Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 7

$(x,y,z)=(5,1,-2)$

Given: $x-3y+z=0; y-z=3; z=-2$ Plug $z=-2$ in the second equation. we get $y-(-2)=3 \implies y=1$ Now, Plug $y=1,z=-2$ in the first equation. we have $x-3(1)+(-2)=0$ or, $x=3+2=5$ Hence, the solution is:$(x,y,z)=(5,1,-2)$