Question
Answer
$$\dfrac{w^2-11w+24}{w^2-18w+80}=\dfrac{w-3}{w-10}$$
Explanation
| $$ \begin{aligned}\frac{w^2-11w+24}{w^2-18w+80}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{w-3}{w-10}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{w^2-11w+24}{w^2-18w+80} $ to $ \dfrac{w-3}{w-10} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{w-8}$. $$ \begin{aligned} \frac{w^2-11w+24}{w^2-18w+80} & =\frac{ \left( w-3 \right) \cdot \color{blue}{ \left( w-8 \right) }}{ \left( w-10 \right) \cdot \color{blue}{ \left( w-8 \right) }} = \\[1ex] &= \frac{w-3}{w-10} \end{aligned} $$ |
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