Question
Answer
$$\dfrac{12h^2-40h+28}{3h^2-7h}=\dfrac{4h-4}{h}$$
Explanation
| $$ \begin{aligned}\frac{12h^2-40h+28}{3h^2-7h}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4h-4}{h}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{12h^2-40h+28}{3h^2-7h} $ to $ \dfrac{4h-4}{h} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{3h-7}$. $$ \begin{aligned} \frac{12h^2-40h+28}{3h^2-7h} & =\frac{ \left( 4h-4 \right) \cdot \color{blue}{ \left( 3h-7 \right) }}{ h \cdot \color{blue}{ \left( 3h-7 \right) }} = \\[1ex] &= \frac{4h-4}{h} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions