$$(\dfrac{1}{3}x-y)(\dfrac{1}{3}x+y)=\dfrac{x^2-9y^2}{9}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(\frac{1}{3}x-y)(\frac{1}{3}x+y)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(\frac{x}{3}-y)(\frac{x}{3}+y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{x-3y}{3}\frac{x+3y}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{x^2-9y^2}{9}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{3} $ by $ x $ to get $ \dfrac{ x }{ 3 } $. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 3 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{1}{3} $ by $ x $ to get $ \dfrac{ x }{ 3 } $. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 3 } \end{aligned} $$ |
| ③ | Subtract $y$ from $ \dfrac{x}{3} $ to get $ \dfrac{ \color{purple}{ x-3y } }{ 3 }$. 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 | Add $ \dfrac{x}{3} $ and $ y $ to get $ \dfrac{ \color{purple}{ x+3y } }{ 3 }$. Step 1: Write $ y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To add raitonal expressions, both fractions must have the same denominator. |
| ⑤ | Multiply $ \dfrac{x-3y}{3} $ by $ \dfrac{x+3y}{3} $ to get $ \dfrac{x^2-9y^2}{9} $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{x-3y}{3} \cdot \frac{x+3y}{3} & \xlongequal{\text{Step 1}} \frac{ \left( x-3y \right) \cdot \left( x+3y \right) }{ 3 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x^2+ \cancel{3xy} -\cancel{3xy}-9y^2 }{ 9 } = \frac{x^2-9y^2}{9} \end{aligned} $$ |