7.2 Add and Subtract Rational Expressions

Adding and Subtracting Rational Expressions

Adding and subtracting rational expressions works just like adding and subtracting numeric fractions: you need a common denominator before you can combine numerators. The difference is that your denominators now contain variables, which makes finding that common denominator a bit more involved.

Adding and Subtracting Rational Expressions

Addition with common denominators

When two rational expressions already share the same denominator, you simply add the numerators and keep the denominator unchanged.

$\frac{A}{C} + \frac{B}{C} = \frac{A + B}{C}$

Numeric example:

$\frac{3}{7} + \frac{2}{7} = \frac{3+2}{7} = \frac{5}{7}$

Algebraic example:

$\frac{4x}{y^2} + \frac{7x}{y^2} = \frac{4x+7x}{y^2} = \frac{11x}{y^2}$

After combining, always simplify the numerator by combining like terms. The denominator stays as-is.

Least common denominator identification

When the denominators are different, you need to find the least common denominator (LCD) before you can add or subtract. The LCD is the least common multiple (LCM) of the denominators.

Finding the LCD — step by step:

1. Factor each denominator completely (into primes and variables).
2. List every distinct factor that appears in any denominator.
3. For each factor, use the highest power that appears in any single denominator.
4. Multiply those together. That product is your LCD.

Once you have the LCD, multiply the numerator and denominator of each fraction by whatever factor is "missing" from its denominator:

$\frac{x}{2y} \cdot \frac{2}{2} = \frac{2x}{4y}, \qquad \frac{3}{4y} \cdot \frac{1}{1} = \frac{3}{4y}$

Now both fractions have the denominator $4y$ and you can combine them.

Subtraction with unlike denominators

Subtraction follows the same process as addition. Find the LCD, rewrite each fraction, then subtract the numerators. A common mistake is forgetting to distribute the negative sign across the entire second numerator.

Example 1 (numeric):

$\frac{5}{6} - \frac{1}{3}$

The LCD of 6 and 3 is 6. Rewrite $\frac{1}{3}$ as $\frac{2}{6}$:

$\frac{5}{6} - \frac{2}{6} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}$

Example 2 (algebraic):

$\frac{3x}{4y} - \frac{2x}{3y}$

The LCD of $4y$ and $3y$ is $12y$:

$\frac{3x \cdot 3}{4y \cdot 3} - \frac{2x \cdot 4}{3y \cdot 4} = \frac{9x}{12y} - \frac{8x}{12y} = \frac{9x - 8x}{12y} = \frac{x}{12y}$

Simplification after combining

After you add or subtract, always check whether the result can be simplified. Factor the numerator and denominator, then cancel any common factors.

Example:

$\frac{6x^2 + 3x}{3x} + \frac{x^2 - x}{3x}$

1. Combine the numerators over the common denominator: $\frac{6x^2 + 3x + x^2 - x}{3x} = \frac{7x^2 + 2x}{3x}$

2. Factor the numerator: $\frac{x(7x + 2)}{3x}$

3. Cancel the common factor of $x$ (noting $x \neq 0$): $\frac{7x + 2}{3}$

Never cancel terms that are added or subtracted inside a numerator or denominator. You can only cancel factors of the entire numerator with factors of the entire denominator.

Techniques for rational functions

The same addition and subtraction rules apply when you're combining rational functions. A rational function is any function of the form $f(x) = \frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials.

Example: Given $f(x) = \frac{3x-1}{2x+1}$ and $g(x) = \frac{2x+3}{x-2}$, find $(f+g)(x)$.

1. Set up the addition: $(f+g)(x) = \frac{3x-1}{2x+1} + \frac{2x+3}{x-2}$

2. The denominators don't share any common factors, so the LCD is $(2x+1)(x-2)$.

3. Rewrite each fraction with the LCD: $\frac{(3x-1)(x-2)}{(2x+1)(x-2)} + \frac{(2x+3)(2x+1)}{(x-2)(2x+1)}$

4. Expand each numerator:

   • $(3x-1)(x-2) = 3x^2 - 6x - x + 2 = 3x^2 - 7x + 2$
   • $(2x+3)(2x+1) = 4x^2 + 2x + 6x + 3 = 4x^2 + 8x + 3$

5. Add the numerators over the LCD: $\frac{3x^2 - 7x + 2 + 4x^2 + 8x + 3}{(2x+1)(x-2)} = \frac{7x^2 + x + 5}{(2x+1)(x-2)}$

$(f+g)(x) = \frac{7x^2 + x + 5}{(2x+1)(x-2)}$

Domain restrictions

The denominator of a rational expression can never equal zero, since division by zero is undefined. Whenever you work with rational expressions, identify the values of the variable that would make any denominator zero and exclude them.

For the example above, $2x + 1 = 0$ gives $x = -\frac{1}{2}$, and $x - 2 = 0$ gives $x = 2$. So the domain of $(f+g)(x)$ is all real numbers except $x = -\frac{1}{2}$ and $x = 2$.

These restrictions apply throughout the entire problem, including to the original expressions and any intermediate steps. Even if a factor cancels during simplification, the restriction it created still holds.