Simultaneous Linear Equations II: Elimination Method – JSS3 Mathematics Lesson Note (Second Term)

The elimination method solves simultaneous linear equations by adding or subtracting the equations to remove one unknown.

Behavioural objectives

By the end of the lesson, students should be able to:

1. state when to add and when to subtract two equations to eliminate an unknown;
2. multiply one or both equations by a number to obtain equal coefficients;
3. solve simultaneous linear equations by the elimination method;
4. substitute a found value into one of the equations to get the other unknown;
5. check an answer in both equations.

What are simultaneous linear equations?

Simultaneous linear equations are two equations with two unknowns that are true for the same pair of values.

• Unknown: a letter that stands for a number we do not yet know, such as x or y.
• Coefficient: the number in front of an unknown. In 6x + 5y = 15, the coefficients are 6 and 5.
• Solution: the pair of values that makes both equations true.
• Eliminate: to remove one unknown so that only one is left.

What is the elimination method?

The elimination method removes one unknown by adding or subtracting the two equations, leaving one equation with one unknown.

• It is very useful when none of the coefficients of the unknowns is 1.
• You solve the one-unknown equation first.
• You then substitute the answer into one of the original equations to find the other unknown.

When do you add or subtract the equations?

You add the equations when the signs of the unknown to be eliminated are opposite, and subtract when the signs are the same.

The unknown you want to remove must first have the same coefficient in both equations.

How do you get equal coefficients?

You multiply one or both equations by a number so that one unknown has the same coefficient in both.

• Already equal: no multiplication is needed, as in 6x + 5y = 15 and 3x + 5y = 12 (y has coefficient 5 in both).
• One multiplication: if one coefficient divides the other, multiply one equation only. For 2x and 4x, multiply the first equation by 2.
• Two multiplications: if neither divides the other, use the LCM. For 2y and 3y, the LCM is 6, so multiply the first equation by 3 and the second by 2.
• Every term: multiply every term on both sides of the equation, including the number on the right.

What are the steps of the elimination method?

There are six steps in the elimination method.

1. Write both equations one under the other and number them (1) and (2).
2. Choose the unknown to eliminate.
3. Multiply one or both equations to make its coefficients equal.
4. Add the equations if the signs are opposite, or subtract if they are the same.
5. Solve the new equation to find the first unknown.
6. Substitute this value into one original equation to find the second unknown, then check in the other equation.

Worked examples

Example III. Solve the following simultaneous equations by the elimination method.

(a) 6x + 5y = 15 and 3x + 5y = 12

6x + 5y = 15 (1)
3x + 5y = 12 (2)

The unknown y has an equal coefficient and the same sign in both equations, so we subtract the two equations to eliminate the y terms.

6x + 5y – (3x + 5y) = 15 – 12
6x + 5y – 3x – 5y = 3
6x – 3x + 5y – 5y = 3
3x + 0 = 3
x = 1

To find y, substitute x = 1 in equation (1) (equation (2) would also work).

6x + 5y = 15
6(1) + 5y = 15
6 + 5y = 15
5y = 15 – 6 = 9
y = 9/5

Answer: x = 1, y = 9/5.

(b) 4c – 4d = 9 and 5c + 4d = 18

4c – 4d = 9 (1)
5c + 4d = 18 (2)

The unknown d has an equal coefficient but different signs, so we add the two equations to eliminate d.

4c – 4d + 5c + 4d = 9 + 18
4c + 5c – 4d + 4d = 27
9c = 27
c = 3

To find d, substitute c = 3 into equation (1).

4c – 4d = 9
4(3) – 4d = 9
12 – 4d = 9
4d = 12 – 9 = 3
d = 3/4

Answer: c = 3, d = 3/4.

(c) 2x + y = 7 and 3x – 2y = 7 (one equation is multiplied)

2x + y = 7 (1)
3x – 2y = 7 (2)

The coefficients of y are 1 and -2. Multiply equation (1) by 2 to make the y coefficient 2.

4x + 2y = 14 (3)

The signs of y in (3) and (2) are opposite, so add them.

4x + 2y + 3x – 2y = 14 + 7
7x = 21
x = 3

Substitute x = 3 into equation (1).

2(3) + y = 7
6 + y = 7
y = 1

Check in (2): 3(3) – 2(1) = 9 – 2 = 7. This is correct.

Answer: x = 3, y = 1.

(d) 5x + 2y = 19 and 3x + 4y = 17 (one equation is multiplied, then subtracted)

5x + 2y = 19 (1)
3x + 4y = 17 (2)

Multiply equation (1) by 2 so that y has coefficient 4 in both equations.

10x + 4y = 38 (3)

The signs of y are the same, so subtract (2) from (3).

10x + 4y – (3x + 4y) = 38 – 17
7x = 21
x = 3

Substitute x = 3 into equation (1).

5(3) + 2y = 19
15 + 2y = 19
2y = 4
y = 2

Check in (2): 3(3) + 4(2) = 9 + 8 = 17. This is correct.

Answer: x = 3, y = 2.

What mistakes should you avoid?

• Wrong operation: same signs need subtraction, opposite signs need addition.
• Sign errors when subtracting: put the second equation in brackets and change every sign inside.
• Partial multiplication: multiply every term, including the right-hand side.
• Stopping early: after finding one unknown, always find the second one.
• No check: substitute both values into the equation you did not use.

Evaluation

Exercise 16.4, Nos 2, 3, 7, 8 and 9. Use the elimination method to solve the following simultaneous equations.

1. 6x + 7y = 15 and 6x – 9y = 31
2. 4x + 3y = 10 and 4x + 5y = 8
3. 2x + 3y = 8 and 3x + 2y = 7
4. 3x + 4y = 10 and 2x + 5y = 9
5. 4x + 3y = 11 and 3x – 4y = 2

Assignment

Solve 4a + 3b = 3 and 3a + 2b = 1 by the elimination method, then check your answer in both equations.