Answer
$ x=7t+1$
$y=4t-1$
$z=t$
Work Step by Step
Express x from the first equation.
$x-2y+z=3 \rightarrow x=2y-z+3$
$2x-5y+6z=7$
$2x-3y-2z=5$
Substitute $ x=2y-z+3$ into the second and the third equation.
$2(2y-z+3)-5y+6z=7$
$2(2y-z+3)-3y-2z=5$
Simplify.
$-y+4z=1$
$y-4z=-1$
Add the equations.
$0=0 \rightarrow$ 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.
$z=t$
$y-4z=-1 \rightarrow y=4t-1$
$x=2y-z+3 \rightarrow x=7t+1$
