Chapter 11, Problem 9CLR

To determine

To calculate:

The system of the following equations

  $\begin{array}{l}x-y-z=5\\ 2x-y+z=3\\ x+2y-2z=4\end{array}$

Answer to Problem 9CLR

The solutions to the system of the equations are $x=-1,z=8,y=3$ .

Explanation of Solution

Given information:

  $\begin{array}{l}x-y-z=5\\ 2x-y+z=3\\ x+2y-2z=4\end{array}$

Calculation:

Consider the equation:

  $\begin{array}{l}x-y-z=5\mathrm{...}\left(1\right)\\ 2x-y+z=3\mathrm{...}\left(2\right)\\ x+2y-2z=4\mathrm{...}\left(3\right)\end{array}$

Equation (1),

  $x-y-z=5$

Subtract $-y-z$ from both the sides:

  $x-y-z-\left(-y-z\right)=5-\left(-y-z\right)$

Simplify the equation:

  $x=5+y+z\mathrm{...}\left(4\right)$

Substitute the value in equation (1):

  $\left[\begin{array}{l}2\left(5+y+z\right)-y+z=3\mathrm{...}\left(5\right)\\ 5+y+z+2y-2z=4\mathrm{...}\left(6\right)\end{array}\right]$

Simplify equation (5) and (6):

  $\begin{array}{l}2\left(5+y+z\right)-y+z=3\\ y+3z+10=3\mathrm{...}\left(7\right)\\ 5+y+z+2y-2z=4\\ 3y-z+5=4\mathrm{...}\left(8\right)\end{array}$

Isolate $y$ for $y+3z+10=3$

Subtract from both the sides:

  $\begin{array}{l}y+3z+10-3z=3-3z\\ y+10=3-3z\end{array}$

Subtract ten from both the sides:

  $\begin{array}{l}y+10-10=3-3z-10\\ y=-3z-7\left(9\right)\end{array}$

Substitute the value of $y=-3z-7$ in (8)

  $\begin{array}{l}3\left(-3z-7\right)-z+5=4\\ -9z-21-z+5=4\\ -10z-16=4\mathrm{...}\left(10\right)\end{array}$

Solve the equation:

  $-10z-16=4$

Add $16$ to both the sides:

  $\begin{array}{l}-10z-16+16=4+16\\ -10z=20\end{array}$

Divide both the sides by $-10$

  $\begin{array}{l}\frac{-10z}{10}=\frac{20}{-10}\\ z=-2\end{array}$

For $y=-3z-7$ in equation (9)

  $\begin{array}{l}y=-3z-7\\ y=-3\left(-2\right)-7\\ y=-1\end{array}$

For $z=-2,y=-1$ in equation of (4):

  $\begin{array}{l}x=5-1-2\\ x=2\end{array}$

The solutions to the system of the equations are $x=2,z=-2,y=-1$ .