Question
Answer
$$\dfrac{x^2+1}{x^2+2x}\dfrac{x^2+x-2}{x^2-1}=\dfrac{x^2+1}{x^2+x}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^2+1}{x^2+2x}\frac{x^2+x-2}{x^2-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+1}{x^2+2x}\frac{x+2}{x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2+1}{x^2+x}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+x-2}{x^2-1} $ to $ \dfrac{x+2}{x+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{x^2+x-2}{x^2-1} & =\frac{ \left( x+2 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+1 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{x+2}{x+1} \end{aligned} $$ |
| ② | Multiply $ \dfrac{x^2+1}{x^2+2x} $ by $ \dfrac{x+2}{x+1} $ to get $ \dfrac{ x^2+1 }{ x^2+x } $. Step 1: Factor numerators and denominators. Step 2: Cancel common factors. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2+1}{x^2+2x} \cdot \frac{x+2}{x+1} & \xlongequal{\text{Step 1}} \frac{ x^2+1 }{ x \cdot \color{red}{ \left( x+2 \right) } } \cdot \frac{ 1 \cdot \color{red}{ \left( x+2 \right) } }{ x+1 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x^2+1 }{ x } \cdot \frac{ 1 }{ x+1 } \xlongequal{\text{Step 3}} \frac{ \left( x^2+1 \right) \cdot 1 }{ x \cdot \left( x+1 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ x^2+1 }{ x^2+x } \end{aligned} $$ |
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