Chapter 1.4, Problem 34PPS

To determine

To solve: the given equation for $z$ . Check the result.

Answer to Problem 34PPS

The value of $z$ in the given equation is $y=-\frac{1}{4}\text{or}y=-\frac{9}{28}$ .

Explanation of Solution

Given information:

Given equation

  $-5\left|3z+8\right|-5=-20$

Concept used:

Absolute value of a function a is given by

  $\left|a\right|=\{\begin{array}{l}aa>0\\ -aa<0\\ 0a=0\end{array}$ \

And

  $\begin{array}{l}\left|-a\right|=a\\ \left|a\right|=a\end{array}$

More over

  $\begin{array}{l}\left|x+a\right|=b\\ \Rightarrow x+a=b\text{or}x+a=-b\end{array}$

Calculation:

Consider the given equation,

  $-5\left|3z+8\right|-5=-20$

Now, solve the equation as shown,

  $\begin{array}{c}-5\left|3z+8\right|-5=-20\\ -5\left|3z+8\right|-5+5=-20+5\\ -5\left|3z+8\right|=-15\\ \frac{-5\left|3z+8\right|}{-5}=\frac{-15}{-5}\\ \left|3z+8\right|=3\\ 3z+8=3\text{or}\text{}3z+8=-3\\ 3z+8-8=3-8\text{or}\text{}3z+8-8=-3-8\\ 3z=-5\text{or}\text{}3z=-11\\ z=-\frac{5}{3}\text{or}z=-\frac{11}{3}\end{array}$

Now, check the result by back substitution.

  $\begin{array}{c}-5\left|3z+8\right|-5=-20\\ -5\left|3\left(-\frac{5}{3}\right)+8\right|-5=-20\\ -5\left|-5+8\right|-5=-20\\ -5\left|3\right|-5=-20\\ -15-5=-20\\ -20=-20\end{array}$

And

  $\begin{array}{c}-5\left|3z+8\right|-5=-20\\ -5\left|3\left(-\frac{11}{3}\right)+8\right|-5=-20\\ -5\left|-11+8\right|-5=-20\\ -5\left|-3\right|-5=-20\\ -15-5=-20\\ -20=-20\end{array}$

Which is true.

Hence, the value of $z$ in the given equation is $z=-\frac{5}{3}\text{or}z=-\frac{11}{3}$ .