## Precalculus (6th Edition) Blitzer

$\left( 13,9,1 \right)$ satisfies the given system.
Substitute $z=1$ in $\left( 12z+1,10z-1,z \right)$ Thus, $\left( 13,9,1 \right)$ Now by substituting $x=13,y=9,z=1$ in the left-hand side of the equation (I), we get, Thus, \begin{align} & 3x-4y+4z=3\times 13-4\times 9+4\times 1 \\ & =39-36+4 \\ & =7 \end{align} The value obtained is equal to the right-hand side value of the equation. Hence, $\left( 13,9,1 \right)$ satisfies equation (I). Now, by substituting $x=13,y=9,z=1$ in the left-hand side of the equation (II), we get, Thus, \begin{align} & x-y-2z=13-9-2\times 1 \\ & =13-9-2 \\ & =2 \end{align} The value obtained is equal to the right hand side value of the equation. Hence, $\left( 13,9,1 \right)$ satisfies equation (II). Now by substituting $x=13,y=9,z=1$ in the left-hand side of the equation (III), we get, Thus, \begin{align} & 2x-3y+6z=2\times 13-3\times 9+6\times 1 \\ & =26-27+6 \\ & =5 \end{align} The value obtained is equal to the right hand side value of the equation. Therefore, $\left( 13,9,1 \right)$ satisfies equation (III). Thus, $\left( 13,9,1 \right)$ satisfies the given system. Hence, $\left( 12z+1,10z-1,z \right)$ for $z=1$ satisfied the given system.