Question

Solve each equation. State any extraneous solutions.

$\frac{\mathbf{r}\mathbf{-}\mathbf{2}}{\mathbf{r}\mathbf{+}\mathbf{2}}\mathbf{-}\frac{\mathbf{3}\mathbf{r}}{\mathbf{r}\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}$

Short answer

The solution of the equation is$\mathbf{-}\frac{\mathbf{2}}{\mathbf{5}}$.

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:

$\begin{array}{c}\frac{r-2}{r+2}-\frac{3r}{r-2}=-2\\ \frac{\left(r-2\right)\left(r-2\right)-3r\left(r+2\right)}{\left(r-2\right)\left(r+2\right)}=-2\\ \left(r-2\right)\left(r-2\right)-3r\left(r+2\right)=-2\left(r-2\right)\left(r+2\right)\\ r^2-2r-2r+4-3r^2-6r=-2r^2+4r-4r+8\\ -2r^2-10r+4=-2r^2+8\\ -10r+4=8\\ -10r=8-4\\ -10r=4\\ r=\frac{-4}{10}\\ r=-\frac{2}{5}\end{array}$

Step 3. Check the solution.

To check the solution, substitute $r=-\frac{2}{5}$for in the original equation,

$\begin{array}{c}\frac{r-2}{r+2}-\frac{3r}{r-2}=-2\\ \frac{\left(\frac{-2}{5}\right)-2}{\left(\frac{-2}{5}\right)+2}-\frac{3\left(\frac{-2}{5}\right)}{\left(\frac{-2}{5}\right)-2}=-2\\ \frac{\frac{-2-2\left(5\right)}{5}}{\frac{-2+2\left(5\right)}{5}}-\frac{\left(\frac{-6}{5}\right)}{\frac{-2-2\left(5\right)}{5}}=-2\\ \frac{\frac{-12}{5}}{\frac{8}{5}}-\frac{\left(\frac{-6}{5}\right)}{\frac{-12}{5}}=-2\\ \frac{-12}{8}-\frac{6}{12}=-2\\ -\frac{3}{2}-\frac{1}{2}=-2\\ \frac{-3-1}{2}=-2\\ \frac{-4}{2}=-2\\ -2=-2\\ LHS=RHS\end{array}$

Since both sides are equal, the equation is correct.