Question

Solve each equation. State any extraneous solutions.

$\frac{\mathbf{4}\mathbf{v}}{\mathbf{2}\mathbf{v}\mathbf{+}\mathbf{3}}\mathbf{-}\frac{\mathbf{2}\mathbf{v}}{\mathbf{2}\mathbf{v}\mathbf{-}\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

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

Step-by-step solution

Step 1. State the concept for observational studies.

The linear equations in one variable is an equation which is expressed in the form of $ax+b=0$, where a and b are two integers, and x is a variable and has only one solution.

Step 2. Solve the equation.

The equation can be solved as follows:

$\begin{array}{c}\frac{4v}{2v+3}-\frac{2v}{2v-3}=1\\ \frac{\left(4v\right)\left(2v-3\right)-2v\left(2v+3\right)}{\left(2v+3\right)\left(2v-3\right)}=1\\ \left(4v\right)\left(2v-3\right)-2v\left(2v+3\right)=\left(2v+3\right)\left(2v-3\right)\\ 8v^2-12v-4v^2-6v={\left(2v\right)}^2-{\left(3\right)}^2\\ 4v^2-18v=4v^2-9\\ -18v=-9\\ v=\frac{-9}{-18}\\ v=\frac{1}{2}\end{array}$

Step 3. Check the solution.

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

role="math" $\begin{array}{c}\frac{4v}{2v+3}-\frac{2v}{2v-3}=1\\ \frac{4\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)+3}-\frac{2\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-3}=1\\ \frac{2}{1+3}-\frac{1}{1-3}=1\\ \frac{2}{4}-\frac{1}{-2}=-1\\ \frac{1}{2}+\frac{1}{2}=1\\ 1=1\\ LHS=RHS\end{array}$

Since both sides are equal, the equation is correct.