Divide $ \dfrac{9x-3}{2x^2} $ by $ \dfrac{3x-1}{8x} $ to get $ \dfrac{ 24x }{ 2x^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.

$$ \begin{aligned} \frac{ \frac{9x-3}{2x^2} }{ \frac{\color{blue}{3x-1}}{\color{blue}{8x}} } & \xlongequal{\text{Step 1}} \frac{9x-3}{2x^2} \cdot \frac{\color{blue}{8x}}{\color{blue}{3x-1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 3 \cdot \color{blue}{ \left( 3x-1 \right) } }{ 2x^2 } \cdot \frac{ 8x }{ 1 \cdot \color{blue}{ \left( 3x-1 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3 }{ 2x^2 } \cdot \frac{ 8x }{ 1 } \xlongequal{\text{Step 4}} \frac{ 3 \cdot 8x }{ 2x^2 \cdot 1 } \xlongequal{\text{Step 5}} \frac{ 24x }{ 2x^2 } \end{aligned} $$