## 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:

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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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## 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 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.

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## 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.

Third Space Learning’s free GCSE maths resource library contains detailed lessons with step-by-step instructions on how to solve ratio problems, as well as worksheets with [topic] practice questions and more GCSE exam questions.

Take a look at the simultaneous equations lessons today. More lessons and resources are added every week!

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Simultaneous equations are multiple equations that share the same variables and which are all true at the same time.

When an equation has 2 variables its much harder to solve, however, if you have 2 equations both with 2 variables, like

2x+y=10\,\,\,\text{ and }\,\,\,x+y=4

then there is a solution for us to find that works for both equations. These equations are called simultaneous for this reason.

There are 2 main types of equation you need to be able to solve.

Make sure you are happy with the following topics before continuing.

- Solving linear equations
- Rearranging formulae
- Solving quadratics by factorising and the quadratic formula

## Type 1: Linear Simultaneous Equations

To do this, we’ll use a process called elimination – we’re going to eliminate one of the variables by subtracting one equation from the other. We will write one equation on top of the other and draw a line underneath, as with normal subtraction.

Example: Find the solution to the following simultaneous equations.

4x + 3y = 14 \,\,\,\,,\ 5x+7y=11

Step 1: Write one equation above the other.

Both equations need to be in the form ax+by=c , so rearrange if needed.

\begin{aligned}4x + 3y &= 14 \\ 5x+7y &= 11 \end{aligned}

Step 2: Get the coefficients to match

The coefficients are the numbers before x and y , make the x coefficients the same by scaling up both equations

(\times5) \,\,\,\,\,\,\,\,\,4x + 3y = 14\,\,\, gives \,\,\, 20x + 15y = 70

\,\, (\times4)\,\,\,\,\,\,\,\,\, 5x+7y = 11 \,\,\, gives \,\,\, 20x+28y=44

Step 3: Add or subtract the equations to eliminate terms with equal coefficients.

As both equations are +20x we must subtract the equations.

\begin{aligned}20x + 15y &= 70 \\ (-)\,\,\,\,\,\,\,\,\, 20x+28y&=44 \\ \hline-13y&=26\end{aligned}

Step 4: Solve the resulting equation

\begin{aligned}(\div-13)\,\,\,\,\,\,\,\,\,-13y&=26 \\ y &=-2\end{aligned}

Step 5: Substitute the answer into the simplest of the two equations to find the other variable.

\begin{aligned}4x + 3y &= 14 \\ 4x +3(-2) &=14 \\ 4x -6 &= 14 \\ 4x &= 20 \\ x &= 5\end{aligned}

This gives the final answer to be;

x = 5, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, y = -2

## Type 2: Non-linear Simultaneous Equations

Because one of these equations is quadrati c (Non-linear) , we can’t use elimination like before. Instead, we have to use substitution .

Example: Solve the following simultaneous equations.

x^2+2y=9,\,\,\,\,y-x=3

Step 1: Rearrange the linear equation to get one of the unknowns on its own and on one side of the equals sign.

\begin{aligned}(+x)\,\,\,\,\,\,\,\,\,y-x&=3 \\ y &= x+3\end{aligned}

Step 2: Substitute the linear equation into the non-linear.

We know y=x+3 so we can replace the y in the first equation with x+3 :

\begin{aligned} x^2+2y&=9 \\ x^2+2(x+3)&=9\end{aligned}

Step 3: Expand and solve the new quadratic formed.

\begin{aligned}x^2+2(x+3)&=9 \\ x^2 + 2x+6 &= 9 \\ x^2+2x-3&=0 \\ (x-1)(x+3)&=0\end{aligned}

x=1 \,\,\, and \,\,\,x=-3 .

Step 4: Substitute both values back into the simplest equation to find both versions of the other variable.

\text{When }x=1, \,\,\,\,\,\,\,\,\, y=1+3=4,

\text{When }x=-3, \,\,\,\,\,\,\,\,\, y=-3+3=0.

Thus, we have two solution pairs:

x=1, y=4, and x=-3, y=0 .

## Example: Applied Simultaneous Equation Question

A store sells milkshakes and ice creams.

2 milkshakes and 2 ice creams, costs £7

4 milkshakes and 3 ice creams, costs £12

Work out the cost of an individual milkshake and an individual ice cream.

Step 1: we need to do is form the two simultaneous equations.

Let’s say that the price of a milkshake is a , and the price of an ice cream is b .

