Chapter 11, Problem 10CLR

To determine

To calculate:

The system of the following equations

  $\begin{array}{l}5x-3y+z=-6\\ x+3y-z=0\\ 4x+y+z=7\end{array}$

Answer to Problem 10CLR

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

Explanation of Solution

Given information:

  $\begin{array}{l}5x-3y+z=-6\\ x+3y-z=0\\ 4x+y+z=7\end{array}$

Calculation:

Consider the equation:

  $\begin{array}{l}5x-3y+z=-6\mathrm{...}\left(1\right)\\ x+3y-z=0\mathrm{...}\left(2\right)\\ 4x+y+z=7\mathrm{...}\left(3\right)\end{array}$

Equation (1),

  $5x-3y+z=-6$

Subtract $-3y+z$ from both the sides:

  $5x-3y+z-\left(-3y+z\right)=-6-\left(-3y+z\right)$

Simplify the equation:

  $5x=-6+3y-z$

Then divide both the sides by $5$ :

  $\frac{5x}{5}=-\frac{6}{5}+\frac{3y}{5}-\frac{z}{5}$

Simplify:

  $x=\frac{-6+3y-z}{5}\mathrm{...}(4)$

Substitute the value in equation (1):

  $\left[\begin{array}{l}\left(\frac{-6+3y-z}{5}\right)+3y-z=0\left(5\right)\\ 4\left(\frac{-6+3y-z}{5}\right)+y+z=7\left(6\right)\end{array}\right]$

Simplify equation (5) and (6):

  $\begin{array}{l}\left(\frac{-6+3y-z}{5}\right)+3y-z=0:\frac{18-6z-6}{5}=0\\ 4\left(\frac{-6+3y-z}{5}\right)+y+z=7:\frac{17y+z-24}{5}=7\\ \left[\begin{array}{l}\frac{18-6z-6}{5}=0\\ \frac{17y+z-24}{5}=7\end{array}\right]\end{array}$

Isolate $y$ for $\frac{18-6z-6}{5}=0$

Multiply on both the sides by $5$ :

  $\frac{5\left(18-6z-6\right)}{5}=0\mathrm{...}\left(7\right)$

Simplify the equation (7):

  $18y-6z-6=0$

Add $6z$ to both the sides:

  $\begin{array}{l}18y-6z-6+6z=0+6z\\ 18y-6=6z\end{array}$

Add $6$ to both the sides:

  $\begin{array}{l}18y-6+6=6z+6\\ 18y=6z+6\end{array}$

Divide both the sides by $18$ :

  $\begin{array}{l}\frac{18y}{18}=\frac{6z}{18}+\frac{6}{18}\\ y=\frac{z+1}{3}\mathrm{...}\left(8\right)\end{array}$

Substitute $y=\frac{z+1}{3}$

  $\frac{17\left(\frac{z+1}{3}\right)+z+-24}{5}=7$

Simplify the equation:

  $\frac{4z-11}{3}=7$

Isolate $z$ for $\frac{4z-11}{3}=7$

  $\frac{4z-11}{3}=7$

Multiply both the sides by three:

  $\begin{array}{l}\frac{3\left(4z-11\right)}{3}=7\times 3\\ 4z-11=21\end{array}$

Add $11$ to both the sides:

  $\begin{array}{l}4z-11+11=21+11\\ 4z=32\end{array}$

Divide both the sides by $4$

  $\begin{array}{l}\frac{4z}{4}=\frac{32}{4}\\ z=8\end{array}$

Substitute the value of $z=8$ in equation (8)

  $\begin{array}{l}y=\frac{z+1}{3}\\ y=\frac{8+1}{3}\\ y=3\end{array}$

Substitute the value of $z=8$ and $y=3$ in equation (4):

  $\begin{array}{l}x=\frac{-6+3y-z}{5}\\ x=\frac{-6+3\left(3\right)-8}{5}\\ x=-1\end{array}$

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