0580

Simultaneous Equations

Algebra and Sequences

Matching a coefficient

  • 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
Exam tip

If you cannot finish, write down any pair of values that fits one of the equations: 7 of the 43 mark schemes award a special-case mark for exactly that. And label which value is x and which is y — three schemes set aside a special case for candidates who found both and wrote them the wrong way round.

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