Answer
$x=-t+3$
$y=2t-3$
$z=t$
Work Step by Step
Express x from the first equation.
$x+y-z=0 \rightarrow x=-y+z$
$x+2y-3z=-3$
$2x+3y-4z=-3$
Substitute $ x=-y+z$ into the second and the third equation.
$-y+z+2y-3z=-3$
$2(-y+z)+3y-4z=-3$
Simplify.
$y-2z=-3$
$y-2z=-3$
Multiply the first equation by -1.
$y-2z=-3 \rightarrow -y+2z=3$
Add these equations:
$-y+2z=3$
$y-2z=-3$
$0=0 \rightarrow$ 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.
$z=t$
$y-2z=-3 \rightarrow y=2t-3$
$x=-y+z \rightarrow x=-t+3$
