is 25

x + y = 25 —— (i)

The difference of numbers is 5.

x – y = 5 —– (ii)

Add both the equations (i) & (ii)

x + y + x – y = 25 + 5

Putting x = 15 in equation (i)

15 + y = 25

y = 25 – 15

So, the two numbers are 15, 10.

The sum of two numbers is 50. If the larger is doubled and the smaller is tripled, the difference is 25. Find the two numbers.

Let the two numbers be x and y

According to the given question,

The sum of two numbers is 50.

x + y = 50 —— (i)

The larger number is doubled and the smaller number is tripled, the difference is 25.

2x – 3y = 25 —— (ii)

Multiply the first equation by 3.

3(x + y) = 3 x 50

3x + 3y = 150 —— (iii)

Add equation (ii) and equation (iii)

2x – 3y + 3x + 3y = 25 + 150

Substituting the value of x in equation (i)

35 + y = 50

y = 50 – 35

Hence the two numbers are 35, 15.

The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. 70 per kg and the other costing Rs. 85 per kg. They estimated that 34 kg of sweets were needed. If the total money spent on sweets was Rs. 2500, find how much sweets of each kind they purchased?

Let the quantity of sweets purchased be x kg which cost Rs. 70 per kg and sweets purchased y kg which cost Rs. 85 per kg.

x + y = 34 ——- (i)

70x + 85y = 2500 ——– (ii)

Multiplying the equation (i) by 70, we get

70(x + y) = 34 x 70

70x + 70y = 2380 —– (iii)

Subtracting equation (iii) from equation (ii), we get

70x + 70y – (70x + 85y) = 2380 – 2500

70x + 70y – 70x – 85y = -120

-15y = -120

Substitute y = 1.6 in equation (i)

x + 1.6 = 34

x = 34 – 1.6

Hence, the sweets purchased 1 kg 600 grams which cost Rs. 85 per kg and 32 kg 400 grams which cost Rs. 70 per Kg.

A two-digit number is eight times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number?

Let the two-digit number be xy.

Given that, the two digit number is eight times the sum of its digits.

xy = 8(x + y) —– (i)

In the two digit number xy, x is in the tens position and y is in ones position.

xy = 10 . x + 1 . y

xy = 10x + y ——- (ii)

Substitute equation (ii) in equation (i)

10x + y = 8(x + y)

10x + y = 8x + 8y

10x – 8x = 8y – y

2x – 7y = 0 —– (iii)

The number formed by reversing the digits 18 less than the given number.

xy – yx = 18

(10 . x + 1 . y) – (10 . y + 1 . x) = 18

(10x + y) – (10y + x) = 18

10x + y – 10y – x = 18

9x – 9y = 18

9(x – y) = 18

(x – y) = 18/9

x – y = 2 —— (iv)

Multiply equation (iv) by 2.

2(x – y) = 2 x 2

2x – 2y = 4 —— (v)

Subtract equation (v) from equation (iii)

(2x – 2y) – (2x – 7y) = 4 – 0

2x – 2y – 2x + 7y = 4

Substitute y = 4/5 in equation (iv)

x – 4/5 = 2

x = 2 + 4/5

x = (10 + 4)/5

So, the required two digit number is 144/5.

If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to ⅗ and if the numerator and denominator are each diminished by 1, the fraction becomes equal to ⅔, find the fraction.

Let the fraction be x/y.

According the given question,

(x + 2) / (y + 1) = ⅗

5(x + 2) = 3(y + 1)

5x + 10 = 3y + 3

5x – 3y = 3 – 10

5x – 3y = -7 —— (i)

(x – 1) / (y – 1) = ⅔

3(x – 1) = 2(y – 1)

3x – 3 = 2y – 2

3x – 2y = -2 + 3

3x – 2y = 1 —— (ii)

Multiply the equation (i) by 2, equation (ii) by 3.

2(5x – 3y) = 2 x -7 ⇒ 10x – 6y = -14

3(3x – 2y) = 3 x 1 ⇒ 9x – 6y = 3

(10x – 6y) – (9x – 6y) = -14 – 3

10x – 6y – 9x + 6y = -17

Substitute x = -17 in equation (i)

5(-17) – 3y = -7

-85 – 3y = -7

-3y = -7 + 85

y = -78 / 3

Therefore, the fraction is 17/26.

If four times the age of the son is added to the age of the father, the sum is 64. But if twice the age of the father is added to the age of the son, the sum is 82. Find the ages of father and son?

Let the father’s age be x years, the son’s age be y years.

4y + x = 64 —– (i)

2x + y = 82 —— (ii)

Multiply the equation (i) by 2

2(x + 4y) = 64 x 2

2x + 8y = 128

Subtract 2x + 8y = 128 from equation (ii)

(2x + 8y) – (2x + y) = 128 – 82

2x + 8y – 2x – y = 46

y = 6 years 7 months

Substituting y = 46/7 in equation (ii)

2x + 46/7 = 82

2x = 82 – 46/7

2x = (574 – 64)/7

x = 36 years 4 months

Therefore, the father’s age is 36 years 4 months, the son’s age is 6 years 7 months.

