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Simultaneous equations for AQA GCSE Maths

Simultaneous equations

This page covers the following topics:

1. Linear simultaneous equations by elimination
2. Linear simultaneous equations by substitution

To solve a set of linear equations we find the values of their variables. To do this the method of elimination can be used. In this method first we get rid of one of our variables by adding or subtracting the linear equations from each other. The equations can be multiplied beforehand by scalars if necessary. Combining equations leaves a simpler equation used to find the variable that has been not eliminated. Once the variable is found, its value can be substituted into one of the initial equations to find the second missing variable.

Linear simultaneous equations by elimination

To solve a set of linear equations, the method of substitution can be used, in which we take one of the equations and rearrange it to make one of the variables the subject of the equation. Substituting this expression for the unknown variable into the other equation gives an equation with just one variable. Then we can solve this equation and substitute its solution the into any of the initial equations to find the second unknown variable.

Linear simultaneous equations by substitution

1

3 pears and 2 apples cost ยฃ1.50. 2 pears and 4 apples cost ยฃ1.40. Find the cost of 1 pear and the cost of 1 apple.

Let the price of an apple be a and of a pear be b.
for the first situation: 2a + 3b = 1.50
for the second situation: 4a + 2b = 1.40
Then rearranging the first equation gives 2a = 1.50 โˆ’ 3b.
Substituting this into the second equation gives 2b + 2(1.50 โˆ’ 3b) = 1.40.
Expanding the brackets gives 2b + 3.00 โˆ’ 6b = 1.40.
Rearranging this gives โˆ’4b = โˆ’1.60 and multiplying this by โˆ’1 gives 4b = 1.60.
Dividing this by 4 gives b = 0.40; therefore, 1 pear costs 40p.
Now substituting b = 0.40 into the equaton for 2a gives 2a = 1.50 โˆ’ 1.20.
Thus, 2a = 0.30 which gives a = 0.15 and 1 apple costs 15p.

1 pear costs 40p, 1 apple costs 15p.

3 pears and 2 apples cost ยฃ1.50. 2 pears and 4 apples cost ยฃ1.40. Find the cost of 1 pear and the cost of 1 apple.

2

2 apples and 4 pears cost ยฃ2.20. 7 apples and 1 pear cost ยฃ1.85. Find the cost of an apple and the cost of a pear by using the method of elimination.

Let the price of an apple be a and the price of pear be p.
This gives the equations 2a + 4p = 2.20 and 7a + p = 1.85.
Multiply the second equation by 4 to get 28a + 4p = 7.40.
Subtract this from the first equation to get โˆ’26a = โˆ’5.20.
a = 0.2
Substituting a = 0.2 into the second equation gives 7 ร— 0.2 + p = 1.85.
p = 0.45.

1 apple costs 20p, 1 pear costs 45p.

2 apples and 4 pears cost ยฃ2.20. 7 apples and 1 pear cost ยฃ1.85. Find the cost of an apple and the cost of a pear by using the method of elimination.

3

Solve the following simultaneous equations by using the method of substitution.

4x โˆ’ y = โˆ’11
โˆ’2x + 5y = 19

Rearranging the first equation gives y = 4x + 11.
Substituting this into the second equation gives โˆ’2x + 5(4x + 11) = 19.
Expanding the brackets gives โˆ’2x + 20x + 55 = 19.
Simplifying this gives 18x + 55 = 19 and rearranging this gives 18x = โˆ’36 and from this x = โˆ’2.
Now substituting x = โˆ’2 into the equation for y above gives y = 4(โˆ’2) + 11.
After simplifying y = โˆ’8 + 11 which gives y = 3.

x = โˆ’2, y = 3

Solve the following simultaneous equations by using the method of substitution. 

4x โˆ’ y = โˆ’11 
โˆ’2x + 5y = 19

4

Use the method of elimination to solve the simultaneous equations provided.

5x + 1/2 = y โˆ’ 10
2x โˆ’ 5 = โˆ’6y โˆ’ 6

Multiply the first equation by 6 to give 30x + 3 = 6y โˆ’ 60.
Add the second equation to the first equation to give 32x โˆ’ 2 = โˆ’66.
Add 2 to both sides of the equation to give 32x = โˆ’64.
Divide both sides by 32 to get x = โˆ’2.
Substitute x = โˆ’2 into the first equation to give โˆ’8.5 = y โˆ’ 10.
Add 10 to both sides of the equation to give 1/2 = y.

x = โˆ’2, y = 1/2

Use the method of elimination to solve the simultaneous equations provided. 

5x + 1/2 = y โˆ’ 10 
2x โˆ’ 5 = โˆ’6y โˆ’ 6

5

Using the method of substitution solve the following simultaneous equations.

4x โˆ’ 5 = 2y โˆ’ 7
โˆ’2x + 3 = 3y โˆ’ 4

Rearranging the second equation gives 2x = โˆ’3y + 7.
Substituting this into the first equation gives 2(โˆ’3y + 7) โˆ’ 5 = 2y โˆ’ 7.
Expanding the brackets gives โˆ’6y + 14 โˆ’ 5 = 2y โˆ’ 7.
Simplifying this gives โˆ’6y + 9 = 2y โˆ’ 7.
Rearranging this gives 8y = 16 and so y = 2.
Substituting this into the equation for 2x above we have 2x = โˆ’3 ร— 2 + 7.
Simplification gives 2x = โˆ’6 + 7 and so 2x = 1 and from this x = 1/2.

x = 0.5, y = 2

Using the method of substitution solve the following simultaneous equations. 

4x โˆ’ 5 = 2y โˆ’ 7 
โˆ’2x + 3 = 3y โˆ’ 4

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