How Should Solutions to a System of Linear Equations Be Expressed for Clarity?

Kiefer
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Find all solutions to the following system of linear equations:
(x1) – 2(x2) – (x3)+(x4)=1
2(x1) – 3(x2) + (x3) – (x4)=6
3(x1) – 3(x2) + 6(x3))=15
(x1) + 5(x3)+(x4)=9

Using a system of linear equations, I found:
1 -2 -1 0 1
0 1 3 -3 4
0 0 0 6 0
0 0 0 0 0

so three solutions are:
(x1)=9, (x2)=4, (x3)=0, (x4)=0
(x1)=4, (x2)=1, (x3)=1, (x4)=0
(x1)=-1, (x2)=-2, (x3)=2, (x4)=0

How do I write my final solution (ie:what form)?
 
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Kiefer said:
Find all solutions to the following system of linear equations:
(x1) – 2(x2) – (x3)+(x4)=1
2(x1) – 3(x2) + (x3) – (x4)=6
3(x1) – 3(x2) + 6(x3))=15
(x1) + 5(x3)+(x4)=9

Using a system of linear equations, I found:
1 -2 -1 0 1
0 1 3 -3 4
0 0 0 6 0
0 0 0 0 0

so three solutions are:
(x1)=9, (x2)=4, (x3)=0, (x4)=0
(x1)=4, (x2)=1, (x3)=1, (x4)=0
(x1)=-1, (x2)=-2, (x3)=2, (x4)=0

How do I write my final solution (ie:what form)?

Every solution is a point on a line that goes through <9, 4, 0, 0>. Your book should have some examples of representing lines with a parameter.

I would advise finishing your row reduction to get the matrix in reduced row-echelon form.
 
Express it into its nullspace (all the special solutions) and particular solution, if you've done that in linear algebra? The sum of the nullspace and particular solution gives the complete solution.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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