Answer
$$\frac{49-y^2}{6y^2+42y}\frac{48y^3}{y^2-10y+21}=-\frac{48y^3}{6y^2-18y}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{49-y^2}{6y^2+42y}\frac{48y^3}{y^2-10y+21}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-y+7}{6y}\frac{48y^3}{y^2-10y+21} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-\frac{48y^3}{6y^2-18y}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{49-y^2}{6y^2+42y} $ to $ \dfrac{-y+7}{6y} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y+7}$. $$ \begin{aligned} \frac{49-y^2}{6y^2+42y} & =\frac{ \left( -y+7 \right) \cdot \color{blue}{ \left( y+7 \right) }}{ 6y \cdot \color{blue}{ \left( y+7 \right) }} = \\[1ex] &= \frac{-y+7}{6y} \end{aligned} $$ |
| ② | Multiply $ \dfrac{-y+7}{6y} $ by $ \dfrac{48y^3}{y^2-10y+21} $ to get $ \dfrac{ -48y^3 }{ 6y^2-18y } $. Step 1: Factor numerators and denominators. Step 2: Cancel common factors. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} \frac{-y+7}{6y} \cdot \frac{48y^3}{y^2-10y+21} & \xlongequal{\text{Step 1}} \frac{ \left( -1 \right) \cdot \color{blue}{ \left( y-7 \right) } }{ 6y } \cdot \frac{ 48y^3 }{ \left( y-3 \right) \cdot \color{blue}{ \left( y-7 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ -1 }{ 6y } \cdot \frac{ 48y^3 }{ y-3 } \xlongequal{\text{Step 3}} \frac{ \left( -1 \right) \cdot 48y^3 }{ 6y \cdot \left( y-3 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ -48y^3 }{ 6y^2-18y } \end{aligned} $$ |
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