91Ó°ÊÓ

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.

The equation can be solved as follows:

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

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.