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Examples of simultaneous equations word problems (step by step)

  • November 3, 2021
  • Abdulrazaq Yahaya

Examples of simultaneous equations word problems and how to solve very fast

Simultaneous equations are two or more algebraic equations that share variables, e.g. x and y. In today’s article, I will show you different examples of how to solve any given type of simultaneous equations word problems step by step very, very fast.

We can consider each equation in a simultaneous equation as a function which when displayed graphically may intersect at a specific point. This point of intersection gives the solution to the simultaneous equations.

Examples of simultaneous equations word problems

Some examples of word problems of simultaneous equations are:

  • A jet weighs 225Kg when it’s half full of passengers and 380Kg when it’s three quarters full of passengers. Calculate the weight of the jet when it’s empty
  • In Obafemi Awolowo University, two students went into the bookshop and bought the books. One student bought 5 different Chemistry text books and 3 different Maths text books at a price of N150 while the second student bought 2 different Chemistry text books and 1 Maths text books at a price of N90. Calculate the price of Chemistry text books and that of Maths text books.

Read this also: top best reasons why many people didn’t gain admission into the university to study their dream course in Nigeria.

Solution to #1

Solutions of number one of the example of simultaneous equation words problems listed above are as follows

Let J and P represent the weight of jet and passengers respectively.

Now, at the time when the jet is half full of passengers, the best equation we can use is J + 1/2(P) = 225Kg when the best equation we can use when the jet contains three quarters of passengers is J + 3/4(P) = 380Kg.

Meanwhile, the weight of the jet when it’s not filled with anything is unknown and we’re being asked to find it out from the example of simultaneous word problem question given above.

Let’s start simultaneously solving this question.

J + 1/2(P) = 225Kg …. equation one

J + 3/4(P) = 380Kg …. equation two

Now, let’s solve it simultaneously using elimination method such that equation two substract equation one.

3/4(P) – 1/2(P) = 380Kg – 225Kg

2P/4 = 155Kg

Now, let’s cross multiply

4(155Kg) = 2P

Since we have just known the weight of passengers only as 310Kg, now let’s substitute the value of the weight of passengers into equation one, J + 1/2(P) = 225Kg.

J + 1/2(310Kg) = 225Kg

J + 155Kg = 225Kg

J = 225Kg – 155Kg

From the solution above, it’s now known that the weight of jet only (J) is 70Kg while that of passengers only (P) is 3310Kg.

To check whether this solution is correct or not, just simply add weight of jet and weight of passengers altogether and you will have 380Kg, i.e, 70Kg + 310Kg = 380Kg which is the weight of the jet when it’s three quarters filled with passengers.

Solution to #2

Let C and M represent the price of Chemistry text books and Maths text books,. respectively.

Therefore, from the question given in number above, we can deduce that:

5C+ 3M = 150 … equation one

2C + M = 90 … equation two.

From equation two, let us make M the subject of formula, hence:

2C + M = 90

M = 90 – 2C … equation three.

Now, let’s substitute the value of M in equation one

Therefore, 5C + 3(90 – 2C) = 150

5C + 270 – 6C = 150

Now, let’s substitute the value of C, 120 into equation three, M = 90 – 2C

M = 90 – 2(120)

M = 90 – 240

Therefore, the price of Chemistry text book is N120 and that of Maths is N30.

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VIDEO

  1. SIMULTANEOUS EQUATION: WORD PROBLEMS

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  3. S2 Ex7C Solving Simultaneous Equations by Algebraic Methods 04B: Elimination

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  5. Word problems simultaneous equations

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COMMENTS

  1. Word problems that lead to simultaneous equations. Examples

    Section 2: Problems. H ERE ARE SOME EXAMPLES of problems that lead to simultaneous equations. Example 1. Andre has more money than Bob. If Andre gave Bob $20, they would have the same amount. While if Bob gave Andre $22, Andre would then have twice as much as Bob.

  2. PDF Simlutaneous Equations Word Problems

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  3. Word Problems on Simultaneous Linear Equations

    Solved Examples - Word Problems on Simultaneous Linear Equations. Q.1: A man buys postage stamps of denominations \(25\) paise and \(50\) ... In this article, we have discussed word problems and how to solve a word problem. Simultaneous equations are those that must be solved at the same time. Word problems or applied problems are real-life ...

  4. Word Problems on Simultaneous Equations

    WORD PROBLEMS ON SIMULTANEOUS EQUATIONS. Problem 1 : In a school, there are 880 students in total. If there is 20% more boys than girls, find the number of boys and girls in the school. Solution : Let "x" and "y" be the no. of boys and girls respectively. Then, we have.

  5. 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. ... 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.

