
3 Variable Equation
I've been looking on youtube to try and learn how to do these more complex equations but all the examples use small numbers that easily fall into place.
Ive got
Equation One: 48A  32B  2C = 0
Equation Two: 32A +46B  11C = 0
Equation Three: 2A  11B + 13C = 54
Ive mulitplied equation three by 16 which gives me:
NewThree: 32A + 176B  208C = 864
Then I added equations Two and NewThree:
Equation Two: 32A + 46B  11C = 0
NewThree: 32A + 176B  208C = 864
Which cancelled out the 32A and gave me:
222B 219C = 864
I then took Equation Three and mulitplied it by 24 which gave me:
Anothernewthree: 48A  264B + 312C = 1296
Then added Equation One and Three together:
Equation One: 48A  32B  2C = 0
Anothernewthree: 48A  264B + 312C = 1296
Which cancelled out the 48A leaving me with:
296B + 310C = 1296
Which left me with:
222B 219C = 864
296B + 310C = 1296
And it's at this point that I get stuck :/
Is this the best route to take? Is there another method that is better to take for these larger numbers? The examples on youtube use examples that have small numbers and are easily solved, but I don't know what step to take next for this one :/
If anyone could help me I'ld be very grateful :)

Re: 3 Variable Equation
You can make the arithmetic easier to handle by simplifying some of the equations.
For example your equation 1 could have been divided throughout by 2.
With the two equations you are left with, the first can be divided throughout by 3 and the second by 2.
Also, the calculation becomes easier with a better choice of which equations to combine.
For example you could have started by adding eq. 2 to eq. 1 to get you 16a + 14b  13c = 0, (with the general idea of reducing the size of the coefficients), and then adding twice this to eq. 2.
That gets you 74b  37c = 0, and that can be divided throughout by 37, yielding 2b  c = 0.
There are other methods, but when dealing with integers like this, this is as good as any.

Re: 3 Variable Equation
Well I've solved it :) Many thanks to you for helping me :)