## Precalculus (6th Edition) Blitzer

To satisfy the system of linear equations, an ordered triple must satisfy all equations of the system. Example: Let us take an example of a system of equations as, \left\{ \begin{align} & x+2y+3z=-2 \\ & 3x+3y+10z=-2 \\ & 2y-5z=6 \end{align} \right. Equation $2y-5z=6$ has only two variables. Now, consider equation $x+2y+3z=-2$ and $3x+3y+10z=-2$, and eliminate z, to get $2y-5z=6$ and then eliminate another variable. Then, multiply equation $x+2y+3z=-2$ by $3$ and equation $3x+3y+10z=-2$ by $1$ and get; \begin{align} & \left( x+2y+3z \right)\times 3=-2\times 3 \\ & 3x+6y+9z=-6 \end{align} And, $3x+3y+10z=-2$. Now, subtract equation $3x+6y+9z=-6$ and $3x+3y+10z=-2$, to get; \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 equations $2y-5z=6$ and $3y-z=-4$ and obtain the values of z and y. The values so obtained: \begin{align} & z=-2 \\ & y=-2 \end{align} Put the values of z and y in equation $x+2y+3z=-2$ and get the value of $x$ as, \begin{align} & x+2y+3z=-2 \\ & x+2\left( -2 \right)+3\left( -2 \right)=-2 \\ & x-4-6=-2 \\ & x=8 \end{align} Values of x, y, and z are $\left( 8,-2,-2 \right)$. Get the ordered triple solution set $\left( x,y,z \right)$ is $\left( 8,-2,-2 \right)$. Since, the solution set $\left( 8,-2,-2 \right)$ is calculated by the use of the given system of equation, so it must satisfy the system of equations.