Hey kjtruitt.

By using MATLAB, we get the following row-reduced echelon matrix:

>> A = [0.3, 0.4, 3.5; 0.4, 0.2, 0.3; 0.3, 0.4, 3.5]

A =

0.3000 0.4000 3.5000

0.4000 0.2000 0.3000

0.3000 0.4000 3.5000

>> rref(A)

ans =

1.0000 0 -5.8000

0 1.0000 13.1000

0 0 0

This confirms your solution of needing an extra parameter. To solve this we solve the augmented system and we get the following solution:

b = 0.6667 0.6667 0.6667

>> B = [A , transpose(b)]

B =

0.3000 0.4000 3.5000 0.6667

0.4000 0.2000 0.3000 0.6667

0.3000 0.4000 3.5000 0.6667

>> rref(B)

ans =

1.0000 0 -5.8000 1.3333

0 1.0000 13.1000 0.6667

0 0 0 0

This translates into x - 5.8z = 4/3 and y + 13.1z = 2/3. If you set z = t as the book has done you get:

x = 4/3 + 5.8t

y = 2/3 - 13.1t

z = t

This is closer to what you said but its not exactly the same. Given what you said though, this should be answer (this was calculated by a computer and not by hand).