Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{y^2-2y}{y^2+7y-18}\frac{y^2-81}{y^2-11y+18}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{y}{y+9}\frac{y+9}{y-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{y}{y-2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{y^2-2y}{y^2+7y-18} $ to $ \dfrac{y}{y+9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y-2}$. $$ \begin{aligned} \frac{y^2-2y}{y^2+7y-18} & =\frac{ y \cdot \color{blue}{ \left( y-2 \right) }}{ \left( y+9 \right) \cdot \color{blue}{ \left( y-2 \right) }} = \\[1ex] &= \frac{y}{y+9} \end{aligned} $$Simplify $ \dfrac{y^2-81}{y^2-11y+18} $ to $ \dfrac{y+9}{y-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y-9}$. $$ \begin{aligned} \frac{y^2-81}{y^2-11y+18} & =\frac{ \left( y+9 \right) \cdot \color{blue}{ \left( y-9 \right) }}{ \left( y-2 \right) \cdot \color{blue}{ \left( y-9 \right) }} = \\[1ex] &= \frac{y+9}{y-2} \end{aligned} $$ |
| ② | Multiply $ \dfrac{y}{y+9} $ by $ \dfrac{y+9}{y-2} $ to get $ \dfrac{ y }{ y-2 } $. Step 1: Cancel $ \color{red}{ y+9 } $ in first and second fraction. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{y}{y+9} \cdot \frac{y+9}{y-2} & \xlongequal{\text{Step 1}} \frac{y}{\color{red}{1}} \cdot \frac{\color{red}{1}}{y-2} \xlongequal{\text{Step 2}} \frac{ y \cdot 1 }{ 1 \cdot \left( y-2 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ y }{ y-2 } \end{aligned} $$ |