Chapter 1.4, Problem 36E

To determine

To solve: the given equation and check the solution.

Answer to Problem 36E

The solution is $c=1\text{ or }c=-\frac{2}{3}$ .

Explanation of Solution

Given:

  $\left|2c+8\right|=\left|10c\right|$

Calculation:

  $\begin{array}{l}\left|2c+8\right|=\left|10c\right|\text{ Given}\\ 2c+8=10c\text{ or }2c+8=-10c\text{ Write two related linear equations}\end{array}$

Solve equation 1:

  $\begin{array}{l}2c+8=10c\\ 2c-2c+8=10c-2c\text{ Subtract both sides by 2}c\\ \frac{8}{8}=\frac{8c}{8}\text{ Divide both sides by 8}\\ 1=c\end{array}$

Solve equation 2:

  $\begin{array}{l}2c+8=-10c\\ 2c-2c+8=-10c-2c\text{ Subtract both sides by 2}c\\ \frac{8}{-12}=\frac{-12c}{-12}\text{ Divide both sides by -12}\\ -\frac{2}{3}=c\end{array}$

Check $\left(n=1\right)$ :

  $\begin{array}{l}\left|2c+8\right|=\left|10c\right|\\ \left|2\left(1\right)+8\right|=\left|10\left(1\right)\right|\text{ Substitute}\\ \left|2+8\right|=\left|10\right|\text{ Simplify}\\ \left|10\right|=\left|10\right|\text{ Simplify}\\ \text{10}=10\text{ Get absolute value}\end{array}$

Check $\left(n=-\frac{2}{3}\right)$ :

  $\begin{array}{l}\left|2c+8\right|=\left|10c\right|\\ \left|2\left(-\frac{2}{3}\right)+8\right|=\left|10\left(-\frac{2}{3}\right)\right|\text{ Substitute}\\ \left|\left(-\frac{4}{3}\right)+\frac{24}{3}\right|=\left|-\frac{20}{3}\right|\text{ Simplify}\\ \left|\frac{20}{3}\right|=\left|\frac{20}{3}\right|\text{ Simplify}\\ \frac{20}{3}=\frac{20}{3}\text{ Get absolute value}\end{array}$

Conclusion:

Therefore, the solution is $c=1\text{ or }c=-\frac{2}{3}$ .