This creates the following two equations.

Step 2: Now we must get the coefficients to match, in this case we can multiply the first equation by 2

(\times 2) \,\,\,\,\,\,\,\,\, 2a+2b=7 \,\,\, \text{Gives} \,\,\, 4a + 4b = 14

Step 3: Subtract the equations to eliminate terms with equal coefficient.

\begin{aligned}4a + 4b &= 14 \\ (-)\,\,\,\,\,\,\,\,\, 4a+3b&=12 \\ \hline b&=2\end{aligned}

Step 4: Substitute the answer into the simplest of the two equations to find the other variable.

\begin{aligned}2a+2b&=7 \\ 2a + 2(2) &=7 \\ 2a+4&=7 \\ 2a &=3 \\ a& = 1.5\end{aligned}

This gives the final answer to be

Milkshake ( a ) = £1.50

Ice Cream ( b ) = £2.00

## Simultaneous Equations Example Questions

Question 1: Solve the following simultaneous equations,

\begin{aligned}y&=2x-6\\y&=\dfrac{1}{2}x+6\end{aligned}

Straight away we can subtract equation 2 from equation 1 so that,

\begin{aligned}y&=2x-6\\ y&=\dfrac{1}{2}x+6 \\ \\ (y-y)&=(2x-\dfrac{1}{2}x)-6-6 \\ 0&= \dfrac{3}{2}x-12 \end{aligned}

If we rearrange to make x the subject we find,

x=\dfrac{2\times12}{3}=\dfrac{24}{3}=8

Substituting x=8 back into the original first equation,

\begin{aligned}y&=2(8)-6 \\ y&=10\end{aligned}

Hence, the solution is,

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Question 2: Solve the following simultaneous equations,

\begin{aligned}2x-3y&=16\\x+2y&=-6\end{aligned}

If we multiply the second equation by 2 , we have two equations both with a 2x term, hence subtracting our new equation 2 from equation 1 we get,

\begin{aligned}2x-3y&=16\\ 2x+4y&=-12 \\ \\ (2x-2x)+(-3y-4y)&= 16-(-12) \\ 0x -7y &=28\end{aligned}

If we rearrange to make y the subject we find,

y=\dfrac{28}{-7}=-4

Substituting y=-4 back into the original second equation,

\begin{aligned}x+2(-4)&=-6 \\ x-8&=-6 \\ x&=2\end{aligned}

Question 3: Solve the following simultaneous equations,

\begin{aligned}x+2y+10&=0\\3x-5y-14&=0\end{aligned}

If we multiply the first equation by 3 , we have two equations both with a 3x term, hence subtracting our new equation 2 from equation 1 we get,

\begin{aligned}3x+6y+30&=0 \\ 3x-5y-14&=0 \end{aligned}

(3x-3x)+(6y-(-5y))+(30-(-14)) = 0

\begin{aligned} 11y &= -44 \\ y &= - 4\end{aligned}

Substituting y=-4 back into the original first equation,

\begin{aligned}x+2(-4)+10&=0\\ x-8+10 &=0 \\ x&=-2 \end{aligned}

Question 4: There are two types of ticket on a train: an adult ticket and a child ticket. The cost of buying 2 adult tickets and 3 child tickets would be £20 . The cost of buying one of each type would be £8.50 .

Work out the cost of a single adult ticket and a single child ticket.

Let A be the cost of an adult ticket and let C be the cost of a child ticket, thus we have two simultaneous equations,

\begin{aligned} 2A+3C&=20 \\ A+C&=8.5 \end{aligned}

If we multiply the second equation by 2 , we have two equations both with a 2A term, hence subtracting our new equation 2 from equation 1 we get,

\begin{aligned} 2A+3C&=20 \\ 2A+2C&=17 \\ \\ (2A-2A)+(3C-2C) &=(20-17) \\ C&=3\end{aligned}

Then, substituting this value back into the original equation 2 , we get,

\begin{aligned} A+3&=8.5 \\ A & =5.5\end{aligned}

Therefore, the cost of a child ticket is £3 , and the cost of an adult ticket is £5.50 .

