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Mathematics for science and technology
Mathematics for science and technology

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5 Simultaneous equations

Simultaneous equations are pairs of equations that are both true (i.e. they are simultaneously true). They are both expressed as equations with two unknowns. By making one of these unknowns the subject of both equations, you can then substitute the subject in one equation and then solve for the other unknown. Then you can substitute back into the equation and solve for the first unknown.

This is easier to see in this example.

Here is a pair of simultaneous equations to solve for x and y.

  • 2y = x + 7
  • y = x + 2

First, you need to rearrange these equations to make either x or y the subject. The order doesn’t matter, as the answers that you will get must be the same. So, starting by making x the subject.

2y = x + 7 subtract 7 from both sides 2y − 7 = x + 7 − 7 So, 2y − 7 = x
y = x + 2 subtract 2 from both sides y − 2 = x + 2 − 2 So, y − 2 = x

Therefore, x = 2y − 7 and x = y − 2

Now there are two expressions for x, so the right-hand sides must be equal, and you can equate them:

So, 2y – 7 = y – 2 deduct y from both sides 2y – y – 7 = y – y – 2
So, y – 7 = –2 add 7 to both sides y – 7 + 7 = –2 + 7

Therefore, y = 5

You can then substitute this back into either of the original equations to find x.

2y = x + 7

So, 10 = x + 7

Therefore, x = 3

As a last check, make sure that this is true of the other equation.

x + 2 = y

x = y – 2 = 5 – 2 = 3

So the solution is x = 3, y = 5. Finally, check that these values satisfy both the original equations.

Have a go yourself in this final activity for the week.

Activity 6 Simultaneous equations

Timing: Allow about 10 minutes

Solve the following pairs of simultaneous equations.

(a) y = x + 10 3y = 2x + 5
(b) y = 4x y = 3x + 5
(c) y = 7x + 4 3y = x + 7
(d) y = 2x + 4 3y = x + 7

Answer

Equations

Step 1: rearrange

to make y the

subject for both

equations

Step 2: solve for x Step 3: solve for y Solution
(a) y = x + 10

y = x + 10

x plus 10 equals left parenthesis two times x plus five right parenthesis divided by three

y = x + 10 x = –25
3y = 2x + 5

y equals left parenthesis two times x plus five right parenthesis divided by three

3x + 30 = 2x + 5 y = –25 + 10 = –15 y = –15
x = –25
(b) y = 4x 4x = 3x + 5 y = 4x = 20 x = 5
y = 3x + 5 x = 5 y = 20
(c) y = 7x + 4 y = 7x + 4

seven times x plus four equals left parenthesis x plus seven right parenthesis divided by three

y = 7x + 4 x = –0.25
3y = x + 7

y equals left parenthesis x plus seven right parenthesis divided by three

21x + 12 = x + 7 y = –1.75 + 4 = 2.25 y = 2.25
20x = –5
x = –0.25
(d) y = 2x + 4 y = 2x + 4

two times x plus four equals left parenthesis x plus seven right parenthesis divided by three

y = 2x + 4 x = –1
3y = x + 7

y equals left parenthesis x plus seven right parenthesis divided by three

6x + 12 = x + 7 y = –2 + 4 = 2 y = 2
5x = – 5
x = –1

There has been a lot to take on board this week. As with any maths skill the more you practice the easier algebra will become, so before completing this week you’ll now have the chance to practice the topics in the end-of-week quiz.