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Question

Answer

$$\dfrac{x}{6x+8}-\dfrac{9}{3x+4}=\dfrac{x-18}{6x+8}$$

Explanation
$$ \begin{aligned}\frac{x}{6x+8}-\frac{9}{3x+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-18}{6x+8}\end{aligned} $$
①

Subtract $ \dfrac{9}{3x+4} $ from $ \dfrac{x}{6x+8} $ to get $ \dfrac{ \color{purple}{ x-18 } }{ 6x+8 }$.

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

$$ \begin{aligned} \frac{x}{6x+8} - \frac{9}{3x+4} & = \frac{ x }{ 6x+8 } - \frac{ 9 \cdot \color{blue}{ 2 }}{ \left( 3x+4 \right) \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ x } }{ 6x+8 } - \frac{ \color{purple}{ 18 } }{ 6x+8 }=\frac{ \color{purple}{ x-18 } }{ 6x+8 } \end{aligned} $$
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