## Algebra and Trigonometry 10th Edition

$x=5; y=-3, z=3$
Add the first equation and the third equation to eliminate the value of $z$. So, we get: $3x+5y=0...(a)$ Next, multiply the first equation by $3$ and add it to the second equation to eliminate the value of $z$. $3(2x+4y+z)+x-2y-3z=(3)(1)+2 \\ \implies 7x+10y=5 ...(b)$ The new system of equations is: $3x+5y=0...(a) \\ 7x+10y=5 ...(b)$ Now, we will multiply the first equation by $-2$ and add it to the second equation to eliminate $y$: $-2(3x+5y)+7x+10y =(-2)(0)+5 \\ x=5$ and $3x+5y=0 \\ (3)(5)+5y=0 \\ y=-3$ Now, we have: $x+y-z=-1 \\ 5-3-z=-1 \\ z=3$ Thus, $x=5; y=-3, z=3$