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Question

Answer

$$\dfrac{x}{x-4}+\dfrac{1}{2x-8}=\dfrac{2x+1}{2x-8}$$

Explanation
$$ \begin{aligned}\frac{x}{x-4}+\frac{1}{2x-8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x+1}{2x-8}\end{aligned} $$
①

Add $ \dfrac{x}{x-4} $ and $ \dfrac{1}{2x-8} $ to get $ \dfrac{ \color{purple}{ 2x+1 } }{ 2x-8 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} \frac{x}{x-4} + \frac{1}{2x-8} & = \frac{ x \cdot \color{blue}{ 2 }}{ \left( x-4 \right) \cdot \color{blue}{ 2 }} + \frac{ 1 }{ 2x-8 } = \\[1ex] &=\frac{ \color{purple}{ 2x } }{ 2x-8 } + \frac{ \color{purple}{ 1 } }{ 2x-8 }=\frac{ \color{purple}{ 2x+1 } }{ 2x-8 } \end{aligned} $$
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