Multiply $ \dfrac{x-2}{2} $ by $ 8x-20 $ to get $ \dfrac{8x^2-36x+40}{2} $.

Step 1: Write $ 8x-20 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x-2}{2} \cdot 8x-20 & \xlongequal{\text{Step 1}} \frac{x-2}{2} \cdot \frac{8x-20}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x-2 \right) \cdot \left( 8x-20 \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 8x^2-20x-16x+40 }{ 2 } = \frac{8x^2-36x+40}{2} \end{aligned} $$

Add $ \dfrac{5}{3} $ and $ x $ to get $ \dfrac{ \color{purple}{ 3x+5 } }{ 3 }$.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{3}$.

$$ \begin{aligned} \frac{5}{3} +x & \xlongequal{\text{Step 1}} \frac{5}{3} + \frac{x}{\color{red}{1}} = \frac{ 5 }{ 3 } + \frac{ x \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 5 } }{ 3 } + \frac{ \color{purple}{ 3x } }{ 3 }=\frac{ \color{purple}{ 3x+5 } }{ 3 } \end{aligned} $$

Multiply $ \dfrac{25}{3} $ by $ x $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{25}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Multiply $ \dfrac{8x^2-36x+40}{2} $ by $ x-2 $ to get $ \dfrac{8x^3-52x^2+112x-80}{2} $.

Step 1: Write $ x-2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{8x^2-36x+40}{2} \cdot x-2 & \xlongequal{\text{Step 1}} \frac{8x^2-36x+40}{2} \cdot \frac{x-2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 8x^2-36x+40 \right) \cdot \left( x-2 \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 8x^3-16x^2-36x^2+72x+40x-80 }{ 2 } = \frac{8x^3-52x^2+112x-80}{2} \end{aligned} $$

Subtract $2$ from $ \dfrac{3x+5}{3} $ to get $ \dfrac{ \color{purple}{ 3x-1 } }{ 3 }$.

Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{3}$.

$$ \begin{aligned} \frac{3x+5}{3} -2 & \xlongequal{\text{Step 1}} \frac{3x+5}{3} - \frac{2}{\color{red}{1}} = \frac{ 3x+5 }{ 3 } - \frac{ 2 \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x+5 } }{ 3 } - \frac{ \color{purple}{ 6 } }{ 3 }=\frac{ \color{purple}{ 3x-1 } }{ 3 } \end{aligned} $$

Multiply $ \dfrac{25}{3} $ by $ x $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{25}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Multiply $ \dfrac{8x^2-36x+40}{2} $ by $ x-2 $ to get $ \dfrac{8x^3-52x^2+112x-80}{2} $.

Step 1: Write $ x-2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{8x^2-36x+40}{2} \cdot x-2 & \xlongequal{\text{Step 1}} \frac{8x^2-36x+40}{2} \cdot \frac{x-2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 8x^2-36x+40 \right) \cdot \left( x-2 \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 8x^3-16x^2-36x^2+72x+40x-80 }{ 2 } = \frac{8x^3-52x^2+112x-80}{2} \end{aligned} $$

Multiply $ \dfrac{25}{2} $ by $ \dfrac{3x-1}{3} $ to get $ \dfrac{ 75x-25 }{ 6 } $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25}{2} \cdot \frac{3x-1}{3} & \xlongequal{\text{Step 1}} \frac{ 25 \cdot \left( 3x-1 \right) }{ 2 \cdot 3 } \xlongequal{\text{Step 2}} \frac{ 75x-25 }{ 6 } \end{aligned} $$

Multiply $ \dfrac{25}{3} $ by $ x $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{25}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Add $ \dfrac{8x^3-52x^2+112x-80}{2} $ and $ \dfrac{75x-25}{6} $ to get $ \dfrac{ \color{purple}{ 24x^3-156x^2+411x-265 } }{ 6 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 3 }$.

$$ \begin{aligned} \frac{8x^3-52x^2+112x-80}{2} + \frac{75x-25}{6} & = \frac{ \left( 8x^3-52x^2+112x-80 \right) \cdot \color{blue}{ 3 }}{ 2 \cdot \color{blue}{ 3 }} + \frac{ 75x-25 }{ 6 } = \\[1ex] &=\frac{ \color{purple}{ 24x^3-156x^2+336x-240 } }{ 6 } + \frac{ \color{purple}{ 75x-25 } }{ 6 }=\frac{ \color{purple}{ 24x^3-156x^2+411x-265 } }{ 6 } \end{aligned} $$

Multiply $ \dfrac{25}{3} $ by $ x $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{25}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Subtract $ \dfrac{25x}{3} $ from $ \dfrac{24x^3-156x^2+411x-265}{6} $ to get $ \dfrac{ \color{purple}{ 24x^3-156x^2+361x-265 } }{ 6 }$.

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

$$ \begin{aligned} \frac{24x^3-156x^2+411x-265}{6} - \frac{25x}{3} & = \frac{ 24x^3-156x^2+411x-265 }{ 6 } - \frac{ 25x \cdot \color{blue}{ 2 }}{ 3 \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ 24x^3-156x^2+411x-265 } }{ 6 } - \frac{ \color{purple}{ 50x } }{ 6 }=\frac{ \color{purple}{ 24x^3-156x^2+361x-265 } }{ 6 } \end{aligned} $$