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How I Was Taught How To Do Simultaneous Equations

Algebra - simultaneous equations

Date : 06/10/2020

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Peter

Uploaded by : Peter
Uploaded on : 06/10/2020
Subject : Maths

I have been tutoring for around 18 months, and every time I have a session which involves solving simultaneous equations at GCSE level my students always say `why are we not shown this way at school?`

Let`s look at the way my maths teacher Mr Duggan taught me ...

  • Below is a typical set of simultaneous equations:
6x + 3y = 21 - let`s call this equation (1)2x + 3y = 17 - let`s call this equation (2)
  • Here we can see that both equation (1) and equation (2) have a common coefficient for y which is 3. So let`s rearrange these two equations as follows:
3y = 21 - 6x & (1)3y = 17 - 2x & (2)
  • Now if 3y = 21 - 6x and also 3y = 17 - 2x it follows that:
21 - 6x = 17 - 2x - let`s add 6x to both sides of this equation:21 - 6x + 6x = 17 - 2x + 6x21 = 17 + 4x - let`s subtract 17 from both sides of this equation:21 - 17 = 17 - 17 + 4x
4 = 4x or put another way around 4x = 4 - let`s divide both sides of this equation by 4:
4x/4 = 4/4
x = 1
  • So, now all we need do is substitute this value of x into either of the above equations to find the value of y:
  • Taking equation (1) we would get:
6x + 3y = 21 - substituting in the value of x = 4 gives us:
(6 x 1) + 3y = 216 + 3y = 21 - let`s subtract 6 from both sides of this equation:
6 - 6 + 3y = 21 - 6
3y = -15 - let`s divide both sides of this equation by 3:
3y/3 = 15/3y = 5
  • But what if we had taken equation (2). In this case we would get:
2x + 3y = 17 - substituting in the value of x = 4 gives us:(2 x 1) + 3y = 172 + 3y = 17 - let`s subtract 2 from both sides of this equation:
2 - 2 + 3y = 17 - 2
3y = 15 - let`s divide both sides of this equation by 3:
3y/3 = 15/3y = 5
  • So the solution to this pair of simultaneous equations is x = 1 and y = 5
Now, admittedly this can look a bit long winded but I wanted to show each step of the process. Where this really comes into its own is with simultaneous equations where there is no common coefficient or indeed where one of the equations is linear and the other is a quadratic.
The numbering of the equations comes into its own here. For example if you had to solve the following pair of simultaneous equations:
4x - y = 30 & & - equation (1)3x + 2y = 17 - equation (2)
You would simply multiply equation (1) by 2 and leave equation (2) as it is to give:
8x - 2y = 60 - therefore 2y = 8x - 60
3x + 2y = 17 - therefore 2y = 17 - 3x
Therefore 8x - 60 = 17 - 3x11x = 77x = 7
Substituting x = 7 into (1) gives us:28 - y = 30y = 28 - 30y = -2
  • The final answer is x = 7, y = -2
I always liked this neat way of solving a pair of simultaneous equations - and on speaking with my students they said they had been struggling with the elimination method and saw this as being much easier to understand.

Anyway, I hope you find this to be useful.
Peter

This resource was uploaded by: Peter