  6. Simultaneous Equations Practice Questions

    Previous: Non-linear Simultaneous Equations Practice Questions Next: Similar Shapes Sides Practice Questions GCSE Revision Cards

  7. Simultaneous Equations

    Simultaneous equations are two or more algebraic equations that share variables such as x and y. They are called simultaneous equations because the equations are solved at the same time. The number of variables in simultaneous equations must match the number of equations for it to be solved. An example of simultaneous equations is 2 x + 4 y ...

  8. 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. 2x + y = 6. x + 2y = 9. Above equations are simultaneous equations in unknown variables 'x' and 'y'. Both equations are true for x = 1 and y = 4.

  9. Word problems that lead to simultaneous equations

    Let x be the present age of the woman. Let y be the present age of her son. y + 30. 2 ( y − 15). To eliminate x, simply substitute equation 1) in equation 2), and solve for y. You should find: y = 45 years. Therefore x, the age of the woman, is 75 years. Problem 2. A total of 925 tickets were sold for $5,925.

  10. Linear simultaneous equations

    Unit 1: Linear simultaneous equations. 1,400 possible mastery points. Mastered. Proficient. Familiar. Attempted. Not started. Quiz. Unit test. ... Word problems: Writing equations reducible to linear form Get 3 of 4 questions to level up! Linear equations word problems. Learn. Age word problem: Ben & William (Opens a modal)

  11. GCSE Maths Takeaway

    Ideally, the question itself shouldn't mention too many numbers. The set of questions as a whole is in some sense 'complete' i.e. covers 'all kinds' of linear simultaneous equation word problems 2. No two questions are of the same 'formula'. There are no more than 15 questions. There are questions of varying difficulty.

  12. Simultaneous Equations: Word Problems

    The video covers word problems leading to simultaneous equations.Watch, learn, like and share.Click the link below for other videos on Simultaneous Equations...

  13. 15 Simultaneous Equations Questions And Practice Problems (KS3 & KS4

    Step 1: Use multiplication to make the coefficient of one of the variables the same in both equations. Multiplying the first equation by 3, 3, we get 9a+6b=57. 9a + 6b = 57. Multiplying the second equation by 2, 2, we get 8a+6b=54. 8a + 6b = 54. The coefficient of b b for both equations is now 6. 6.

  14. Simultaneous Equations

    Simultaneous Equations Examples. Example 1: Solve the simultaneous equations 2x - y = 5 and y - 4x = 1 using the appropriate method. Solution: To solve 2x - y = 5 and y - 4x = 1, we will use the elimination method as it is easy to eliminate the variable y by adding the two equations. So, we have.

  15. Systems of equations word problems

    Systems of equations word problems. Google Classroom. You might need: Calculator. Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h . If doubled, Malcolm's maximum speed would be 80 km/h more than Ravi's maximum speed.

  16. Mr Barton Maths

    Solving Simultaneous Equations: Worksheets with Answers. Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you. And best of all they all (well, most!) come with answers. ... Word Problems : 1: 2: 3: Simultaneous Equations - Graphical : 1: 2: 3:

  17. PDF simultaneous equations word problems

    Simultaneous equations - word problems. Set up simultaneous equations for each of the following problems, then solve them. The length of a rectangle is twice its width. The perimeter is 30. Find its dimensions. The difference of two numbers is 3, and the sum of three times the larger one and twice the smaller one is 19. Find the two numbers.

  18. Simultaneous Equations

    y = 2. The cost of a book and a pencil is £2.00 each. E.g.3. If I double a number and add three times a second number, the answer is 1. If I multiply the first number by 3 and take away twice the second number, the answer is 8. Find the numbers. Let the numbers be x and y. 2x + 3y = 1 1. 3x - 2y = 8 2.

  19. Solving Problems Using Simultaneous Equations

    Here are some examples of how you can use simultaneous equations to solve word problems. Make sure to get all the information you need prior to creating the ...

  20. Word Problems on Simultaneous Linear Equations

    Here those simultaneous linear equations are in the form of word problems. So, you can solve different word problems with the help of linear equations. ... Simultaneous Equations Word Problems. Example 1. One number is greater than thrice the other number by 6. If 4 times the smaller number exceeds the greater by 7, find the numbers?

  21. Examples of simultaneous equations word problems (step by step)

    Therefore, from the question given in number above, we can deduce that: 5C+ 3M = 150 … equation one. 2C + M = 90 … equation two. From equation two, let us make M the subject of formula, hence: 2C + M = 90. M = 90 - 2C … equation three. Now, let's substitute the value of M in equation one. Therefore, 5C + 3 (90 - 2C) = 150.

  22. SIMULTANEOUS EQUATIONS

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  23. PDF Simultaneous Equations

    The solution obtained by graphing two or simultaneous solution equations in the number plane and eg The equations x + y = 10 and x − y = 6 observing the point of intersection. have many solutions but the only. If the point of intersection is (3, −2), simultaneous solution is x = 8 and y = 2. then the solution is x = 3 and y = −2 ...