Chapter 11, Problem 4CT

To determine

To find: The solution of equations $\left\{\begin{array}{l}3x+2y-8z=-3\\ -x-\frac{2}{3}y+z=1\\ 6x-3y+15z=8\end{array}\right\}$ .

Answer to Problem 4CT

The solution of the equation is $\left(\frac{1}{3},-2,0\right)$ .

Explanation of Solution

Given information:

The given equation is $\left\{\begin{array}{l}3x+2y-8z=-3\\ -x-\frac{2}{3}y+z=1\\ 6x-3y+15z=8\end{array}\right\}$

Calculation:

Calculate the expression,

  $\left\{\begin{array}{l}3x+2y-8z=-3\\ -x-\frac{2}{3}y+z=1\\ 6x-3y+15z=8\end{array}\right\}$

Multiply in first equation by 2 and the second equation by -6. The equation obtained are given below

  $\left\{\begin{array}{c}6x+4y-16z=-6\\ 6x+4y-6z=-6\\ 6x-3y+15z=8\end{array}\right\}$

Subtract first equation from the second equation and by subtracting first equation from the third equation.

  $\begin{array}{l}\left\{\begin{array}{l}6x-6x+4y-4y-6z-\left(16z\right)=-6-\left(-6\right)\\ 6x-6x-3y-4y+15z-\left(-16z\right)=8-\left(-6\right)\end{array}\right\}\\ \left\{\begin{array}{c}10z=0\\ -7y+31z=14\end{array}\right\}\end{array}$

The first equation can be multiply by $\frac{31}{10}$

  $\begin{array}{l}\left\{\begin{array}{l}\left(10z=0\right)\times \frac{31}{10}\\ \left(-7+31z=14\right)\times 1\end{array}\right\}\\ \left\{\begin{array}{c}314z=0\\ -7y+31z=14\end{array}\right\}\end{array}$

The value can now be solved,

  $\begin{array}{c}-7y+31z-31z=14-0\\ y=-2\end{array}$

The value of $z$ can b e determine by substituting the obtained value of $y$

  $\begin{array}{c}-7\left(-20\right)+31z=14-0\\ -7y=14\\ y=-2\end{array}$

The value of $z$ can be determined by substituting the obtained bvalue of $y$

  $\begin{array}{c}-7\left(-2\right)+31z=14\\ z=0\end{array}$

The value of $x$ can be obtained,

  $\begin{array}{c}3x+2y-8z=-3\\ 3x+2\left(-2\right)-8\left(0\right)=-3\\ x=\frac{1}{3}\end{array}$

Therefore, the solution of the equation is $\left(\frac{1}{3},-2,0\right)$ .