How do you solve the system of equations algebraically $x-3z=7,2x+y-2z=11,-x-2y+9z=13$?

Hint: To solve the given system of equations algebraically, we first need to reduce them from three variable systems to two variable systems. For that we have to eliminate the variable $x$ by adding first and the second equation. The second eliminated equation will be obtained by multiplying the first equation by $2$ and subtracting the obtained equation from the second equation. On solving the two equations obtained, we will get the values of $y$ and $z$. On substituting these in any one of the three given equations, we will get the value of $x$.

Complete step by step solution:
The equations given in the above question are
$\begin{align}
  & \Rightarrow x-3z=7........\left( i \right) \\
 & \Rightarrow 2x+y-2z=11........\left( ii \right) \\
 & \Rightarrow -x-2y+9z=13........\left( iii \right) \\
\end{align}$
Adding the equations (i) and (iii) we get
$\begin{align}
  & \Rightarrow x-3z-x-2y+9z=7+13 \\
 & \Rightarrow -2y+6z=20........\left( iv \right) \\
\end{align}$
Multiplying the equation (i) by $2$ we get
$\Rightarrow 2x-6z=14.......\left( v \right)$
Subtracting the above equation from the equation (ii) we get
\[\begin{align}
  & \Rightarrow 2x+y-2z-\left( 2x-6z \right)=11-14 \\
 & \Rightarrow 2x+y-2z-2x+6z=-3 \\
 & \Rightarrow y+4z=-3 \\
\end{align}\]
Subtracting \[4z\] from both the sides we get
$\begin{align}
  & \Rightarrow y+4z-4z=-3-4z \\
 & \Rightarrow y=-3-4z........\left( vi \right) \\
\end{align}$
Putting this in the equation (iv) we get
$\begin{align}
  & \Rightarrow -2\left( -3-4z \right)+6z=20 \\
 & \Rightarrow 6+8z+6z=20 \\
 & \Rightarrow 6+14z=20 \\
\end{align}$
Subtracting $6$ from both sides
\[\begin{align}
  & \Rightarrow 6+14z-6=20-6 \\
 & \Rightarrow 14z=14 \\
\end{align}\]
Dividing both sides by \[14\] we get
$\begin{align}
  & \Rightarrow \dfrac{14z}{14}=\dfrac{14}{14} \\
 & \Rightarrow z=1 \\
\end{align}$
Substituting this value in the equation (vi) we get
$\begin{align}
  & \Rightarrow y=-3-4\left( 1 \right) \\
 & \Rightarrow y=-3-4 \\
 & \Rightarrow y=-7 \\
\end{align}$
Putting the value of $z$ in the equation (v) we get
$\begin{align}
  & \Rightarrow 2x-6\left( 1 \right)=14 \\
 & \Rightarrow 2x-6=14 \\
\end{align}$
Adding $6$ both sides, we get
$\begin{align}
  & \Rightarrow 2x-6+6=14+6 \\
 & \Rightarrow 2x=20 \\
\end{align}$
Finally, dividing both sides by $2$ we get
$\begin{align}
  & \Rightarrow \dfrac{2x}{2}=\dfrac{20}{2} \\
 & \Rightarrow x=10 \\
\end{align}$

Hence, the solution of the given system of equations is $x=10,y=-7,z=1$.

Note: The algebraic solution of the system of linear solution in three variables involves many calculations, as can be seen in the above solution. So there are huge chances of committing calculation mistakes. Therefore, after obtaining the solution, we must substitute it back in the equations and confirm whether the LHS is coming equal to the RHS.