### Theory:

Subtraction of Rational Expressions
Let us discuss the subtraction of two rational expressions in two cases.

Case 1: Like Denominators

If the rational numbers are of the form $$\frac{p(x)}{q(x)}$$ and $$\frac{r(x)}{q(x)}$$, the subtraction is performed as follows.

Working Rule:
Step 1: Subtract the numerators $$p(x)$$ and $$r(x)$$.

Step 2: Write the difference of the numerators found in the previous step over the common denominator $$q(x)$$.

Step 3: Reduce the resulting rational expression $$\frac{p(x) - r(x)}{q(x)}$$ into its lowest form.
Example:
Subtract: $$\frac{2x + 1}{x + 1}$$ and $$\frac{5x + 4}{x + 1}$$

Solution:

Step 1: Subtract the numerators.

$$2x + 1$$ $$-$$ $$5x + 4$$ $$=$$ $$(2 - 5)x + (1 - 4)$$

$$=$$ $$-3x - 3$$

Step 2: Write the difference of the numerators found in the previous step over the common denominator.

$$\frac{2x + 1}{x + 1}$$ $$-$$ $$\frac{5x + 4}{x + 1}$$ $$=$$ $$\frac{-3x - 3}{x + 1}$$

Step 3: Reduce the resulting rational expression into its lowest form.

$$\frac{-3x - 3}{x + 1}$$ $$=$$ $$\frac{-3(x + 1)}{x + 1}$$

$$=$$ $\frac{-3\overline{)\left(x+1\right)}}{\overline{)\left(x+1\right)}}$

$$=$$ $$-3$$

Therefore, $$\frac{2x + 1}{x + 1}$$ $$-$$ $$\frac{5x + 4}{x + 1}$$ $$=$$ $$-3$$.
Case 2: Unlike Denominators

If the rational numbers are of the form $$\frac{p(x)}{q(x)}$$ and $$\frac{r(x)}{s(x)}$$, the subtraction is performed as follows.

Working Rule:
Step 1: Determine the Least Common Multiple of the denominator $$q(x)$$ and $$s(x)$$.

Step 2: Rewrite each fraction as an equivalent fraction with the $$LCM$$ obtained in the previous step by multiplying both the numerators and denominator of each expression by any factors needed to obtain the $$LCM$$.

Step 3: Follow the same steps given for doing subtraction of the rational expression with like denominators.
Example:
Subtract: $$\frac{2x}{x + 1}$$ and $$\frac{5x}{x + 2}$$

Solution:

Step 1: Determine the Least Common Multiple of the denominator.

$$LCM \left((x+1)(x+2) \right)$$ $$=$$ $$(x+1)(x+2)$$

Step 2: Rewrite each fraction as an equivalent fraction with the $$LCM$$ obtained in the previous step by multiplying both the numerators and denominator of each expression by any factors needed to obtain the $$LCM$$.

$$\frac{2x}{x + 1}$$ $$-$$ $$\frac{5x}{x + 2}$$ $$=$$ $$\frac{2x \times (x+2)}{(x + 1) \times (x+2)}$$ $$-$$ $$\frac{5x \times (x+1)}{(x + 2) \times (x+1)}$$

$$=$$ $$\frac{2x^2 + 4x}{(x + 1)(x+2)}$$ $$-$$ $$\frac{5x^2 + 5x}{(x + 2)(x+1)}$$

Step 3: Follow the same steps given for doing subtraction of the rational expression with like denominators.

$$\frac{2x^2 + 4x}{(x + 1)(x+2)}$$ $$-$$ $$\frac{5x^2 + 5x}{(x + 2)(x+1)}$$ $$=$$ $$\frac{(2 - 5)x^2 + (2 - 5)x}{(x + 1)(x+2)}$$

$$=$$ $$\frac{-3x^2 - 3x}{(x + 1)(x+2)}$$

$$=$$ $$\frac{-3x(x + 1)}{(x + 1)(x+2)}$$

$$=$$ $\frac{-3x\overline{)\left(x+1\right)}}{\overline{)\left(x+1\right)}\left(x+2\right)}$

$$=$$ $$\frac{-3x}{x+2}$$

Therefore, $$\frac{2x}{x + 1}$$ $$-$$ $$\frac{5x}{x + 2}$$ $$=$$ $$\frac{-3x}{x+2}$$.