GCSE Maths · Algebra

Simultaneous Equations

Solving pairs of linear equations, plus one-linear-one-quadratic systems at Higher.

200 GCSE-style practice questions

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Simultaneous Equations, explained point by point

Everything the GCSE specification expects you to be able to do, and how to actually do it - the same lesson a signed-in student studies from.

  1. Solve two linear simultaneous equations by elimination

    Match the coefficients of one letter, then add or subtract the equations to eliminate it. For 3x + 2y = 12 and x − 2y = 4, the y terms already match: adding gives 4x = 16, so x = 4, then y = 0 by substituting back. Same signs subtract, opposite signs add.

  2. Solve two linear simultaneous equations by substitution

    Rearrange one equation to "y = …" (or "x = …") and substitute it into the other. From y = 2x − 1 and 3x + y = 14: 3x + 2x − 1 = 14, so x = 3 and y = 5. Substitution shines when one equation already has a letter on its own.

  3. Solve simultaneous equations graphically, as the point where two lines cross

    Each equation draws a line; the solution is the crossing point, because it is the only point on both. Plot both lines and read off the coordinates. Parallel lines never cross - that is the picture of simultaneous equations with no solution.

  4. Set up simultaneous equations from worded problems

    Two unknowns need two facts. "3 teas and 2 coffees cost £9.40; 1 tea and 2 coffees cost £6.20" becomes 3t + 2c = 9.4 and t + 2c = 6.2. Define your letters clearly, solve, and answer in the units the question asked for.

  5. Solve one linear and one quadratic equation simultaneouslyHigher only

    Substitute the linear equation into the quadratic one and solve the resulting quadratic. For y = x + 1 and x² + y² = 25: x² + (x + 1)² = 25 leads to 2x² + 2x − 24 = 0, giving x = 3 or x = −4. Each x needs its matching y - pair the answers as coordinates.

Practice

Try a Simultaneous Equations question

A GCSE-style original question from this topic. Have a go before you open the working - deciding on an answer first is what makes the working stick.

Solve the simultaneous equations 2x + 3y = 16 and x + y = 6. Find the value of y.

  1. y = 10
  2. y = 4
  3. y = 3
  4. y = 2
Show the answer and the working

Answer: y = 4

Doubling x + y = 6 gives 2x + 2y = 12; subtracting this from the first equation leaves y = 4.

  1. Multiply x + y = 6 by 2: 2x + 2y = 12.
  2. Subtract from 2x + 3y = 16: (2x + 3y) − (2x + 2y) = 16 − 12.
  3. This gives y = 4.

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