Simultaneous Equations Worksheets | Questions and Revision | MME

Simultaneous Equations Questions, Revision and Worksheets

Level 4-5
Ultimate GCSE Maths Bundle (Guaranteed Pass)

Simultaneous Equations 

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.

Level 4-5

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

Level 4-5
Level 6-7

Type 2: Non-linear Simultaneous Equations

Because one of these equations is quadratic (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.

Level 8-9
Level 4-5

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.

[4 marks]

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. 

2a+2b=7

4a+3b=12

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 

Level 4-5
GCSE Maths Revision Cards

GCSE Maths Revision Cards

(236 Reviews) £8.99
View Buy this product on Amazon Buy this product on Amazon

Take an Online Exam

Simultaneous Equations (Linear) Online Exam

Simultaneous Equations (Non-Linear) Online Exam

Example Questions

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,

x=8, y=10

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}

 

Hence, the solution is,

x=2, y=-4

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}

Hence, the solution is,

x=-2, y=-4

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.

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.

 

When x=8

\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

Maths Exam Worksheets

Worksheets and Exam Questions

mme logo

(NEW) Simultaneous Equations (Linear) Exam Style Questions - MME

Level 4-5 New Official MME
mme logo

(NEW) Simultaneous Equations (Non-linear) Exam Style Questions - MME

Level 8-9 New Official MME

Drill Questions

mme logo

Solving Simultaneous - Drill Questions

mme logo

Simultaneous Equations (Linear) - Drill Questions

mme logo

Simultaneous Equations (Linear and Non-linear) - Drill Questions

mme logo

Simultaneous Equations (Linear and Non-linear) 2 - Drill Questions

mme logo

Simultaneous Equations (Mixed)

Try a revision card

Try a revision card on this topic.