## College Algebra 7th Edition

$(x,y,z)=(4,3,4)$
Given: $x-2y+3z=10$ ; $2y-z=2; 3z=12$ Here $3z=12 \implies z=4$ Plug $z=4$ in the second equation. we get $2y-4=2 \implies y=3$ Now, Plug $y=3,z=4$ in the first equation. we have $x-2(3)+3(4)=10$ or, $x=10-6 \implies x=4$ Hence, the solution is:$(x,y,z)=(4,3,4)$