Chapter 12, Problem 13GR

Solution

To Solve: The System of equations

The values are $x=7,y=5,z=0$

Given:

  $\left\{\begin{array}{rr}\hfill -2x+3y+z=& \hfill 1\\ \hfill x-3z=& \hfill 7\\ \hfill -y+z=& \hfill -5\end{array}\right.$

Explanation:

Given that, $\left\{\begin{array}{rr}\hfill -2x+3y+z=& \hfill 1\\ \hfill x-3z=& \hfill 7\\ \hfill -y+z=& \hfill -5\end{array}\right.$

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

  $\begin{array}{l}z=1+2x-3y\hfill \\ x-3z=7\hfill \\ -y+z=-5\hfill \end{array}$

Replace all occurrences of $z$ in $x-3z=7$ with $1+2x-3y$ :

  $\begin{array}{c}x-3\left(1+2x-3y\right)=7\\ z=1+2x-3y\\ -y+z=-5\end{array}$

Simplify the left side:

  $\begin{array}{l}-5x-3+9y=7\hfill \\ z=1+2x-3y\hfill \\ -y+z=-5\hfill \end{array}$

Replace all occurrences of $z$ in $-y+z=-5$ with $1+2x-3y$ :

  $\begin{array}{c}-y+1+2x-3y=-5\\ -5x-3+9y=7\\ z=1+2x-3y\end{array}$

Simplify the left side:

  $\begin{array}{c}-4y+1+2x=-5\\ -5x-3+9y=7\\ z=1+2x-3y\end{array}$

Solve for $y$ in $-4y+1+2x=-5$ :

  $\begin{array}{c}-4y=-6-2x\\ y=\frac{-6-2x}{-4}\\ y=\frac{3}{2}+\frac{x}{2}\end{array}$

Thus, the equations are:

  $\begin{array}{c}y=\frac{3}{2}+\frac{x}{2}\\ -5x-3+9y=7\\ z=1+2x-3y\end{array}$

Replace all occurrences of $y$ in $-5x-3+9y=7$ with $\frac{3}{2}+\frac{x}{2}$ :

  $\begin{array}{c}-5x-3+9\left(\frac{3}{2}+\frac{x}{2}\right)=7\\ y=\frac{3}{2}+\frac{x}{2}\\ z=1+2x-3y\end{array}$

Simplify the left side:

  $\begin{array}{c}-\frac{x-21}{2}=7\\ y=\frac{3}{2}+\frac{x}{2}\\ z=1+2x-3y\end{array}$

Replace all occurrences of $y$ in $z=1+2x-3y$ with $\frac{3}{2}+\frac{x}{2}$ :

  $\begin{array}{c}z=1+2x-3\left(\frac{3}{2}+\frac{x}{2}\right)\\ -\frac{x-21}{2}=7\\ y=\frac{3}{2}+\frac{x}{2}\end{array}$

Simplify the right side.

  $\begin{array}{c}z=\frac{x-7}{2}\\ -\frac{x-21}{2}=7\\ y=\frac{3}{2}+\frac{x}{2}\end{array}$

Solve for $x$ in $-\frac{x-21}{2}=7$ : Multiply both sides of the equation by $-2$ :

  $\begin{array}{c}-2\left(-\frac{x-21}{2}\right)=-2\cdot 7\\ x-21=-14\\ x=\frac{-14}{21}\\ x=7\end{array}$

Thus, the equation as are:

  $\begin{array}{c}x=7\\ z=\frac{x-7}{2}\\ y=\frac{3}{2}+\frac{x}{2}\end{array}$

Replace all occurrences of $x$ in $z=\frac{x-7}{2}$ with $7$ :

  $\begin{array}{c}z=\frac{\left(7\right)-7}{2}\\ x=7\\ y=\frac{3}{2}+\frac{x}{2}\end{array}$

Simplify the right side:

  $\begin{array}{c}z=0\\ x=7\\ y=\frac{3}{2}+\frac{x}{2}\end{array}$

Replace all occurrences of $x$ in $y=\frac{3}{2}+\frac{x}{2}$ with $7$ :

  $\begin{array}{c}y=\frac{3}{2}+\frac{7}{2}\\ z=0\\ x=7\end{array}$

Simplify the right side.

  $\begin{array}{c}y=\frac{10}{2}\\ z=0\\ x=7\end{array}$

Simplify $y$ term again:

  $\begin{array}{c}y=5\\ z=0\\ x=7\end{array}$

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

  $\left(7,5,0\right)$