How do you solve the system of equations algebraically $ - x - 3y + z = 54 $ , $ 4x + 2y - 3z = - 32 $ , $ 2y + 8z = 78 $ ?

Hint: We have to solve the system of given three equations. Solving these equations means finding the values of the variables $ x $ , $ y $ and $ z $ which satisfies all the given three equations. We can solve this by substitution method where we convert one variable in terms of other variables using one equation and then use it in other equations to solve them.

Complete step by step solution:
We have been given a system of three equations in three variables,
$
   - x - 3y + z = 54\;\;\;\;\;\;...\left( 1 \right) \\
  4x + 2y - 3z = - 32\;\;...\left( 2 \right) \\
  2y + 8z = 78\;\;\;\;\;\;\;\;\;\;\;...\left( 3 \right) \;
 $
First we will try to eliminate one variable from two equations using the third equation.
From third equation we can write $ z $ in terms of $ y $ as follows,
$
  2y + 8z = 78 \\
   \Rightarrow 8z = 78 - 2y \\
   \Rightarrow z = \dfrac{{78 - 2y}}{8} = \dfrac{{39 - y}}{4}\;\;\;...\left( 4 \right) \;
 $
Now we substitute this value of $ z $ in the equations $ \left( 1 \right) $ and $ \left( 2 \right) $ so that we can eliminate the variable $ z $ from these equations.
From equation $ \left( 1 \right) $ we get,
$
   - x - 3y + z = 54 \\
   \Rightarrow - x - 3y + \left( {\dfrac{{39 - y}}{4}} \right) = 54 \\
   \Rightarrow - x - 3y + \dfrac{{39}}{4} - \dfrac{y}{4} = 54 \\
   \Rightarrow - x - \left( {3 + \dfrac{1}{4}} \right)y = 54 - \dfrac{{39}}{4} \\
   \Rightarrow - x - \left( {\dfrac{{12 + 1}}{4}} \right)y = \dfrac{{216 - 39}}{4} \\
   \Rightarrow - x - \dfrac{{13}}{4}y = \dfrac{{177}}{4}\;\;\;\;\;\;\;\;\;\;\;...\left( 5 \right) \;
 $
Similarly from equation $ \left( 2 \right) $ we get,
\[
  4x + 2y - 3z = - 32 \\
   \Rightarrow 4x + 2y - 3\left( {\dfrac{{39 - y}}{4}} \right) = - 32 \\
   \Rightarrow 4x + 2y - \dfrac{{117}}{4} + \dfrac{{3y}}{4} = - 32 \\
   \Rightarrow 4x + \left( {2 + \dfrac{3}{4}} \right)y = - 32 + \dfrac{{117}}{4} \\
   \Rightarrow 4x + \left( {\dfrac{{8 + 3}}{4}} \right)y = \left( {\dfrac{{ - 128 + 117}}{4}} \right) \\
   \Rightarrow 4x + \dfrac{{11}}{4}y = - \dfrac{{11}}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 6 \right) \;
 \]
Equations $ \left( 5 \right) $ and $ \left( 6 \right) $ now become two equations in two variables. To solve these we can again use the substitution method.
From equation $ \left( 5 \right) $ we get,
$
   - x - \dfrac{{13}}{4}y = \dfrac{{177}}{4} \\
   \Rightarrow x = - \dfrac{{13}}{4}y - \dfrac{{177}}{4}\;\;\;\;\;...\left( 7 \right) \;
 $
Substituting this in equation $ \left( 6 \right) $ we get,
\[
  4x + \dfrac{{11}}{4}y = - \dfrac{{11}}{4} \\
   \Rightarrow 4\left( { - \dfrac{{13}}{4}y - \dfrac{{177}}{4}} \right) + \dfrac{{11}}{4}y = - \dfrac{{11}}{4} \\
   \Rightarrow - 13y - 177 + \dfrac{{11}}{4}y = - \dfrac{{11}}{4} \\
   \Rightarrow \left( { - 13 + \dfrac{{11}}{4}} \right)y = - \dfrac{{11}}{4} + 177 \\
   \Rightarrow \left( {\dfrac{{ - 52 + 11}}{4}} \right)y = \left( {\dfrac{{ - 11 + 708}}{4}} \right) \\
   \Rightarrow \left( {\dfrac{{ - 41}}{4}} \right)y = \left( {\dfrac{{697}}{4}} \right) \\
   \Rightarrow y = \dfrac{{697}}{{ - 41}} = - 17 \;
 \]
Thus, we get the value \[y = - 17\].
Now we use this value to find the value of $ x $ and $ z $ .
From equation $ \left( 7 \right) $ we get,
$
  x = - \dfrac{{13}}{4}y - \dfrac{{177}}{4} \\
   \Rightarrow x = \left( { - \dfrac{{13}}{4} \times - 17} \right) - \dfrac{{177}}{4} \\
   \Rightarrow x = \dfrac{{221}}{4} - \dfrac{{177}}{4} = \dfrac{{44}}{4} = 11 \;
  $
And from equation $ \left( 4 \right) $ we get,
 $ z = \dfrac{{39 - y}}{4} = \dfrac{{39 - \left( { - 17} \right)}}{4} = \dfrac{{39 + 17}}{4} = \dfrac{{56}}{4} = 14 $
Hence, we get the values $ x = 11 $ , $ y = - 17 $ and $ z = 14 $ as the solution of the given system of equations.
So, the correct answer is “ $ x = 11 $ , $ y = - 17 $ and $ z = 14 $ ”.

Note: When we are solving a system of equations, we have to find the values of the variables which satisfies all the given equations. We have used substitute and eliminate methods to find the values of the variables one-by-one. We can also check our solution by putting the values in the given equations.
For $ - x - 3y + z = 54 $ ,
$ LHS = - x - 3y + z = - \left( {11} \right) - \left( {3 \times - 17} \right) + 14 = - 11 + 51 + 14 = 54 = RHS $
For $ 4x + 2y - 3z = - 32 $ ,
$ LHS = 4x + 2y - 3z = \left( {4 \times 11} \right) + \left( {2 \times - 17} \right) - \left( {3 \times 14} \right) = 44 - 34 - 42 = - 32 = RHS $
For $ 2y + 8z = 78 $ ,
$ LHS = 2y + 8z = \left( {2 \times - 17} \right) + \left( {8 \times 14} \right) = - 34 + 112 = 78 = RHS $
Thus, our solution satisfies all the given three equations.