1. ## Back-substitution help

The answer to this system is probably really simple but the book doesn't explain it very well, so I'm asking you guys.

x+y+z=0
y=0

I'm not sure if the answer is no solution or if there is a way I can go further by maybe doing something like x+z=0-y. Any help would be much appreciated.

2. Originally Posted by learningguy
The answer to this system is probably really simple but the book doesn't explain it very well, so I'm asking you guys.

x+y+z=0
y=0

I'm not sure if the answer is no solution or if there is a way I can go further by maybe doing something like x+z=0-y. Any help would be much appreciated.
What you have is an under determined system.

If you write this as an augmented matrix you get

$\begin{bmatrix} 1 & 1 & 1 &0 \\ 0 & 1& 0 & 0 \end{bmatrix}$

Since the z column does not have a leading 1 it is you free parameter.
So lets set $z=t, \, \, t \in \mathbb{R}$
We know that $y=0$ and that $x=-z=-t$

So this gives the solution to the system as

$\begin{pmatrix} -t \\ 0 \\t\end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\1\end{pmatrix}t$

So every solution to this system is one the above line.

3. Thanks alot, sometimes I think I make stuff too hard after you explained I now see how the book explained it after looking through some of the examples again but it didn't make it very obvious. I have another small question too though. There is another question that wants me to solve a system.

1/2x - 1/3y = 1
-2x + 4/3y = -4

In order to solve that system I think I would have to multiply the top equation by 4 but then that would make the answer 0 = 0. Would the answer to that system be no solution?