Linear Algebra
posted by Linus .
Express the solutions of the following systems in terms of the free variables:
x1 + 3*(x2)  2*(x3) + 2*(x5) = 0
2*(x1) + 6*(x2)  5*(x3)  2*(x4) + 4*(x5)  3*(x6) = 1
5*(x3) + 10*(x4) + 15*(x6) = 5
2*(x1) + 6*(x2) + 8*(x4) + 4*(x5) + 18*(x6) = 6

Hey Linus,
First step: simplify where you can. For example, equation 3 can be divided by 5, so you end up with (x3) + 2(x4) + 3(x6) = 1. Equation 4 can also be simplified. This will make the algebra later a little easier.
Second step: Pick an x (x1 seems to be a good one) to start solving, and begin moving around your equations.
Do you know where to go from here? (Or are you onto matrices already? If so ignore step 2 and build your matrix). If you're stuck show me what you've done so far and I'll help out! 
I got:
x1 = 3*(x2) 2*(x5) + 66*(x6)  22
x2 = free
x3 = 33*(x6)  11
x4 = 18*(x6) + 6
x5 = free
x6 = free
Is that correct?