Question 5: Solve the following simultaneous equations,

\begin{aligned}x^2-y&=14\\2y-4&=12x\end{aligned}

If we multiply the first equation by 2 , we have two equations both with a 2y term, hence adding our new equation 1 and equation 2 we get,

\begin{aligned} 2x^2-2y&=28 \\ 2y-4&=12x \\ \\ 2x^2+(-4)+(-2y+2y)&=28+12x \\ 2x^2-12x-32&=0\end{aligned}

After rearranging to form a quadratic we can solve for x ,

\begin{aligned} 2x^2-12x-32&=0 \\ 2(x^2-6x-16)&=0 \\ (x-8)(x+2)&=0\end{aligned}

Therefore, the 2 solutions for x are x=8, x=-2 . To find the two y solutions, we can substitute these values back into the original second equation.

\begin{aligned}2y-4&=96 \\ y&=50 \end{aligned}

When x=-2 ,

\begin{aligned}2y-4&=-24 \\ y &= -10\end{aligned}

Thus, the two pairs of solutions are,

x=8,y=50 and x=-2,y=-10

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Worksheet on Simultaneous Linear Equations

Practice each pair of the equation problems from the worksheet on simultaneous linear equations with the two variables and two linear equations. Solving simultaneous linear equations with two variables by using the method of substitution to solve each pair of the equations and also solve the equations by using the method of elimination.

1. Use the method of substitution to solve each other of the pair of simultaneous equations: (a) x + y = 15 x - y = 3

(b) x + y = 0 x - y = 2

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

(d) 2x - 9y = 9 5x + 2y = 27

(e) x + 4y = -4 3y - 5x = -1

(f) 2x - 3y = 2 x + 2y = 8

(g) x + y = 7 2x - 3y = 9

(h) 11y + 15x = -23 7y - 2x = 20

(i) 5x - 6y = 2 6x - 5y = 9

2. Solve each other pair of equation given below using elimination method: (a) x + 2y = -4 3x - 5y = -1

(b) 4x + 9y = 5 -5x + 3y = 8

(c) 9x - 6y = 12 4x + 6y = 14

(d) 2y - (3/x) = 12 5y + (7/x) = 1

(e) (3/x) + (2/y) = (9/xy) (9/x) + (4/y) = (21/xy)

(f) (4/y) + (3/x) = 8 (6/y) + (5/x) = 13

(g) 5x + (4/y) = 7 4x + (3/y) = 5

(h) x + y = 3 -3x + 2y = 1

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

3. Solve the following simultaneous equations: (a) 3a + 4b = 43 -2a + 3b = 11

(b) 4x - 3y = 23 3x + 4y = 11

(c) 5x + (4/y) = 7 4x + (3/y) = 5

(d) 4/(p - 3) + 6/(q - 4) = 5 5/(p - 3) - 3/(q - 4) = 1

(e) (l/6) - (m/15) = 4 (l/3) - (m/12) = 19/4

(f) 3x + 2y = 8 4x + y = 9

(g) x - y = -1 2y + 3x = 12

(h) (3y/2) - (5x/3) = -2 (y/3) + (x/3) = 13/6

(i) x - y = 3 (x/3) + (y/2) = 6

(j) (2x/3) + (y/2) = -1 (-x/3) + y = 3

(k) 5x + 8y = 9 2x + 3y = 4

(l) 3 - 2(3a - 4b) = -59 (a - 3)/4 - (b - 4)/5 = 2¹/₁₀

Answers for the worksheet on simultaneous linear equations are given below to check the exact answers of the above questions on system of linear equations.

1. (a) x = 9, y = 6

(b) x = 1, y = -1

(c) x = 1, y = -1

(d) x = 261/49, y = 9/49

(e) x = -8/23, y = -21/23

(f) x = 4, y = 2

(g) x = 6, y = 1

(h) x = -3, y = 2

(i) x = 4, y = 3

2. (a) x = -2, y = -1

(b) x = -1, y = 1

(c) x = 2, y = 1

(d) x = -1/2, y = 3

(e) x = 3, y = 1

(f) x = 1/2, y = 2

(g) x = -1, y = 1/3

(h) x = 1, y = 2

(i) x = -21/23, y = 26/23

3. (a) a = 5, b = 7

(b) x = 5, y = -1

(c) x = -1, y = 1/3

(d) p = 5, q = 6

(e) l = -2, m = -65

(f) x = 2, y = 1

(g) x = 2, y = 3

(h) x = 141/38, y = 53/19

(i) x = 9, y = 6

(j) x = -3, y = 2

(k) x = 5, y = -2

(l) a = 5, b = -4

● Simultaneous Linear Equations

Simultaneous Linear Equations

Comparison Method

Elimination Method

Substitution Method

Cross-Multiplication Method

Solvability of Linear Simultaneous Equations

Pairs of Equations

Word Problems on Simultaneous Linear Equations

Practice Test on Word Problems Involving Simultaneous Linear Equations

● Simultaneous Linear Equations - Worksheets

Worksheet on Problems on Simultaneous Linear Equations

8th Grade Math Practice From Worksheet on Simultaneous Linear Equations to HOME PAGE

