Chapter 12, Problem 12GR

Solution

To Solve: The System of equations

The values are $x=0,y=0,z=8$

Given:

  $\left\{\begin{array}{rr}\hfill -x+y+2z=& \hfill 16\\ \hfill 2x-2y-2z=& \hfill -16\\ \hfill x+y=& \hfill 0\end{array}\right.$

Explanation:

Given that, $\left\{\begin{array}{rr}\hfill -x+y+2z=& \hfill 16\\ \hfill 2x-2y-2z=& \hfill -16\\ \hfill x+y=& \hfill 0\end{array}\right.$

Move all terms not containing $y$ to the right side of the equation:

  $\begin{array}{l}y=16+x-2z\hfill \\ 2x-2y-2z=-16\hfill \\ x+y=0\hfill \end{array}$

Replace all occurrences of $y$ in $2x-2y-2z=-16$ with $16+x-2z$ :

  $\begin{array}{c}2x-2\left(16+x-2z\right)-2z=-16\\ y=16+x-2z\\ x+y=0\end{array}$

Simplify the left side:

  $\begin{array}{l}-32+2z=-16\hfill \\ y=16+x-2z\hfill \\ x+y=0\hfill \end{array}$

Replace all occurrences of $y$ in $x+y=0$ with $16+x-2z$ :

  $\begin{array}{c}x+16+x-2z=0\\ -32+2z=-16\\ y=16+x-2z\end{array}$

Simplify the left side:

  $\begin{array}{l}2x+16-2z=0\hfill \\ -32+2z=-16\hfill \\ y=16+x-2z\hfill \end{array}$

Solve for $z$ in $-32+2z=-16$ :

  $\begin{array}{l}2x+16-2z=0\hfill \\ z=8\hfill \\ y=16+x-2z\hfill \end{array}$

Replace all occurrences of $z$ with $8$ in each equation:

  $\begin{array}{l}y=x\hfill \\ 2x=0\hfill \\ z=8\hfill \end{array}$

Divide each term in $2x=0$ by $2$ and simplify:

  $\begin{array}{l}x=0\hfill \\ y=x\hfill \\ z=8\hfill \end{array}$

Replace all occurrences of $x$ with $0$ in each equation:

  $\begin{array}{l}y=0\hfill \\ x=0\hfill \\ z=8\hfill \end{array}$

The solution to the system is the complete set of ordered pairs that are valid solutions.

  $\left(0,0,8\right)$