$$2\cdot\dfrac{y}{y^2+8y+7}-4\dfrac{y}{5y^2-20}=\dfrac{6y^3-32y^2-68y}{5y^4+40y^3+15y^2-160y-140}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2 \cdot \frac{y}{y^2+8y+7}-4\frac{y}{5y^2-20}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2y}{y^2+8y+7}-\frac{4y}{5y^2-20} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6y^3-32y^2-68y}{5y^4+40y^3+15y^2-160y-140}\end{aligned} $$ | |
| ① | Multiply $2$ by $ \dfrac{y}{y^2+8y+7} $ to get $ \dfrac{ 2y }{ y^2+8y+7 } $. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{y}{y^2+8y+7} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{y}{y^2+8y+7} \xlongequal{\text{Step 2}} \frac{ 2 \cdot y }{ 1 \cdot \left( y^2+8y+7 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2y }{ y^2+8y+7 } \end{aligned} $$ |
| ② | Multiply $4$ by $ \dfrac{y}{5y^2-20} $ to get $ \dfrac{ 4y }{ 5y^2-20 } $. Step 1: Write $ 4 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 4 \cdot \frac{y}{5y^2-20} & \xlongequal{\text{Step 1}} \frac{4}{\color{red}{1}} \cdot \frac{y}{5y^2-20} \xlongequal{\text{Step 2}} \frac{ 4 \cdot y }{ 1 \cdot \left( 5y^2-20 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4y }{ 5y^2-20 } \end{aligned} $$ |
| ③ | Subtract $ \dfrac{4y}{5y^2-20} $ from $ \dfrac{2y}{y^2+8y+7} $ to get $ \dfrac{ \color{purple}{ 6y^3-32y^2-68y } }{ 5y^4+40y^3+15y^2-160y-140 }$. To subtract raitonal expressions, both fractions must have the same denominator. |