Chapter 12.1, Problem 37SR

To determine

To solve: The given equation and state any extraneous solutions.

Answer to Problem 37SR

The solution of the equation is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The following equation:

  $\frac{4v}{2v+3}-\frac{2v}{2v-3}=1$

Calculation:

The equation is solved as follows:

  $\begin{array}{l}\frac{4v}{2v+3}-\frac{2v}{2v-3}=1 [\text{Original equation]}\\ \frac{4v(2v-3)-2v(2v+3)}{(2v+3)(2v-3)}=1 [\text{Subtract fractions]}\\ 4v(2v-3)-2v(2v+3)=1(2v+3)(2v-3) [\text{Form cross product]}\\ 8v^2-12v-4v^2-6v=4v^2-6v+6v-9 [\text{Distributive property]}\\ \text{ 4}{v}^2-18v=4v^2-9 [\text{Combine like terms]}\\ \underset{\_}{-4v^2=-4v^2} [\text{Subtract 4}{v}^2\text{ from both sides]}\\ -18v=-\text{9} [\text{Simplify]}\\ \frac{-18v}{-18}=\frac{-9}{-18} [\text{Divide both sides by }-18]\\ v=\frac{1}{2} [\text{Simplify]}\end{array}$

To check the solution, substitute $\frac{1}{2}$ for $v$ in the original equation

  $\begin{array}{l}\frac{4v}{2v+3}-\frac{2v}{2v-3}=1 [\text{Original equation}]\\ \frac{4\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)+3}-\frac{2\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-3}=1\left[\text{Substitute }\frac{1}{2}\text{ for }v\right]\\ \frac{2}{1+3}-\frac{1}{1-3}=1\text{ [Multiply]}\\ \frac{2}{4}+\frac{1}{2}=1 [\text{Simplify]}\\ \frac{1}{2}+\frac{1}{2}=1 [\text{Divide common factors]}\\ \frac{2}{2}=1 [\text{Add]}\\ 1=1 [\text{Simplify]}\end{array}$

Since both sides are equal, the solution is correct.