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How I Was Taught How To Do Simultaneous Equations
Algebra - simultaneous equations
Date : 06/10/2020
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
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
Anyway, I hope you find this to be useful.
Peter
This resource was uploaded by: Peter