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## Simultaneous Equations Questions

Simultaneous equations questions with solutions are given here for practice and understanding how to determine the solution of given equations. Simultaneous equations are the system of two or more independent equations which satisfies a common solution. Simultaneous equations could have,

- Linear Equation
- Quadratic Equation
- Polynomial Equations

There are several methods of solving simultaneous equations, and they are:

Learn more about Simultaneous Equations .

## Video Lesson on Consistent and Inconsistent Simultaneous Equations

## Simultaneous Equations Questions with Solutions

Let us solve a few simultaneous equations questions.

Question 1:

Solve the following:

2x + 3y = –2

5x + 4y + 2 = 0

2x + 3y = –2 ….(i)

5x + 4y = –2 ….(ii)

Multiply (i) by 5 and (ii) 2 on both sides, subtracting (ii) from (i), we get

10x + 15y = –10

10x + 8y = –4

(–)_ (–)___(+)___

7y = –6

⇒ y = –6/7

Substituting the value of y in (i), we get

∴ the solution of the given simultaneous equation is (2/7, –6/7)

Question 2:

Solve the following equations:

2a – b = 4 and 2b – 3a = 2

2a – b = 4 …..(i)

2b – 3a = 2

⇒ –3a + 2b = 2 ….(ii)

a = ½ (b + 4)

Putting this value in (ii)

–3/2(b + 4) + 2b = 2

⇒ –3b – 12 + 4b = 4

⇒ b = 16

Thus, a = ½(16 + 4) = 10

∴ the solution of the given simultaneous equation is a = 10 and b = 16.

Question 3:

Solve the following simultaneous equations:

7m – 9n = 3 and 2m – 3n = 1

7m – 9n = 3 ….(i)

2m – 3n = 1 ….(ii)

Multiplying equation (ii) by 3 on both sides, then subtracting from (i), we get,

(7m – 6m) + ( –9n + 9n) = 3 – 3

⇒ m = 0

Consequently, n = –⅓

∴ the solution of the given simultaneous equation is m = 0 and n = –⅓.

Question 4:

a 2 + 2b = 9 and b – a = 3

a 2 + 2b = 9 …(i)

b – a = 3 …(ii)

From (ii) b = 3 + a

Substituting the value of b in (i), we get

a 2 + 2(3 + a) = 9

⇒ a 2 + 2a + 6 = 9

⇒ a 2 + 2a + – 3 = 0

⇒ (a + 3)(a – 1) = 0

⇒ a = 1 and –3

When a = 1, b = 4, and when a = –3, b = 0

∴ the solution of the given simultaneous equation is a = 1, b = 4 and a = –3, b = 0.

Also, learn about solving simultaneous equations by Cramer’s rule .

Question 5:

a 2 – b = 14 and 2b – 4 = 12a

a 2 – b = 14 ….(i)

2b – 4 = 12a

⇒ 12a – 2b = –4

⇒ 6a – b = –2

⇒ b = 6a + 2

a 2 – 6a – 2 = 14

⇒ a 2 – 6a – 16 = 0

⇒ (a – 8)(a + 2) = 0

⇒ a = 8 and –2

When a = 8, b = 50 and when a = –2, b = 10

∴ the solution of the given simultaneous equation is a = 8, b = 50 and a = –5, b = 11.

