Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 9

$(x,y,z)=(4,0,3)$

Given: $x+2y+z=7; -y+3z=9; 2z=6$ Here $2z=6 \implies z=3$ Plug $z=3$ in the second equation. we get $-y+3(3)=9 \implies y=0$ Now, Plug $y=0,z=3$ in the first equation. we have $x+2(0)+3=7$ or, $x=7-3 \implies x=4$ Hence, the solution is:$(x,y,z)=(4,0,3)$