Answer
$$\frac{25r^2-4}{5r^2-2r}=\frac{5r+2}{r}$$
Explanation
| $$ \begin{aligned}\frac{25r^2-4}{5r^2-2r}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5r+2}{r}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{25r^2-4}{5r^2-2r} $ to $ \dfrac{5r+2}{r} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{5r-2}$. $$ \begin{aligned} \frac{25r^2-4}{5r^2-2r} & =\frac{ \left( 5r+2 \right) \cdot \color{blue}{ \left( 5r-2 \right) }}{ r \cdot \color{blue}{ \left( 5r-2 \right) }} = \\[1ex] &= \frac{5r+2}{r} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions