## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 2,-1,1 \right)$.
Let us consider the given equation: $4x-y+2z=11$ …… (I) $x+2y-z=-1$ …… (II) $2x+2y-3z=-1$ …… (III) $2\times$ equation (II), add equation (1) \begin{align} & 4x-y+2z=11 \\ & 2x+4y-2z=-2 \\ & \overline{6x+3y=9\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \end{align} $2x+y=3$ …… (IV) Equation (III) subtract $3\times$ Equation (II): \begin{align} & 3x+6y-3z=-3 \\ & 2x+2y-3z=-1 \\ & -\text{ }-\text{ }+\text{ }+ \\ & \overline{\,\,\,\,\,\,\,x+4y=-2\,\,\,\,\,\,} \\ \end{align} $x+4y=-2$ …… (V) Now equation (IV) subtract $2\times$ equation (V): \begin{align} & 2x+y=3 \\ & 2x+8y=-4 \\ & -\text{ }-\text{ }+ \\ & \overline{\begin{align} & \,\,\,\,\,\,\,-7y=7\,\,\,\,\,\, \\ & y=-1 \\ \end{align}} \\ \end{align} Now, putting the value of $y=-1$ in equation (IV): \begin{align} & 2x-1=3 \\ & 2x=4 \\ & x=2 \end{align} And, putting the value of x and y in equation (I): \begin{align} & 4\left( 2 \right)-\left( -1 \right)+2z=11 \\ & 8+1+2z=11 \\ & 2z=2 \\ & z=1 \end{align} Thus, the order triple $\left( 2,-1,1 \right)$ satisfies the three equations.