Question
Answer
$$\dfrac{49y^3-25y}{49y+35}=\dfrac{7y^2-5y}{7}$$
Explanation
| $$ \begin{aligned}\frac{49y^3-25y}{49y+35}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7y^2-5y}{7}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{49y^3-25y}{49y+35} $ to $ \dfrac{7y^2-5y}{7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{7y+5}$. $$ \begin{aligned} \frac{49y^3-25y}{49y+35} & =\frac{ \left( 7y^2-5y \right) \cdot \color{blue}{ \left( 7y+5 \right) }}{ 7 \cdot \color{blue}{ \left( 7y+5 \right) }} = \\[1ex] &= \frac{7y^2-5y}{7} \end{aligned} $$ |
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Simplify Rational Expressions