I've noticed a couple of tweets recently about the importance of being playful when doing mathematics and this has brought to mind an insight I had about simultaneous linear equations that occurred when I was being playful with them in technology. As a result of this I now have a different method for solving that I prefer to the standard textbook approaches.

### Playing with simultaneous equations

In trying to construct a pair of linear simultaneous equations where the solution went through a given point. I can't remember the point but I'll use (3,2) for the example. I did this by creating a point A at (3,2) then two more points B and C and finding the lines that went through these and A. This gave me the simultaneous equations:3

*x*+

*y*= 11

*x*+ 5

*y*= 13

I then moved the points B and C around to find some different equations that would have the same point of intersection. This is probably best displayed here by showing all the different lines I got in a different colour.

The equations for these lines are:

4*x*–

*y*= 10

3

*x*+

*y*= 11

2

*x*+ 3

*y*= 12

*x*+ 5

*y*= 13

*y*= 2

By playing around with these I noticed some interesting features:

- There are infinitely many lines through the point A.
- The coefficients of these lines are related linearly.
- One of these lines will be horizontal and one will be vertical (and hence have a simple equation just in terms of
*x*or*y*)*.*

*x*+ 7

*y*= 14 makes the linear relationship more obvious.

### A simple method for solving linear simultaneous equations

Putting all this together I realised this could be used to form a simple method for solving simultaneous linear equations. They can be solved by moving linearly along the coefficients until one of the coefficients of either*x*or

*y*is 0. For example:

3

*x*+

*y*= 11

*x*+ 5

*y*= 13

The gap from 3

*x*to

*x*is –2

*x*, so if you move half this gap again (–

*x*) from

*x*you'll have 0

*x*.

Moving an equivalent amount for the other terms means you need to move half the gap from

*y*to 5*y*, i.e. +2*y*, on from 5*y*to get 7*y*and half the gap from 11 to 13, i.e. +1, to get 14.This gives 0

*x*+ 7

*y*= 14 and hence

*y*= 2. Then using the first equation 3

*x*= 9 gives

*x =*3.

I've tried this method with lots of linear simultaneous equations and I think it's quicker and makes more sense to me. It's the method I know use if I need to solve them.

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