$$(2y+\dfrac{3}{4}x)(2y-\dfrac{3}{4}x)=\dfrac{-9x^2+64y^2}{16}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2y+\frac{3}{4}x)(2y-\frac{3}{4}x)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(2y+\frac{3x}{4})(2y-\frac{3x}{4}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3x+8y}{4}\frac{-3x+8y}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-9x^2+64y^2}{16}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{3}{4} $ by $ x $ to get $ \dfrac{ 3x }{ 4 } $. 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. $$ \begin{aligned} \frac{3}{4} \cdot x & \xlongequal{\text{Step 1}} \frac{3}{4} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 3 \cdot x }{ 4 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 3x }{ 4 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{3}{4} $ by $ x $ to get $ \dfrac{ 3x }{ 4 } $. 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. $$ \begin{aligned} \frac{3}{4} \cdot x & \xlongequal{\text{Step 1}} \frac{3}{4} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 3 \cdot x }{ 4 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 3x }{ 4 } \end{aligned} $$ |
| ③ | Add $2y$ and $ \dfrac{3x}{4} $ to get $ \dfrac{ \color{purple}{ 3x+8y } }{ 4 }$. Step 1: Write $ 2y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To add raitonal expressions, both fractions must have the same denominator. |
| ④ | Subtract $ \dfrac{3x}{4} $ from $ 2y $ to get $ \dfrac{ \color{purple}{ -3x+8y } }{ 4 }$. Step 1: Write $ 2y $ 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{3x+8y}{4} $ by $ \dfrac{-3x+8y}{4} $ to get $ \dfrac{-9x^2+64y^2}{16} $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{3x+8y}{4} \cdot \frac{-3x+8y}{4} & \xlongequal{\text{Step 1}} \frac{ \left( 3x+8y \right) \cdot \left( -3x+8y \right) }{ 4 \cdot 4 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ -9x^2+ \cancel{24xy} -\cancel{24xy}+64y^2 }{ 16 } = \frac{-9x^2+64y^2}{16} \end{aligned} $$ |