Teach Me Maths

Simultaneous Equations

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How to solve a pair of simultaneous equations by adding, subtracting or multiplying them, or by substitution, how the answer shows on a graph, and how to form the equations from a problem in words.

What simultaneous equations are

Simultaneous equations are two equations with the same two unknowns, such as x+y=10 and x-y=2. Solving them means finding the one pair of values that makes both equations true at the same time.

Lots of pairs fit x+y=10 on its own: x=1 and y=9, or x=5 and y=5, or x=6 and y=4. Only one of these also fits x-y=2. It is x=6 and y=4, because 6-4=2

In the exam you are usually asked to solve the simultaneous equations, for 3 marks, often as the last question on the paper. Give both values: x=\dots and y=\dots. Some questions are written in words instead, such as the prices of two things in a café, for up to 5 marks. Now and then you read the answer off a graph, for 1 mark.

Write down your method. Guessed answers score nothing if they are wrong, and only 1 mark out of 3 if they are right with no working. Check your pair in both of the original equations at the end.

The solution on a graph

Every point on a straight line fits that line's equation. The point where two lines cross is on both lines, so it fits both equations. Its coordinates are the solution of the simultaneous equations.

These lines cross at (2, 4), so the solution of y=x+2 and x+y=6 is x=2, y=4. Check: 2+2=4 and 2+4=6

If a question gives you the two lines drawn on a grid and asks you to use them, just read off the x and y values where they cross. You don't need any algebra.

Adding or subtracting the equations

The usual method is elimination: you add or subtract the two equations to get rid of one letter. That leaves one equation with one letter, which you can solve.

Number the equations (1) and (2), so you can say which one you are using. Then look for a letter with the same number in front of it in both equations. In 3x+y=14 and x+y=6, both have one y.

  • If the signs in front of that letter are the same, as in +y and +y, subtract one equation from the other.
  • If the signs are different, as in +3y and -3y, add the equations.

For 3x+y=14 and x+y=6, subtract: 3x-x=2x, y-y=0 and 14-6=8. This leaves

2x=8

so x=4. Then put x=4 into either equation to find y: 4+y=6, so y=2

Worked example 1: Subtracting the equations

Solve the simultaneous equations

4x+y=23

2x+y=13

  1. Call the equations (1) and (2). Both have +y. The signs are the same, so subtract (2) from (1).
  2. 4x-2x=2x and 23-13=10, so 2x=10 and x=5

  3. Your turn. Put x=5 into (2): 2\times5+y=13. What is y?

    10+y=13, so y=13-10=3

  4. Check in (1): 4\times5+3=20+3=23
  5. Answer: x=5, y=3

Worked example 2: Adding the equations

Solve the simultaneous equations

4x+3y=6

2x-3y=12

  1. The y-terms are +3y in (1) and -3y in (2). The signs are different, so add the equations.
  2. 4x+2x=6x, 3y-3y=0 and 6+12=18, so 6x=18 and x=3

  3. Your turn. Put x=3 into (1): 12+3y=6. What is y?

    3y=6-12=-6, so y=-6\div3=-2

  4. Check in (2): 2\times3-3\times(-2)=6+6=12
  5. Answer: x=3, y=-2

When no letter matches, multiply first

Often no letter has the same number in front of it in both equations. In 5x+2y=16 and 3x+y=9, the y-terms are 2y and y. Multiply every term of the second equation by 2 to get 6x+2y=18. Now both equations have 2y.

Multiply every term, including the number after the equals sign. 3x+y=9 times 2 is 6x+2y=18, not 6x+2y=9

Then add or subtract, using the same rule about signs. Here both have +2y, so subtract: 6x-5x=x and 18-16=2, so x=2. Then 3\times2+y=9, so y=3

Don't add or subtract the equations before a letter matches. No letter would disappear, and you would still have two unknowns.

Worked example 3: Multiplying one equation first

Solve the simultaneous equations

3x+2y=9

x+4y=8

  1. No letter matches. Multiply every term of (1) by 2: 6x+4y=18. Call this (3).
  2. Equations (3) and (2) both have +4y, so subtract (2) from (3): 6x-x=5x and 18-8=10, so 5x=10 and x=2

  3. Your turn. Put x=2 into (2): 2+4y=8. What is y? Give your answer as a decimal.

    4y=8-2=6, so y=6\div4=1.5

  4. Check in (1): 3\times2+2\times1.5=6+3=9
  5. Answer: x=2, y=1.5. An answer does not have to be a whole number.

When one equation gives a letter

Sometimes one equation already says what a letter equals, such as y=x+4. Then you can use substitution: in the other equation, write what that letter equals in its place.

