Mathematics (Core)Exam code: 0580

Simultaneous Equations

Algebra and Sequences

Elimination removes one unknown by adding or subtracting the two equations
Multiply one or both equations so that one unknown has coefficients of equal size
Subtract when those coefficients have the same sign, and add when the signs are opposite
Solve the resulting single-unknown equation, then substitute back to find the other

Worked example

Elimination with both equations scaled

Solve 3x + 5y = 31 and 2x + 3y = 20.

Solution:

Elimination needs one letter to have the same coefficient in both equations, and here neither does
Match the y terms: 5 and 3 both divide into 15, so multiply the first equation by 3 and the second by 5
Multiply every term, the constant included, or the equation stops being true
9x + 15y = 93 and 10x + 15y = 100
The matched terms now have the same sign, so subtracting removes them
Subtracting the first from the second: 10x − 9x = 1x, 15y − 15y = 0, 100 − 93 = 7, so x = 7
Substitute back into either original equation: 3 × 7 + 5y = 31
21 + 5y = 31, so 5y = 10 and y = 2
The solution is x = 7, y = 2
Check in the equation not used for the substitution: 2 × 7 + 3 × 2 = 14 + 6 = 20
Had the matched terms carried opposite signs, they would be added instead

Worked example

Adding to eliminate, with a negative answer

Solve 2x + 3y = −1 and 5x − 6y = 38.

Solution:

Match the y terms: multiply every term of the first equation by 2, giving 4x + 6y = −2
Now +6y and −6y have opposite signs, so adding the two equations removes them
4x + 5x = 9x, 6y − 6y = 0 and −2 + 38 = 36, so 9x = 36 and x = 4
Substitute back into the first equation: 2 × 4 + 3y = −1
8 + 3y = −1, so 3y = −1 − 8 = −9 and y = −3
The solution is x = 4, y = −3
Check in the second equation: 5 × 4 − 6 × (−3) = 20 + 18 = 38

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