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Rational Expressions: (lesson 3 of 3)

## Adding and Subtracting Rational Expressions

### Rational Expressions with the Same Denominator

To add/subtract rational expressions with the same denominator

1. Add/subtract the numerators. Write this sum/difference as the numerator over the common denominator.

2. Reduce to lowest terms.

Example 1

Simplify the following:

$\frac{{4x}}{{5y}} + \frac{{6x}}{{5y}}$

Solution

These fractions already have a common denominator

1: Write this sum as the numerator over the common denominator:

$\frac{{4x}}{{5y}} + \frac{{6x}}{{5y}} = \frac{{4x + 6x}}{{5y}}$

2: Reduce to lowest terms:

$\frac{{4x}}{{5y}} + \frac{{6x}}{{5y}} = \frac{{4x + 6x}}{{5y}} = \frac{{10x}}{{5y}} = \frac{{2x}}{y}$

Example 2

Simplify the following:

$\frac{{4x - 1}}{{x + 4}} - \frac{{2x - 9}}{{x + 4}}$

Solution

Again, these already have a common denominator

1: Write this sum as the numerator over the common denominator:

$\frac{{4x - 1}}{{x + 4}} - \frac{{2x - 9}}{{x + 4}} = \frac{{(4x - 1) - (2x - 9)}}{{x + 4}}$

2: Reduce to lowest terms:

$$\frac{{4x - 1}}{{x + 4}} - \frac{{2x - 9}}{{x + 4}} = \frac{{(4x - 1) - (2x - 9)}}{{x + 4}} =$$ $$= \frac{{4x - 1 - 2x + 9}}{{x + 4}} =$$ $$= \frac{{2x + 8}}{{x + 4}} =$$ $$= \frac{{2\cancel{{(x + 4)}}}}{{\cancel{{x + 4}}}} = 2$$

Exercise 1: Simplify the following expression

Level 1

 $$\color{blue}{\frac{{2y}}{x} - \frac{{y - 1}}{x} = }$$ $\frac{{y + 1}}{x}$ $\frac{{3y - 1}}{x}$ $\frac{{3y + 1}}{x}$ $\frac{{y - 1}}{x}$

Level 2

 $$\color{blue}{\frac{{a + b}}{{2a - b}} - \frac{{2a - 3b}}{{2a - 5}} = }$$ $\frac{{a - 4b}}{{2a - b}}$ $\frac{{- a + 4b}}{{2a - b}}$ $\frac{{a + 4b}}{{2a - b}}$ $\frac{{2a + 4b}}{{2a - b}}$

### Adding or Subtracting Rational Expressions with Different Denominators

1. Factor each denominator completely.

2. Build the LCD of the denominators.

3. Rewrite each rational expression with the LCD as the denominator.

Example 3:

Simplify the following:

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

Solution 3:

1: Factor each denominator completely.

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

2: Build the LCD of the denominators.

$LCD = 2(x - 1)(x - 2)$

3: Rewrite each rational expression with the LCD as the denominator.

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

$$\frac{{5x - 1}}{{{x^2} - 3x + 2}} + \frac{3}{{2x - 4}} = \frac{{5x - 1}}{{(x - 1)(x - 2)}} + \frac{3}{{2(x - 2)}} =$$ $$= \frac{{2(5x - 1)}}{{2(x - 1)(x - 2)}} + \frac{{3(x - 1)}}{{2(x - 1)(x - 2)}} =$$ $$= \frac{{2(5x - 1) + 3(x - 1)}}{{2(x - 1)(x - 2)}} =$$ $$= \frac{{13x - 5}}{{2(x - 1)(x - 2)}}$$

Example 4:

Simplify the following:

$\frac{{5x + 1}}{{{x^2} - 2x - 3}} - \frac{{5x - 3}}{{{x^2} - x - 6}}$

Solution 4:

1: Factor each denominator completely.

$\frac{{5x + 1}}{{{x^2} - 2x - 3}} - \frac{{5x - 3}}{{{x^2} - x - 6}} = \frac{{5x + 1}}{{(x - 3)(x + 1)}} - \frac{{5x - 3}}{{(x - 3)(x + 2)}}$

2: Build the LCD of the denominators.

$LCD = (x - 3)(x + 1)(x + 2)$

3: Rewrite each rational expression with the LCD as the denominator.

$$\frac{{5x + 1}}{{{x^2} - 2x - 3}} - \frac{{5x - 3}}{{{x^2} - x - 6}} = \frac{{5x + 1}}{{(x - 3)(x + 1)}} - \frac{{5x - 3}}{{(x - 3)(x + 2)}} =$$ $$= \frac{{(5x + 1)(x + 2)}}{{(x - 3)(x + 1)(x + 2)}} - \frac{{(5x - 3)(x + 1)}}{{(x - 3)(x + 1)(x + 2)}}$$

4: Subtract the numerators.

$$\frac{{5x + 1}}{{{x^2} - 2x - 3}} - \frac{{5x - 3}}{{{x^2} - x - 6}} = \frac{{5x + 1}}{{(x - 3)(x + 1)}} - \frac{{5x - 3}}{{(x - 3)(x + 2)}} =$$ $$= \frac{{(5x + 1)(x + 2)}}{{(x - 3)(x + 1)(x + 2)}} - \frac{{(5x - 3)(x + 1)}}{{(x - 3)(x + 1)(x + 2)}} =$$ $$= \frac{{(5x + 1)(x + 2) - (5x - 3)(x + 1)}}{{(x - 3)(x + 1)(x + 2)}} =$$ $$= \frac{{(5{x^2} + 10x + x + 2) - (5{x^2} + 5x - 3x - 3)}}{{(x - 3)(x + 1)(x + 2)}} =$$ $$= \frac{{5{x^2} + 10x + x + 2 - 5{x^2} - 5x + 3x + 3)}}{{(x - 3)(x + 1)(x + 2)}} =$$ $$= \frac{{9x + 5}}{{(x - 3)(x + 1)(x + 2)}}$$

Exercise 2: Simplify the following expression

Level 1

 $$\color{blue}{\frac{1}{{{x^2} - x}} - \frac{1}{x} = }$$ $\frac{{2 + x}}{{{x^2} - x}}$ $\frac{{2 - x}}{{{x^2} - x}}$ $\frac{{2 - x}}{{{x^2} + x}}$ $\frac{{2 + x}}{{{x^2} + x}}$

Level 2

 $$\color{blue}{\frac{{2x + 3}}{{x + 3}} - \frac{{4{x^2}}}{{2{x^2} + 5x - 3}} = }$$ $\frac{{4x + 3}}{{2{x^2} + 5x - 3}}$ $\frac{{4x - 3}}{{2{x^2} + 5x - 3}}$ $\frac{{3x + 4}}{{2{x^2} + 5x - 3}}$ $\frac{{3x - 4}}{{2{x^2} + 5x - 3}}$
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