$$(x-4)\dfrac{x+2}{8(x+3)}(x+3)\dfrac{x+4}{2(x-4)}=\dfrac{x^2+6x+8}{16}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-4)\frac{x+2}{8(x+3)}(x+3)\frac{x+4}{2(x-4)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x-4)\frac{x+2}{8x+24}(x+3)\frac{x+4}{2x-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{x^2-2x-8}{8x+24}\frac{x^2+7x+12}{2x-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{x^2+6x+8}{16}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{8} $ by $ \left( x+3\right) $ $$ \color{blue}{8} \cdot \left( x+3\right) = 8x+24 $$ |
| ② | Multiply $ \color{blue}{2} $ by $ \left( x-4\right) $ $$ \color{blue}{2} \cdot \left( x-4\right) = 2x-8 $$ |
| ③ | Multiply $x-4$ by $ \dfrac{x+2}{8x+24} $ to get $ \dfrac{x^2-2x-8}{8x+24} $. Step 1: Write $ x-4 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} x-4 \cdot \frac{x+2}{8x+24} & \xlongequal{\text{Step 1}} \frac{x-4}{\color{red}{1}} \cdot \frac{x+2}{8x+24} \xlongequal{\text{Step 2}} \frac{ \left( x-4 \right) \cdot \left( x+2 \right) }{ 1 \cdot \left( 8x+24 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2+2x-4x-8 }{ 8x+24 } = \frac{x^2-2x-8}{8x+24} \end{aligned} $$ |
| ④ | Multiply $x+3$ by $ \dfrac{x+4}{2x-8} $ to get $ \dfrac{x^2+7x+12}{2x-8} $. Step 1: Write $ x+3 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} x+3 \cdot \frac{x+4}{2x-8} & \xlongequal{\text{Step 1}} \frac{x+3}{\color{red}{1}} \cdot \frac{x+4}{2x-8} \xlongequal{\text{Step 2}} \frac{ \left( x+3 \right) \cdot \left( x+4 \right) }{ 1 \cdot \left( 2x-8 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2+4x+3x+12 }{ 2x-8 } = \frac{x^2+7x+12}{2x-8} \end{aligned} $$ |
| ⑤ | Multiply $ \dfrac{x^2-2x-8}{8x+24} $ by $ \dfrac{x^2+7x+12}{2x-8} $ to get $ \dfrac{x^2+6x+8}{16} $. Step 1: Factor numerators and denominators. Step 2: Cancel common factors. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2-2x-8}{8x+24} \cdot \frac{x^2+7x+12}{2x-8} & \xlongequal{\text{Step 1}} \frac{ \left( x+2 \right) \cdot \color{blue}{ \left( x-4 \right) } }{ 8 \cdot \color{red}{ \left( x+3 \right) } } \cdot \frac{ \left( x+4 \right) \cdot \color{red}{ \left( x+3 \right) } }{ 2 \cdot \color{blue}{ \left( x-4 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x+2 }{ 8 } \cdot \frac{ x+4 }{ 2 } \xlongequal{\text{Step 3}} \frac{ \left( x+2 \right) \cdot \left( x+4 \right) }{ 8 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ x^2+4x+2x+8 }{ 16 } = \frac{x^2+6x+8}{16} \end{aligned} $$ |