Simultaneous Equations

Simultaneous equations are pairs of equations that share the same unknowns. You need to find values of the unknowns that satisfy both equations at the same time ("simultaneously"). This topic is a key part of the AQA GCSE Mathematics specification and appears regularly on both Foundation and Higher tier papers.

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Why Simultaneous Equations?

A single equation with two unknowns (e.g. x + y = 10) has infinitely many solutions. To find a unique solution, you need a second equation. Together, the two equations give one pair of values (or occasionally more) that satisfy both.

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Method 1: Elimination

The elimination method works by adding or subtracting the two equations to eliminate one of the variables.

Steps

1. Label the equations (1) and (2).
2. Make the coefficients of one variable the same in both equations (by multiplying one or both equations).
3. Add or subtract the equations to eliminate that variable.
4. Solve for the remaining variable.
5. Substitute back to find the other variable.

Worked Example 1

Solve:

Equation (1): 2x + y = 7

Equation (2): x + y = 4

The coefficient of y is the same in both equations (1), so subtract equation (2) from equation (1):

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

2x - x + y - y = 3

x = 3

Substitute x = 3 into equation (2):

3 + y = 4

y = 1

Answer: x = 3, y = 1

Worked Example 2

Solve:

Equation (1): 3x + 2y = 16

Equation (2): x + 2y = 8

Subtract (2) from (1):

(3x + 2y) - (x + 2y) = 16 - 8

2x = 8

x = 4

Substitute into (2): 4 + 2y = 8, so 2y = 4, y = 2

Answer: x = 4, y = 2

Exam Tip: If the coefficients of one variable are the same, subtract the equations. If the coefficients have opposite signs, add the equations. This is the key decision in the elimination method.

Worked Example 3 — Multiplying First

Solve:

Equation (1): 2x + 3y = 12

Equation (2): 5x + y = 13

To eliminate y, multiply equation (2) by 3:

Equation (2) x 3: 15x + 3y = 39

Now subtract equation (1) from this:

(15x + 3y) - (2x + 3y) = 39 - 12

13x = 27

x = 27/13 — this is tricky, so let us try eliminating x instead.

To eliminate x, multiply (1) by 5 and (2) by 2:

Equation (1) x 5: 10x + 15y = 60

Equation (2) x 2: 10x + 2y = 26

Subtract: 15y - 2y = 60 - 26

13y = 34

y = 34/13 — still messy.

Actually, let us reconsider the original equations. Let us use neater numbers.

Solve:

Equation (1): 2x + 3y = 12

Equation (2): 5x - 3y = 9

The y terms have opposite signs and the same magnitude, so add:

(2x + 3y) + (5x - 3y) = 12 + 9

7x = 21

x = 3

Substitute into (1): 2(3) + 3y = 12, so 6 + 3y = 12, 3y = 6, y = 2

Answer: x = 3, y = 2

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Method 2: Substitution

The substitution method works by rearranging one equation to express one variable in terms of the other, then substituting into the second equation.

Steps

1. Rearrange one equation to make one variable the subject.
2. Substitute this expression into the other equation.
3. Solve the resulting equation.
4. Substitute back to find the other variable.

Worked Example 4

Solve:

Equation (1): y = 2x + 1

Equation (2): 3x + y = 11

Substitute equation (1) into equation (2):

3x + (2x + 1) = 11

5x + 1 = 11

5x = 10

x = 2

Substitute into (1): y = 2(2) + 1 = 5

Answer: x = 2, y = 5

Worked Example 5

Solve:

Equation (1): x + 2y = 10

Equation (2): 3x - y = 4

Rearrange (1): x = 10 - 2y

Substitute into (2): 3(10 - 2y) - y = 4

30 - 6y - y = 4

30 - 7y = 4

-7y = -26

y = 26/7 — again messy. Let us use cleaner numbers.

Solve:

Equation (1): x + 2y = 9

Equation (2): 3x - y = 13

Rearrange (1): x = 9 - 2y

Substitute into (2): 3(9 - 2y) - y = 13

27 - 6y - y = 13

27 - 7y = 13

-7y = -14

y = 2

Substitute back: x = 9 - 2(2) = 5

Answer: x = 5, y = 2

Exam Tip: Use substitution when one equation already has a variable as the subject (e.g. y = ...). Use elimination when both equations are in the form ax + by = c. The exam will sometimes tell you which method to use.

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Simultaneous Equations from Context

Worked Example 6

Two cinema tickets and one drink cost 19 pounds. One cinema ticket and three drinks cost 17 pounds. Find the cost of one ticket and one drink.

Let t = cost of a ticket, d = cost of a drink.

Equation (1): 2t + d = 19

Equation (2): t + 3d = 17

Multiply (1) by 3: 6t + 3d = 57

Subtract (2): 6t + 3d - t - 3d = 57 - 17

5t = 40

t = 8

Substitute into (2): 8 + 3d = 17, 3d = 9, d = 3

Answer: A ticket costs 8 pounds and a drink costs 3 pounds.

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One Linear and One Quadratic [H]

On the Higher tier, you may need to solve a pair of simultaneous equations where one is linear and one is quadratic. This requires the substitution method.

Worked Example 7

Solve:

Equation (1): y = x + 3

Equation (2): y = x^2 + 1

Substitute (1) into (2):

x + 3 = x^2 + 1

0 = x^2 - x - 2

0 = (x - 2)(x + 1)

x = 2 or x = -1

When x = 2: y = 2 + 3 = 5

When x = -1: y = -1 + 3 = 2

Answer: x = 2, y = 5 or x = -1, y = 2

Worked Example 8

Solve:

Equation (1): y = 2x - 1

Equation (2): x^2 + y^2 = 10

Substitute (1) into (2):

x^2 + (2x - 1)^2 = 10

x^2 + 4x^2 - 4x + 1 = 10

5x^2 - 4x + 1 = 10

5x^2 - 4x - 9 = 0

(5x - 9)(x + 1) = 0

x = 9/5 = 1.8 or x = -1