Chapter 1.4, Problem 42E

To determine

To solve: the given equation and check the solution.

Answer to Problem 42E

The solution is $h=\frac{1}{4}$ .

Explanation of Solution

Given:

  $\left|3h+1\right|=7h$

Calculation:

  $\begin{array}{l}\left|3h+1\right|\text{=7}h\text{ Given}\\ 3h+1=7h\text{ or }3h+1=-7h\text{ Write two linear equations}\end{array}$

Solve equation 1:

  $\begin{array}{l}3h+1=7h\\ 3h-3h+1=7h-3h\text{ Subtract both sides by 3}h\\ \frac{1}{4}=\frac{4h}{4}\text{ Divide both sides by 4}\\ \frac{1}{4}=h\end{array}$

Solve equation 2:

  $\begin{array}{l}3h+1=-7h\\ 3h-3h+1=-7h-3h\text{ Subtract both sides by 3}h\\ \frac{1}{-10}=\frac{-10h}{-10}\text{ Divide both sides by -10}\\ \frac{1}{-10}=h\end{array}$

Check $\left(h=\frac{1}{4}\right)$ :

  $\begin{array}{l}\left|3h+1\right|\text{=7}h\\ \left|3\left(\frac{1}{4}\right)+1\right|\text{=7}\left(\frac{1}{4}\right)\text{ Substitute}\\ \left|\frac{3}{4}+\frac{4}{4}\right|=\frac{7}{4}\text{ Simplify}\\ \frac{3}{4}+\frac{4}{4}=\frac{7}{4}\text{ Get absolute values}\\ \frac{7}{4}=\frac{7}{4}\text{ Simplify}\end{array}$

Check $\left(h=-\frac{1}{10}\right)$ :

  $\begin{array}{l}\left|3h+1\right|\text{=7}h\\ \left|3\left(-\frac{1}{10}\right)+1\right|\text{=7}\left(-\frac{1}{10}\right)\text{ Substitute}\\ \left|-\frac{3}{10}+\frac{10}{10}\right|=-\frac{7}{10}\text{ Simplify}\\ \left|\frac{7}{10}\right|=-\frac{7}{10}\text{ Simplify}\\ \frac{7}{10}\ne -\frac{7}{10}\text{ Get absolute values}\end{array}$

Since $h=-\frac{1}{10}$ does not satisfy the original equation, we must reject it since it is an extraneous root.

Conclusion:

Therefore, the solution is $h=\frac{1}{4}$ .