A2400 ch3a | Version 1.1 | September 2020
Solving linear simultaneous equations by
elimination
A LEVEL LINKS
Scheme of work: 1c. Equations – quadratic/linear simultaneous
Key points
• Two equations are simultaneous when they are both true at the same time.
• Solving simultaneous linear equations in two unknowns involves finding the value of each
unknown which works for both equations.
• Make sure that the coefficient of one of the unknowns is the same in both equations.
• Eliminate this equal unknown by either subtracting or adding the two equations.
Example 1 Solve the simultaneous equations 3x + y = 5 and x + y = 1
3x + y = 5
– x + y = 1
2x = 4
So x = 2
Using x + y = 1
2 + y = 1
So y = −1
Check:
equation 1: 3 × 2 + (−1) = 5 YES
equation 2: 2 + (−1) = 1 YES
1 Subtract the second equation from
the first equation to eliminate the y
term.
2 To find the value of y, substitute
x = 2 into one of the original
equations.
3 Substitute the values of x and y into
both equations to check your
answers.
A2400 ch3a | Version 1.1 | September 2020
Example 2 Solve x + 2y = 13 and 5x − 2y = 5 simultaneously.
x + 2y = 13
+ 5x − 2y = 5
6x = 18
So x = 3
Using x + 2y = 13
3 + 2y = 13
So y = 5
Check:
equation 1: 3 + 2 × 5 = 13 YES
equation 2: 5 × 3 − 2 × 5 = 5 YES
1 Add the two equations together to
eliminate the y term.
2 To find the value of y, substitute
x = 3 into one of the original
equations.
3 Substitute the values of x and y into
both equations to check your
answers.
Practice questions
Solve these simultaneous equations.
1 4x + y = 8 2 3x + y = 7
x + y = 5 3x + 2y = 5
3 4x + y = 3 4 3x + 4y = 7
3x – y = 11 x – 4y = 5
5 2x + y = 11 6 2x + 3y = 11
x – 3y = 9 3x + 2y = 4
7 4x + y = 25
x – 3y = 16
A2400 ch3a | Version 1.1 | September 2020
Answers
1 x = 1, y = 4
2 x = 3, y = – 2
3 x = 2, y = –5
4 x = 3, y = –
5 x = 6, y = –1
6 x = – 2, y = 5
7 x = 7, y = – 3
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