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Simultaneous equations (systems of equations)

We solve simultaneous equations by turning two equations into one with only one unknown. With substitution we put the expression for y into the other equation. With addition we add the equations so that one unknown disappears. Finally, we work out the other unknown.

Year 9 and upper secondary · Updated

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What are simultaneous equations?

Two equations with two unknowns that must be true at the same time. x + y = 10 has many solutions. Together with x − y = 2, only x = 6 and y = 4 fit.

x + y = 10

x − y = 2

Two methods

We want one equation with only one unknown.

  • Substitution: get y on its own and put the expression, in brackets, into the other equation.
  • Addition: add the equations one side at a time so that one unknown disappears, as when +2y and −2y cancel each other out.
  • If needed, multiply a whole equation, every term on both sides, by a number.
  • Put in the value you found and work out the other unknown.

Adding is allowed because we add the same amount to both sides.

On a graph: where the lines cross

Each equation is a straight line. The solution is the point where the lines cross. y = 2x − 1 and y = 14 − 3x cross at (3, 5).

The linesNumber of solutionsWhat happens when you work it out
cross each otherone solutionyou get one value for x and one for y
are parallelno solutionx and y disappear and something false is left, like 0 = 5
lie on top of each otherinfinitely manyx and y disappear and something true is left, like 0 = 0

Example

Example 1: substitution, y = 2x − 1 and 3x + y = 14

We want an equation with only x. We replace y with 2x − 1.

  1. We put 2x − 1, in brackets, in place of y.

    3x + (2x − 1) = 14

  2. A plus in front of the bracket changes nothing. 3x + 2x = 5x.

    5x − 1 = 14

  3. We add 1 to both sides.

    5x = 15

  4. We divide both sides by 5.

    x = 3

  5. We put x = 3 into y = 2x − 1.

    y = 2 · 3 − 1 = 5

First: 2 · 3 − 1 = 5. Second: 3 · 3 + 5 = 14. Both are true, so x = 3 and y = 5.

Example 2: addition, 3x + 2y = 18 and 5x − 2y = 14

We want to get rid of y. With +2y and −2y the sum is zero y.

  1. We add the left-hand sides together and the right-hand sides together.

    3x + 2y + 5x − 2y = 18 + 14

  2. 3x + 5x = 8x and 2y − 2y = 0.

    8x = 32

  3. We divide both sides by 8.

    x = 4

  4. We put x = 4 into the first equation.

    12 + 2y = 18

  5. We subtract 12 from both sides and then divide by 2.

    2y = 6, so y = 3

First: 3 · 4 + 2 · 3 = 18. Second: 5 · 4 − 2 · 3 = 14. Both are true, so x = 4 and y = 3.

Example 3: cinema tickets

2 adults and 3 children cost 520 kr. 1 adult and 1 child cost 220 kr. Adult price x, child price y.

  1. We write each sentence as an equation.

    2x + 3y = 520 and x + y = 220

  2. We multiply the whole second equation by 2, every term on both sides.

    2x + 2y = 440

  3. We subtract the new equation from the first, one side at a time. 3y − 2y = y and 520 − 440 = 80.

    y = 80

  4. We put y = 80 into x + y = 220 and subtract 80 from both sides.

    x = 220 − 80 = 140

2 · 140 + 3 · 80 = 520 kr and 140 + 80 = 220 kr. An adult ticket costs 140 kr and a child ticket 80 kr.

Pitfalls

  • Stopping once x is found and forgetting to work out y.

    The solution is a pair. Put in x to get y, and answer x = … and y = ….

  • Multiplying only one term, so that x + y = 220 times two becomes 2x + y = 440.

    Every term on both sides is multiplied: 2x + 2y = 440.

  • Forgetting the bracket: 3x − y = 4 with y = 2x − 1 becomes 3x − 2x − 1 = 4.

    Write 3x − (2x − 1) = 4. The minus changes the sign of both terms: 3x − 2x + 1 = 4, so x = 3.

Practise on your own

0 of 4 correct

  1. Solve the simultaneous equations y = x + 3 and x + y = 11. Answer in the form x = …, y = ….

  2. Solve the simultaneous equations x + y = 9 and x − y = 3. Answer in the form x = …, y = ….

  3. Solve the simultaneous equations y = 3x − 2 and 2x + y = 13. Answer in the form x = …, y = ….

  4. Solve the simultaneous equations 3x + 2y = 16 and x + y = 6. Answer in the form x = …, y = ….

Common questions

Which method should I choose?

If x or y is already on its own, substitution is a good choice. If an unknown has opposite numbers in front of it, like +2y and −2y, addition is a good choice. If the numbers are the same, we subtract instead. Both methods give the same answer.

How do I check the answer?

Put both x and y into both equations. The solution must be true in both at the same time.

How do I write the answer?

Write both values, for example x = 4 and y = 2. As a point it is (4, 2), with the x value first.

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