## Precalculus (6th Edition) Blitzer

The solution of the equation is $\left\{ \left( 8,-2,-2 \right) \right\}$.
The system of equations is given below: \left\{ \begin{align} & x+2y+3z=-2 \\ & 3x+3y+10z=-2 \\ & 2y-5z=6 \end{align} \right. And the equation $2y-5z=6$ has only two variables x and y: Now, consider the equations $x+2y+3z=-2$ and $3x+3y+10z=-2$; eliminate $z$ and obtain the equation containing the y and z variables in equation $2y-5z=6$. Multiply the equation $x+2y+3z=-2$ by 3 and equation $3x+3y+10z=-2$ by 1. Therefore, \begin{align} & \left( x+2y+3z \right)\times 3=-2\times 3 \\ & 3x+6y+9z=-6 \end{align} And $3x+3y+10z=-2$ And subtract the equation $3x+6y+9z=-6$ from $3x+3y+10z=-2$ to obtain: \begin{align} & \left( 3x+6y+9z \right)-\left( 3x+3y+10z \right)=\left( -6 \right)-\left( -2 \right) \\ & 3y-z=-4 \end{align} Now, solve the equations $2y-5z=6$ and $3y-z=-4$ ; the values of $z$ and $y$ are \begin{align} & z=-2 \\ & y=-2 \end{align} Put values of z and y in equation $x+2y+3z=-2$, to obtain the value of $x$ as given below: \begin{align} & x+2y+3z=-2 \\ & x+2\left( -2 \right)+3\left( -2 \right)=-2 \\ & x-4-6=-2 \\ & x=8 \end{align} Thus, the values of x, y, z are $\left\{ \left( 8,-2,-2 \right) \right\}$.