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

Step 1: Write $ x $ 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 first fraction by $\color{blue}{ 5 }$.

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

Subtract $ \dfrac{1}{9} $ from $ x $ to get $ \dfrac{ \color{purple}{ 9x-1 } }{ 9 }$.

Step 1: Write $ x $ 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 first fraction by $\color{blue}{ 9 }$.

$$ \begin{aligned} x- \frac{1}{9} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} - \frac{1}{9} = \frac{ x \cdot \color{blue}{ 9 }}{ 1 \cdot \color{blue}{ 9 }} - \frac{ 1 }{ 9 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 9x } }{ 9 } - \frac{ \color{purple}{ 1 } }{ 9 }=\frac{ \color{purple}{ 9x-1 } }{ 9 } \end{aligned} $$

Multiply $x-1$ by $ \dfrac{5x-1}{5} $ to get $ \dfrac{5x^2-6x+1}{5} $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} x-1 \cdot \frac{5x-1}{5} & \xlongequal{\text{Step 1}} \frac{x-1}{\color{red}{1}} \cdot \frac{5x-1}{5} \xlongequal{\text{Step 2}} \frac{ \left( x-1 \right) \cdot \left( 5x-1 \right) }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5x^2-x-5x+1 }{ 5 } = \frac{5x^2-6x+1}{5} \end{aligned} $$

Subtract $ \dfrac{1}{9} $ from $ x $ to get $ \dfrac{ \color{purple}{ 9x-1 } }{ 9 }$.

Step 1: Write $ x $ 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 first fraction by $\color{blue}{ 9 }$.

$$ \begin{aligned} x- \frac{1}{9} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} - \frac{1}{9} = \frac{ x \cdot \color{blue}{ 9 }}{ 1 \cdot \color{blue}{ 9 }} - \frac{ 1 }{ 9 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 9x } }{ 9 } - \frac{ \color{purple}{ 1 } }{ 9 }=\frac{ \color{purple}{ 9x-1 } }{ 9 } \end{aligned} $$

Multiply $ \dfrac{5x^2-6x+1}{5} $ by $ \dfrac{9x-1}{9} $ to get $ \dfrac{45x^3-59x^2+15x-1}{45} $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5x^2-6x+1}{5} \cdot \frac{9x-1}{9} & \xlongequal{\text{Step 1}} \frac{ \left( 5x^2-6x+1 \right) \cdot \left( 9x-1 \right) }{ 5 \cdot 9 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 45x^3-5x^2-54x^2+6x+9x-1 }{ 45 } = \frac{45x^3-59x^2+15x-1}{45} \end{aligned} $$