$$\dfrac{3y-3\dfrac{x^2}{y}}{x-\dfrac{y^2}{x}}=\dfrac{-3x^3+3xy^2}{x^2y-y^3}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3y-3\frac{x^2}{y}}{x-\frac{y^2}{x}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{(y-\frac{x^2}{y})\cdot3}{\frac{x^2-y^2}{x}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\frac{-x^2+y^2}{y}\cdot3}{\frac{x^2-y^2}{x}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{\frac{-3x^2+3y^2}{y}}{\frac{x^2-y^2}{x}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-3x^3+3xy^2}{x^2y-y^3}\end{aligned} $$ | |
| ① | Use the distributive property. |
| ② | Subtract $ \dfrac{y^2}{x} $ from $ x $ to get $ \dfrac{ \color{purple}{ x^2-y^2 } }{ x }$. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |
| ③ | Subtract $ \dfrac{x^2}{y} $ from $ y $ to get $ \dfrac{ \color{purple}{ -x^2+y^2 } }{ y }$. Step 1: Write $ y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |
| ④ | Subtract $ \dfrac{y^2}{x} $ from $ x $ to get $ \dfrac{ \color{purple}{ x^2-y^2 } }{ x }$. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |
| ⑤ | Multiply $ \dfrac{-x^2+y^2}{y} $ by $ 3 $ to get $ \dfrac{ -3x^2+3y^2 }{ y } $. Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{-x^2+y^2}{y} \cdot 3 & \xlongequal{\text{Step 1}} \frac{-x^2+y^2}{y} \cdot \frac{3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( -x^2+y^2 \right) \cdot 3 }{ y \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -3x^2+3y^2 }{ y } \end{aligned} $$ |
| ⑥ | Subtract $ \dfrac{y^2}{x} $ from $ x $ to get $ \dfrac{ \color{purple}{ x^2-y^2 } }{ x }$. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |
| ⑦ | Divide $ \dfrac{-3x^2+3y^2}{y} $ by $ \dfrac{x^2-y^2}{x} $ to get $ \dfrac{ -3x^3+3xy^2 }{ x^2y-y^3 } $. Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{ \frac{-3x^2+3y^2}{y} }{ \frac{\color{blue}{x^2-y^2}}{\color{blue}{x}} } & \xlongequal{\text{Step 1}} \frac{-3x^2+3y^2}{y} \cdot \frac{\color{blue}{x}}{\color{blue}{x^2-y^2}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( -3x^2+3y^2 \right) \cdot x }{ y \cdot \left( x^2-y^2 \right) } \xlongequal{\text{Step 3}} \frac{ -3x^3+3xy^2 }{ x^2y-y^3 } \end{aligned} $$ |