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