# A slightly different way to solve systems of equations

I'm sure all of us remember solving systems of equations in high school.  Back then I had the dreams of discovering a new branch of mathematics, earning the right to have some obscure function - or if I was lucky - a constant named after myself.  Like most things when I was that age, I quickly lost interest and picked up something else for the moment.  I did succeed though at coming up with a rather handy way to solve systems of equations.  Of course there are a multitude of ways to do this but I've often found this way to be very quick and easy - much easier at least than the solving/substitution method we learned in school.  I haven't shared this until now but I thought I would throw it up for comments.

To start, we need to know the degree of the equation before solving.  We also need at least two inputs and their corresponding outputs to solve the equation.  So, for example say we have as inputs {5,3} and as outputs {8,4}.  We then write these equations as follows.

5g = 8

3g = 4

Originally I used squigly lines over the 5 and the 3, but with this text editor I find it easier to use g as a variable for the equation.  The first step is to subtract.

5g = 8

3g = 4

2g = 4

Then solve for g.

g = 2

This gives us the first part of the equation - 2x.  Then take the first term and multiply by 2.

5g = 8  10

3g = 4  6

2g = 4

Subtracting the 10 from the 8 gives us -2.  Therefore the equation is 2x - 2.

I used this most often to solve linear systems of equations.  This same procedure can be extended to higher order equations if you know the degree - though I haven't determined if this will always work.

Say we have a second degree equation with the following inputs and outputs.

5g = 93

3g = 37

The first step is to square both inputs (because we know there is an x-squared term).

25g = 93

9g =   37

Then subtract.

25g = 93

9g =   37

16g = 56

Dividing 56 by 16 and discarding the remainder gives us 3.  We now need to remove the squared terms from both sides.

5g = 93 – 75 = 18

3g = 37 – 27 = 10

Then perform the same steps for the new equation to solve for x.  The equation comes out to 3x^2 + 4x - 2

Of course this will only work if the lower degree terms are all low enough to not be greater than the highest degree term for each piece of input.  If this is not the case, more inputs and outputs can be used - not exceeded one more than the degree.

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