Solving a system of linear equations having infinitely many solutions.

Question

\begin{align} x-\hphantom{2}y+2z+2t&=0 \\ 2x-2y+4z+3t&=1 \\ 3x-3y+6z+9t&=-3 \\ 4x-4y+8z+8t&=0 \end{align}

Solve for x,y,z and t

Answer 1

Then treat any two variables as constants and express other two in terms of those constants

For Example if you keep $x$ and $y$ as constants then $$z=\frac{y-x+2}{2}$$ $$t=-1$$

Answer 2

$2x-2y+4z+4t=0$
$2x-2y+4z+3t=1$
Then we get that t=-1 . And after that all equations become :
$x-y+2z=2$
So we have $t=-1$ and $x=2+y-2z$ , where $y$ and $z$ can be everything