## Algebra and Trigonometry 10th Edition

$x=4; y=1; z=2$
Add the first equation to equation $2$, then subtract equation 3 from equation $2$. The new system of equations is as follows: $x+y+z=7 \\ 3x+2z=16 \\3z=6$ The third equation yields $z=2$ Plug the value of $z$ into the second equation, giving: $x=4$ Now, we will plug the value of $x$ and $z$ into the first equation to get the value of $y$. $4+y+2=7 \\ \implies y=1$ Thus, $x=4; y=1; z=2$