Solve the following system of linear equations:
x + y + z = 4
x + y + z = 4
2x + 2y + 2z = 8
I'd like some help understanding how to go about solving this. I know that there are infinitely many solutions with two arbitrary parameters - y and z. But I don't know which method to use to get there and also find the answer to x. I appreciate any advice.
Edit: I apologize to those who down voted my question. I'll explain what I've tried so far. I tried row reduction but I just end up with
1 1 1 | 4
0 0 0 | 0
0 0 0 | 0
I don't know what this means. Quite lost here. Any help on how to start this problem?
Your reduction (because of the two lines of zeros) shows that two equations are redundant. As you only have three, there is really only one equation there. The first two are identical and dividing the third by $2$ makes it the same as the first two. So really you only have the one equation $x+y+z=4$ You can choose any two of the variables to be the arbitrary parameters, and you have chosen $y$ and $z$. If I tell you what $y$ and $z$ are, can you find $x$? The solution you give should just be an equation $x=$ (something involving $y$ and $z$).