## Precalculus (6th Edition) Blitzer

$\left( 1,-1,0 \right)$ satisfies the given system of equations.
Substitute $z=0$ in $\left( 12z+1,10z-1,z \right)$ Thus, $\left( 1,-1,0 \right)$ Now by substituting $x=1,y=-1,z=0$ in the left-hand side of the equation (1), we get, Thus, \begin{align} & 3x-4y+4z=3\times 1-4\times -1+4\times 0 \\ & =3+4 \\ & =7 \end{align} The value obtained is equal to the right-hand side value of the equation. Therefore, $\left( 1,-1,0 \right)$ satisfies equation (1). Now by substituting $x=1,y=-1,z=0$ in the left-hand side of the equation (2), we get, Thus, \begin{align} & x-y-2z=1-1\times -1-2\times 0 \\ & =1+1 \\ & =2 \end{align} The value obtained is equal to the right-hand side value of the equation. Therefore, $\left( 1,-1,0 \right)$ satisfies equation (2). Now, by substituting $x=1,y=-1,z=0$ in the left-hand side of the equation (3), we get, Thus, \begin{align} & 2x-3y+6z=5=2\times 1-3\times -1+6\times 0 \\ & =2+3 \\ & =5 \end{align} The value obtained is equal to the right hand side value of the equation. Therefore, $\left( 1,-1,0 \right)$ satisfies equation (3). Thus, $\left( 1,-1,0 \right)$ satisfies the given system of equations. Hence, $\left( 12z+1,10z-1,z \right)$ for $z=0$ satisfied the given system.