#### Answer

A solution of a system of linear equations in three variables is an ordered triple of real numbers that satisfies all equations of the system.

#### Work Step by Step

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.