Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 12

$(x,y,z)=(\dfrac{-14}{4},1,8)$

Given: $4x+3z=10$ ; $2y-z=-6; \dfrac{z}{2}=4$ Here $\dfrac{z}{2}=4 \implies z=8$ Plug $ z=8$ in the second equation. we get $2y-8=-6 \implies y=1$ Now, Plug $y=1, z=8$ in the first equation. consider first equation $4x+3(8)=10$ or, $4x=10-24 \implies x=\dfrac{-14}{4}$ Hence, the solution is:$(x,y,z)=(\dfrac{-14}{4},1,8)$