Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 8

$(x,y,z)=(-3,-2,3)$

Given: $3x-3y+z=0; y+4z=10; z=3$ Plug $z=3$ in the second equation. we get $y+4(3)=10 \implies y=-2$ Now, Plug $y=-2,z=3$ in the first equation. we have $3x-3(-2)+(3)=0$ or, $3x=-9 \implies x=-3$ Hence, the solution is:$(x,y,z)=(-3,-2,3)$