## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 1,2,3 \right)$.
Let us consider the given equation: $3x+2y-3z=-2$ …… (I) $2x-5y+2z=-2$ …… (II) $4x-3y+4z=10$ …… (III) Calculate $2\times$ equation (II) minus equation (III): \begin{align} & 4x-10y+4z=-4 \\ & 4x-3y+4z=10 \\ & -\text{ }+\text{ }-\text{ }- \\ & \overline{\,\,\,\,\,\,\,\,\,\,-7y=-14\,\,\,\,\,\,\,} \\ \end{align} Further simplify, $y=2$ Apply $2\times$ equation (I) and add $3\times$ equation (II): \begin{align} & 6x+4y-6z=-4 \\ & 6x-15y+6z=-6 \\ & \overline{\,\,\,\,\,\,\,\,\,12x-11y=-10\,\,\,} \\ \end{align} $12x-11y=-10$ …… (IV) Putting the value of y in equation (IV): \begin{align} & 12x-11(2)=-10 \\ & 12x=-10+22 \\ & 12x=12 \\ & x=1 \end{align} Now, putting the value $x=1,y=2$ in equation (I): \begin{align} & 3\left( 1 \right)+2\left( 2 \right)-3z=-2 \\ & 3+4-3z=-2 \\ & -3z=-9 \\ & z=3 \end{align} Thus, the order triple $\left( 1,2,3 \right)$ satisfies the three equations.