Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 11

$(x,y,z)=(5,2,-\dfrac{1}{2})$

Given: $2x-y+6z=5$ ; $y+4z=0; -2z=1$ Here $-2z=1 \implies z=-\dfrac{1}{2}$ Plug $ z=-\dfrac{1}{2}$ in the second equation. we get $y+4( -\dfrac{1}{2})=0 \implies y=2$ Now, Plug $y=2, z=-\dfrac{1}{2}$ in the first equation. consider first equation $2x-2+6(-\dfrac{1}{2})=5$ or, $2x=5+5 \implies x=5$ Hence, the solution is:$(x,y,z)=(5,2,-\dfrac{1}{2})$