Chapter 1.4, Problem 38E

To determine

To solve: the given equation and check the solution.

Answer to Problem 38E

The solution is $\left(k=6\right)\text{ or }\left(k=-\frac{2}{5}\right)$ .

Explanation of Solution

Given:

  $\left|3k-2\right|=2\left|k+2\right|$

Calculation:

  $\begin{array}{l}\left|3k-2\right|=2\left|k+2\right|\text{ Given}\\ 3k-2=2\left(k+2\right)\text{ or }3k-2=-2\left(k+2\right)\text{ Write two linear equations}\end{array}$

Solve equation 1:

  $\begin{array}{l}3k-2=2\left(k+2\right)\\ 3k-2=2k+4\text{ Use distributive Property}\\ 3k-2k-2=2k-2k+4\text{ Subtract both sides by 2}k\\ k-2+2=4+2\text{ Add both sides by 2}\\ k=6\end{array}$

Solve equation 2:

  $\begin{array}{l}3k-2=-2\left(k+2\right)\\ 3k-2=-2k-4\text{ Use distributive Property}\\ 3k+2k-2=-2k+2k-4\text{ Add both sides by 2}k\\ 5k-2+2=-4+2\text{ Add both sides by 2}\\ \frac{5k}{5}=\frac{-2}{5}\text{ Divide both sides by 5}\\ k=\frac{-2}{5}\end{array}$

Check $\left(k=6\right)$ :

  $\begin{array}{l}\left|3k-2\right|=2\left|k+2\right|\\ \left|3\left(6\right)-2\right|=2\left|\left(6\right)+2\right|\text{ Substitute}\\ \left|18-2\right|=2\left|8\right|\text{ Simplify}\\ \left|16\right|=2\left|8\right|\text{ Simplify}\\ 16=2\left(8\right)\text{ Get absolute values}\\ 16=16\text{ Simplify}\end{array}$

Check $\left(k=-\frac{2}{5}\right)$ :

  $\begin{array}{l}\left|3k-2\right|=2\left|k+2\right|\\ \left|3\left(-\frac{2}{5}\right)-2\right|=2\left|\left(-\frac{2}{5}\right)+2\right|\text{ Substitute}\\ \left|-\frac{6}{5}-\frac{10}{5}\right|=2\left|\left(-\frac{2}{5}\right)+\frac{10}{5}\right|\text{ Simplify}\\ \left|-\frac{16}{5}\right|=2\left|\frac{8}{5}\right|\text{ Simplify}\\ \frac{16}{5}=2\left(\frac{8}{5}\right)\text{ Get absolute values}\\ \frac{16}{5}=\frac{16}{5}\text{ Get absolute values}\end{array}$

Conclusion:

Therefore, the solution is $\left(k=6\right)\text{ or }\left(k=-\frac{2}{5}\right)$ .