Chapter 1.4, Problem 22E

To determine

To solve the equation:

$-3\left|1-\frac{2}{3}v\right|=-9$

And also graph the solution.

Answer to Problem 22E

  $v=-6\text{ and }v=6$

Explanation of Solution

Given:

$-3\left|1-\frac{2}{3}v\right|=-9$

Concept Used:

To solve $\left|ax+b\right|=c,c\ge 0$ , solve the related equations:

  $ax+b=c,\text{ or }ax+b=-c$

When $c<0$ the absolute value equation $\left|ax+b\right|=c$ has no solution, since absolute value always indicates a value that is not negative.

Calculation:

In order to solve the equation:

$-3\left|1-\frac{2}{3}v\right|=-9$

First, simplify the equation so that the only absolute value term is on one side and rest is on other side of the equation, as

$\begin{array}{c}-3\left|1-\frac{2}{3}v\right|=-9\\ \Rightarrow \frac{-3\left|1-\frac{2}{3}v\right|}{-3}=\frac{-9}{-3}\\ \Rightarrow \left|1-\frac{2}{3}v\right|=3\end{array}$

Now, form the related equations of above:

  $1-\frac{2}{3}v=3,\text{ and }1-\frac{2}{3}v=-3$

Now, solve above equations first subtract $-1$ from each sides, as

  $\begin{array}{c}1-\frac{2}{3}v-1=3-1,\text{ and }1-\frac{2}{3}v-1=-3-1\\ \Rightarrow -\frac{2}{3}v=2,\text{ and }-\frac{2}{3}v=-4\end{array}$

Now, isolate v as

  $\begin{array}{c}-\frac{2}{3}v=2,\text{ and }-\frac{2}{3}v=-4\\ \Rightarrow \left(-\frac{3}{2}\right)\times -\frac{2}{3}v=\left(-\frac{3}{2}\right)\times 2,\text{ and }\left(-\frac{3}{2}\right)\times -\frac{2}{3}v=\left(-\frac{3}{2}\right)\times -4\\ \Rightarrow \left(-\frac{\cancel{3}}{\cancel{2}}\right)\times -\frac{\cancel{2}}{\cancel{3}}v=-6,\text{ and }\left(-\frac{\cancel{3}}{\cancel{2}}\right)\times -\frac{\cancel{2}}{\cancel{3}}v=\left(-3\right)\times -2\\ \Rightarrow v=-6,\text{ and }v=6\end{array}$

Thus, solutions for given equation are $v=-6\text{ and }v=6$ .

Now, represent the solution of the given equation by representing points $-6$ , and 6 on the number line using a graphing tool.