Answer
$x=-2t+2 $
$ y=\frac{4}{3}t-\frac{2}{3} $
$z=t $;
$t \in R $
Work Step by Step
Express x from the first equation.
$x+3y-2z=0 \rightarrow x=-3y+2z$
$2x+4z=4$
$4x+6y=4$
Substitute $ x=-3y+2z$ into the second and the third equation.
$2(-3y+2z)+4z=4$
$4(-3y+2z)+6y=4$
Simplify.
$-6y+8z=4$
$-6y+8z=4$
Multiply the first equation by -1.
$-6y+8z=4 \rightarrow 6y-8z=-4$
Add these equations:
$6y-8z=-4$
$-6y+8z=4$
$0=0 \rightarrow$ This means that the system has infinitely many solutions, but x,y and z cannot be any random numbers. To describe the complete solution of the system, let $z=t$ where t is any real number. Use back-substitution to express x and y.
$z=t$
$6y-8z=-4 \rightarrow 6y=8t-4 \rightarrow y=\frac{4}{3}t-\frac{2}{3} $
$x=-3y+2z \rightarrow x=-3(\frac{4}{3}t-\frac{2}{3})+2t \rightarrow x=-2t+2 $
