Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{13\frac{x}{4+x^2}-13\frac{a}{4+a^2}}{x-a}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{(\frac{x}{4+x^2}-\frac{a}{4+a^2})\cdot13}{x-a} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{a^2x-ax^2-4a+4x}{a^2x^2+4a^2+4x^2+16}\cdot13}{x-a} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\frac{13a^2x-13ax^2-52a+52x}{a^2x^2+4a^2+4x^2+16}}{x-a} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{13a^2x-13ax^2-52a+52x}{-a^3x^2+a^2x^3-4a^3+4a^2x-4ax^2+4x^3-16a+16x}\end{aligned} $$ | |
① | Use the distributive property. |
② | To subtract raitonal expressions, both fractions must have the same denominator. |
③ | Step 1: Write $ 13 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{a^2x-ax^2-4a+4x}{a^2x^2+4a^2+4x^2+16} \cdot 13 & \xlongequal{\text{Step 1}} \frac{a^2x-ax^2-4a+4x}{a^2x^2+4a^2+4x^2+16} \cdot \frac{13}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( a^2x-ax^2-4a+4x \right) \cdot 13 }{ \left( a^2x^2+4a^2+4x^2+16 \right) \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 13a^2x-13ax^2-52a+52x }{ a^2x^2+4a^2+4x^2+16 } \end{aligned} $$ |
④ | Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{ \frac{13a^2x-13ax^2-52a+52x}{a^2x^2+4a^2+4x^2+16} }{x-a} & \xlongequal{\text{Step 1}} \frac{13a^2x-13ax^2-52a+52x}{a^2x^2+4a^2+4x^2+16} \cdot \frac{\color{blue}{1}}{\color{blue}{x-a}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 13a^2x-13ax^2-52a+52x \right) \cdot 1 }{ \left( a^2x^2+4a^2+4x^2+16 \right) \cdot \left( x-a \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 13a^2x-13ax^2-52a+52x }{ a^2x^3-a^3x^2+4a^2x-4a^3+4x^3-4ax^2+16x-16a } = \frac{13a^2x-13ax^2-52a+52x}{-a^3x^2+a^2x^3-4a^3+4a^2x-4ax^2+4x^3-16a+16x} \end{aligned} $$ |