0580

Simultaneous Equations

Algebra and Sequences

When one unknown is already isolated

  • Rearrange one equation to make an unknown the subject, then substitute that expression into the other
  • This suits an equation such as 2x − y = 1, which rearranges easily to y = 2x − 1
  • Substituting gives one equation in one unknown
    • With 4x + 3y = 27, substituting y = 2x − 1 gives 4x + 3(2x − 1) = 27, so 10x = 30 and x = 3, giving y = 5
  • Substitution is the method that also handles non-linear pairs
Exam tip

Substitute your pair back into the other equation, not the one you rearranged. The one you rearranged will fit whatever you found, so it cannot catch an arithmetic slip; the other one will.

Worked example

Solving by substitution

Solve the simultaneous equations y = 2x + 1 and 3x + y = 11.

Solution:

  • One equation already has y on its own, so substitute that expression into the other
  • 3x + (2x + 1) = 11
  • Collect: 5x + 1 = 11
  • Subtract 1: 5x = 10, so x = 2
  • Substitute back into the isolated equation: y = 2 × 2 + 1 = 5
  • The solution is x = 2, y = 5
  • Check in the equation not used for the substitution: 3 × 2 + 5 = 11