For y=x+4 and 2x+y=13, write x+4 in place of y: 2x+x+4=13. Now there is only one letter: 3x+4=13, so 3x=9 and x=3. Then y=3+4=7

If the letter is multiplied in the other equation, put what it equals in a bracket. With y=x+4, the term 2y becomes 2(x+4), which is 2x+8

Worked example 4: Where two lines cross

The lines y=2x-3 and 4x+y=21 cross at the point P.

Work out the coordinates of P.

  1. P is on both lines, so solve the equations simultaneously. Write 2x-3 in place of y in 4x+y=21: 4x+2x-3=21
  2. Collect the x-terms: 6x-3=21, so 6x=24 and x=4

  3. Your turn. Use y=2x-3 to find y. Write down the coordinates of P.

    Write your answer as (x, y)

    y=2\times4-3=8-3=5, so P is (4, 5)

  4. Check in the other equation: 4\times4+5=16+5=21
  5. Answer: (4, 5)

Forming equations from a problem

Some questions give the facts in words. Choose a letter for each unknown and write down what it stands for, such as s for the cost of one sandwich, in pence.

Turn each fact into an equation. If 3 sandwiches and 2 drinks cost £9, then 3s+2d=900, working in pence so there are no decimals. Then solve the two equations as before.

When the two facts share an amount, you can compare them in words. This is elimination without letters. If 2 pens and 1 ruler cost £1.60, and 2 pens and 3 rulers cost £2.40, the difference is 2 rulers, costing 80p. So 1 ruler costs 40p. Then 2 pens cost £1.20, so 1 pen costs 60p.

At the end, read the question again. Answer exactly what it asks, with units, and check your values fit every fact.

Worked example 5: Two kinds of ticket

A theatre sells adult tickets for £12 each and child tickets for £7 each.

One evening it sells 80 tickets and takes £785.

How many adult tickets and how many child tickets does it sell?

  1. Let x be the number of adult tickets and y the number of child tickets. There are 80 tickets, so x+y=80 (1). The money is £785, so 12x+7y=785 (2).
  2. Multiply every term of (1) by 7: 7x+7y=560. Call this (3).
  3. Subtract (3) from (2): 12x-7x=5x and 785-560=225, so 5x=225

  4. Your turn. How many adult tickets does the theatre sell?

    x=225\div5=45

  5. Then y=80-45=35. Check: 12\times45+7\times35=540+245=785
  6. Answer: 45 adult tickets and 35 child tickets.

Worked example 6: Working out the change

At a bakery, 3 bread rolls and 1 cake cost £1.85

1 bread roll and 2 cakes cost £1.70

Sam buys 2 bread rolls and 3 cakes. She pays with a £5 note.

Work out how much change she should get.

  1. Work in pence. Let r be the cost of a roll and c the cost of a cake: 3r+c=185 (1) and r+2c=170 (2).
  2. Multiply every term of (2) by 3: 3r+6c=510. Call this (3).
  3. Subtract (1) from (3): 6c-c=5c and 510-185=325, so 5c=325 and c=65

  4. Your turn. Put c=65 into (2): r+130=170. What is r, in pence?

    r=170-130=40. Check in (1): 3\times40+65=120+65=185

  5. Sam's order costs 2\times40+3\times65=80+195=275 pence. Her change is 500-275=225 pence.
  6. Answer: \pounds 2.25

Common mistakes

  • Adding or subtracting before a letter matches. Multiply one equation first, so the numbers in front of one letter are the same.
  • Multiplying only some of the terms. Multiply every term, including the number after the equals sign.
  • Adding when you should subtract. Same signs in front of the matching letter, subtract; different signs, add.
  • Losing a minus sign. Take care with negative numbers: 5-(-3)=8
  • Stopping after one letter. Put the value back into one of the equations to find the other letter.
  • Guessing the answers. Show your method, then check your pair in both equations.

Practice questions

More on this topic: Simultaneous Equations worksheet with full solutions