Chapter 9, Problem 19CR

To determine

To solve the system:

  $x+y+z=2$

  $x-2y-z=2$

  $3x+2y+z=2$

Answer to Problem 19CR

The answer is $c.$

  $\left\{\left(1,-2,3\right)\right\}$ .

Explanation of Solution

Given:

  $x+y+z=2$

  $x-2y-z=2$

  $3x+2y+z=2$

Calculation:

Given:

  $x+y+z=2$

  $x-2y-z=2$

  $3x+2y+z=2$

Solving the equation $x+y+z=2$ for value of $y$ , we have:

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

Solving the equation $x-2y-z=2$ for value of $y$ , we have:

  $\begin{array}{l}x-2y-z=2\\ y=\frac{x-z-2}{2}\end{array}$

Solving the equation $3x+2y+z=2$ for value of $y$ , we have:

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

Comparing first two equations, we have:

  $\begin{array}{l}-x-z+2=\frac{x-z-2}{2}\\ -2x-2z+4-x+z+2=0\\ -3x-z+6=0\\ 3x+z=6\end{array}$

Comparing last two equations, we have:

  $\begin{array}{l}\frac{-3x-z+2}{2}=\frac{x-z-2}{2}\\ -3x-z+2-x+z+2=0\\ -4x+4=0\\ x=1\end{array}$

Putting the value of $x=1$ in equation $3x+z=6$ , we have:

  $\begin{array}{l}3\left(1\right)+z=6\\ z=3\end{array}$

Putting the value of $x=1$ and $z=3$ in equation $y=-x-z+2$ , we have:

  $\begin{array}{l}y=-1-3+2\\ y=-2\end{array}$

Thus, we have $x=1$ , $y=-2$ and $z=3$ .

Conclusion:

The answer is $c.$

  $\left\{\left(1,-2,3\right)\right\}$