Question 6:

8q + p + r = 0

q + 2p + r = 0

q + p – 1600 = 0

8q + p + r = 0 …(i)

q + 2p + r = 0 …(ii)

q + p – 1600 = 0 …(iii)

Subtracting (ii) from (i), we get

7q – p = 0 ….(iv)

Solving (iii) and (iv), we get,

q = 200 and p = 1400

And from (i), we get r = –3000

Question 7:

(m + n – 8)/2 = (m + 2n – 14)/3 = (3m + n – 12)/11

Taking the first two equations:

(m + n – 8)/2 = (m + 2n – 14)/3

⇒ 3m + 3n – 24 = 2m + 4n – 28

⇒ m – n = – 28 + 24

⇒ m – n = –4 ….(i)

Taking the last two equations:

(m + 2n – 14)/3 = (3m + n – 12)/11

⇒ 11m + 22n – 154 = 9m + 3n – 36

⇒ 2m + 19n = 154 – 36

⇒ 2m + 19n = 118 ….(ii)

Multiplying both sides of (i) by 2 and then subtracting from (ii), we get

19n + 2n = 118 + 8

⇒ 21n = 126

⇒ n = 126/21 = 6

∴ the solution of the given simultaneous equation is m = 2 and n = 6.

The other methods of solving simultaneous equations are:

- Gaussian Elimination
- Jacobi and Gauss-Seidel Iterative Methods

Question 8:

A man can row 16 km downstream and 8 km upstream in 6 hours. He can row 6 km upstream and 24 km downstream in 6 hours. Find the speed of man in still water.

Let the speed of a man rowing a boat in still water be x km/hr and the speed of the stream be y km/hr.

Speed of rowing in downstream = x + y = 1/u (let)

Speed of rowing in upstream = x – y = 1/v

16/(x + y) + 8/(x – y) = 6

⇒ 16u + 8v = 6

⇒ 8u + 4v = 3 ….(i)

And 24/(x + y) + 6/(x – y) = 6

⇒ 24u + 6v = 6

⇒ 4u + v = 1 ….(ii)

Solving (i) and (ii) we get, u = ⅛ ⇒ x + y = 8 and v = ½ ⇒ x – y = 2

Thus, we get x = 5km/hr and y = 3km/hr.

∴ The speed of the man rowing the boat in still water is 5 km/hr.

Question 9:

If (sin 𝜃, cos 𝜃) satisfies the system of equations mx + ny + a + b = a – b and nx + my + 2b = 0, then find the value of 𝜃 where 0≤ 𝜃 ≤ 90 o .

Given equations,

mx + ny + a + b = a – b

⇒ mx + ny + 2b = 0 …..(i)

And nx + my + 2b = 0 ….(ii)

Multiply (i) by n and (ii) by m on both sides, subtracting (ii) from (i), we get,

(n 2 – m 2 )y + 2nb – 2mb = 0

y = –2/(m + n)

Similarly, x = –2/(m + n)

That is, x = y ⇒ sin 𝜃 = cos 𝜃, for 𝜃 = 45 o , the value of sin 𝜃 is equal to the value of cos 𝜃.

Question 10:

A sum of ₹ 400 was distributed among the students of a class. Each boy received ₹ 8, and each girl received ₹ 4. If each girl had received ₹ 10, then each boy would have received ₹ 5. Find the total number of students in the class.

Let the number of boys be x, and the number of girls be y. Then

8x + 4y = 400 and 5x + 10y = 400

⇒ 2x + y = 100 and x + 2y = 80

Multiplying the second equation by 2 on both sides, then subtracting from the first equation, we get,

–3y = –60 ⇒ y = 20 and x = 40

∴ there are 40 boys and 20 girls in the classroom.

## Practice Questions on Simultaneous Equations

1. Solve the following simultaneous equations:

(i) 3x + 5y = 6 and 2x + 6y = –2

(ii) 2x + y = z, x + 4y = 3z and x – 3y = 16

(iii) 2x 2 – 1 = 5 and x + y = –2

(iv) x/a + y/b = a 2 + b 2 and x/a 2 + y/b 2 = a + b

2. Four years ago, the age of a person was thrice that of his son. Eight years later, the age of the person will be twice that of his son. Find the present ages of the person and his son.

3. For what value of k, the following simultaneous equations will not have any solutions: 2x + 3y = 1 and (3k – 1)x + (1 – 2k)y = 2k + 3?

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## 4.4 Solving simultaneous equations

4.4 solving simultaneous equations (ema38).

Up to now we have solved equations with only one unknown variable. When solving for two unknown variables, two equations are required and these equations are known as simultaneous equations. The solutions are the values of the unknown variables which satisfy both equations simultaneously. In general, if there are \(n\) unknown variables, then \(n\) independent equations are required to obtain a value for each of the \(n\) variables.

An example of a system of simultaneous equations is:

We have two independent equations to solve for two unknown variables. We can solve simultaneous equations algebraically using substitution and elimination methods. We will also show that a system of simultaneous equations can be solved graphically.

