Posted by **Linus** on Monday, January 17, 2011 at 4:34pm.

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

- Linear Algebra -
**EssKay**, Monday, January 17, 2011 at 4:57pm
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!

- Linear Algebra -
**Linus**, Monday, January 17, 2011 at 5:48pm
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?

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