Question
Answer
$$\dfrac{r^2-8r+7}{r^2-3r+2}=\dfrac{r-7}{r-2}$$
Explanation
| $$ \begin{aligned}\frac{r^2-8r+7}{r^2-3r+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{r-7}{r-2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{r^2-8r+7}{r^2-3r+2} $ to $ \dfrac{r-7}{r-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{r-1}$. $$ \begin{aligned} \frac{r^2-8r+7}{r^2-3r+2} & =\frac{ \left( r-7 \right) \cdot \color{blue}{ \left( r-1 \right) }}{ \left( r-2 \right) \cdot \color{blue}{ \left( r-1 \right) }} = \\[1ex] &= \frac{r-7}{r-2} \end{aligned} $$ |
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Simplify Rational Expressions