## Solving by substitution (EMA39)

Use the simplest of the two given equations to express one of the variables in terms of the other.

Substitute into the second equation. By doing this we reduce the number of equations and the number of variables by one.

We now have one equation with one unknown variable which can be solved.

Use the solution to substitute back into the first equation to find the value of the other unknown variable.

The following video shows how to solve simultaneous equations using substitution.

Video: 2FD5

## Worked example 6: Simultaneous equations

Solve for \(x\) and \(y\):

## Use equation \(\left(1\right)\) to express \(x\) in terms of \(y\)

Substitute \(x\) into equation \(\left(2\right)\) and solve for \(y\), substitute \(y\) back into equation \(\left(1\right)\) and solve for \(x\), check the solution by substituting the answers back into both original equations, write the final answer, worked example 7: simultaneous equations.

Solve the following system of equations:

## Use either equation to express \(x\) in terms of \(y\)

Solving by elimination (ema3b), worked example 8: simultaneous equations, make the coefficients of one of the variables the same in both equations.

The coefficients of \(y\) in the given equations are \(\text{1}\) and \(-\text{1}\). Eliminate the variable \(y\) by adding equation \(\left(1\right)\) and equation \(\left(2\right)\) together:

## Simplify and solve for \(x\)

Substitute \(x\) back into either original equation and solve for \(y\), check that the solution \(x=3\) and \(y=-7\) satisfies both original equations, worked example 9: simultaneous equations.

By multiplying equation \(\left(1\right)\) by \(\text{3}\) and equation \(\left(2\right)\) by \(\text{2}\), both coefficients of \(a\) will be \(\text{6}\).

(When subtracting two equations, be careful of the signs.)

## Simplify and solve for \(b\)

Substitute value of \(b\) back into either original equation and solve for \(a\), check that the solution \(a=10\) and \(b=5\) satisfies both original equations, solving graphically (ema3c).

This section can be included in the chapter on functions and graphs with graphs of linear equations. Before beginning this section it may be necessary to revise plotting graphs of linear equations with your learners.

It is also important that learners are either given the graphs or are encouraged to draw accurate graphs on graph paper to help them solve simultaneous equations graphically. Graph sketching software can be used in this section to ensure that graphs are accurate.

Simultaneous equations can also be solved graphically. If the graphs of each linear equation are drawn, then the solution to the system of simultaneous equations is the coordinates of the point at which the two graphs intersect.

For example:

The graphs of the two equations are shown below.

The intersection of the two graphs is \((2;1)\). So the solution to the system of simultaneous equations is \(x=2\) and \(y=1\). We can also check the solution using algebraic methods.

Substitute equation \((1)\) into \((2)\):

Then solve for \(y\):

Substitute the value of \(y\) back into equation \((1)\):

Notice that both methods give the same solution.

You can use an online tool such as graphsketch to draw the graphs and check your solution.

## Worked example 10: Simultaneous equations

Solve the following system of simultaneous equations graphically:

## Write both equations in form \(y=mx + c\)

Sketch the graphs on the same set of axes.

## Find the coordinates of the point of intersection

The two graphs intersect at \(\left(4;22\right)\)

Look at the graph below

Solve the equations \(y = 2x + 1\) and \(y = -x - 5\) simultaneously

From the graph we can see that the lines intersect at \(x = -2\) and \(y = -3\)

Solve the equations \(y = 2x - 1\) and \(y = 2x + 1\) simultaneously

The lines are parallel. Therefore there is no solution to \(x\) and \(y\).

Solve the equations \(y = -2x + 1\) and \(y = -x - 1\) simultaneously

From the graph we can see that the lines intersect at \(x = 2\) and \(y = -3\)

\(- 10 x = -1\) and \(- 4 x + 10 y = -9\).

Solve for \(x\):

Substitute the value of \(x\) into the second equation and solve for \(y\):

Therefore \(x = \frac{1}{10} \text{ and } y = - \frac{43}{50}\).

\(3x - 14y = 0\) and \(x - 4y + 1 = 0\)

Write \(x\) in terms of \(y\):

Substitute value of \(x\) into second equation:

Substitute value of \(y\) back into first equation:

Therefore \(x = -7 \text{ and } y = -\frac{3}{2}\).

\(x + y = 8\) and \(3x + 2y = 21\)

Therefore \(x = 5 \text{ and } y = 3\).

\(y = 2x + 1\) and \(x + 2y + 3 = 0\)

Write \(y\) in terms of \(x\):

Substitute value of \(y\) into second equation:

Substitute value of \(x\) back into first equation:

Therefore \(x = -1 \text{ and } y = -1\).

\(5x-4y = 69\) and \(2x+3y = 23\)

Make \(x\) the subject of the first equation:

Therefore \(x =13 \text{ and } y = -1\).

\(x + 3y = 26\) and \(5x + 4y = 75\)

Therefore \(x =11 \text{ and } y = 5\).

\(3x - 4y = 19\) and \(2x - 8y = 2\)

If we multiply the first equation by 2 then the coefficient of \(y\) will be the same in both equations:

Now we can subtract the second equation from the first:

Substitute the value of \(x\) into the first equation and solve for \(y\):

Therefore \(x = 9 \text{ and } y = 2\).

\(\dfrac{a}{2} + b = 4\) and \(\dfrac{a}{4} - \dfrac{b}{4} = 1\)

Make \(a\) the subject of the first equation:

Substitute value of \(a\) into second equation:

Substitute value of \(b\) back into first equation:

Therefore \(a = \frac{16}{3} \text{ and } b = \frac{4}{3}\).

\(-10x + y = -1\) and \(-10x - 2y = 5\)

If we subtract the second equation from the first then we can solve for \(y\):

Solve for \(y\):

Substitute the value of \(y\) into the first equation and solve for \(x\):

Therefore \(x = \frac{-1}{10} \text{ and } y = -2\).

\(- 10 x - 10 y = -2\) and \(2 x + 3 y = 2\)

Substitute the value of \(y\) in the first equation:

Therefore \(x = - \frac{7}{5} \text{ and } y = \frac{8}{5}\).

\(\dfrac{1}{x} + \dfrac{1}{y} = 3\) and \(\dfrac{1}{x} - \dfrac{1}{y} = 11\)

Rearrange both equations by multiplying by \(xy\):

Add the two equations together:

Therefore \(x = \frac{1}{7} \text{ and } y = -\frac{1}{4}\).

\(y = \dfrac{2(x^2 + 2) - 3}{x^2 + 2}\) and \(y = 2 - \dfrac{3}{x^2 + 2}\)

Since this is true for all \(x\) in the real numbers, \(x\) can be any real number.

Look at what happens to \(y\) when \(x\) is very small or very large:

The smallest \(x\) can be is 0. When \(x=0\), \(y=2-\frac{3}{2}=\frac{1}{2}\).

If \(x\) gets very large, then the fraction \(\dfrac{3}{x^{2}+2}\) becomes very small (think about what happens when you divide a small number by a very large number). Then \(y = 2 - 0 = 2\).

From this we can see that \(\frac{1}{2}\leq y \leq 2\).

Therefore \(x\) can be any real number, \(\frac{1}{2} \leq y < 2\).

\(3a + b=\dfrac{6}{2a}\) and \(3a^2 = 3 - ab\)

Note \(a \neq 0\)

Look at the first equation

Note that this is the same as the second equation

\(a\) and \(b\) can be any real number except for \(\text{0}\).

Solve graphically and check your answer algebraically:

\(y + 2x = 0\) and \(y - 2x - 4 = 0\)

First write the equations in standard form:

Draw the graph:

The graphs intersect at \((-1;2)\) so \(x = -1\) and \(y=2\).

Checking algebraically we get:

Substitute the value of \(x\) back into the first equation:

\(x + 2y = 1\) and \(\dfrac{x}{3} + \dfrac{y}{2} = 1\)

The graphs intersect at \((9;-4)\) so \(x = 9\) and \(y=-4\).

Substitute value of \(x\) into first equation:

Substitute the value of \(y\) back into the first equation:

\(y - 2 = 6x\) and \(y - x = -3\)

The graphs intersect at \((-1;-4)\) so \(x = -1\) and \(y=-4\).

Substitute value of \(y\) into first equation:

\(2x + y = 5\) and \(3x - 2y = 4\)

The graphs intersect at \((2;1)\) so \(x = 2\) and \(y=1\).

\(5 = x + y\) and \(x = y - 2\)

The graphs intersect at \((\text{1,5};\text{3,5})\) so \(x = \text{1,5}\) and \(y=\text{3,5}\).

## IMAGES

## VIDEO

## COMMENTS

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