Chapter 10 - Systems of Equations and Inequalities - Section 10.1 Systems of Linear Equations: Substitution and Elimination - 10.1 Assess Your Understanding - Page 734: 43

$(2,-1,1)$

1. Add up the first and the third equations to get $-2x+z=-3$ or $2x-z=3$ 2. Multiply 2 to the second equation and add the result to the first equation, we can get $5x+5z=15$ or $x+z=3$ 3. Add the above two equations to get $3x=6$, thus $x=2$ 4. Back substitute to get $z=3-x=1$ then $y=4-2x-z=4-2(2)-1=-1$ 5. The solution is $(2,-1,1)$