# Sums and Differences of Rational Expressions with Unequal Denominators

## Combine fractions including unlike denominators with variables

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Adding and Subtracting Rational Expressions with Unlike Denominators

One part of a line segment measures \begin{align*}\frac{3}{x-2}\end{align*}. The other part of the segment measures \begin{align*}\frac{2}{x+1}\end{align*}. What is the total length of the line segment?

### Adding and Subtracting Rational Expressions

Now, we will add two rational expressions where were you will have to multiply both fractions by a constant in order to get the Lowest Common Denominator or LCD. Recall how to add fractions where the denominators are not the same.

\begin{align*}\frac{4}{15} + \frac{5}{18}\end{align*}

Find the LCD. \begin{align*}15=3 \cdot 5\end{align*} and \begin{align*}18=3 \cdot 6\end{align*}. So, they have a common factor of 3. Anytime two denominators have a common factor, it only needs to be listed once in the LCD. The LCD is therefore \begin{align*}3 \cdot 5 \cdot 6=90\end{align*}.

\begin{align*}\frac{4}{15} + \frac{5}{18} &= \frac{4}{3 \cdot 5} + \frac{5}{3 \cdot 6} \\ &= {\color{red}\frac{6}{6}} \cdot \frac{4}{3 \cdot 5} + \frac{5}{3 \cdot 6} \cdot {\color{blue}\frac{5}{5}} \\ &= \frac{24}{90} + \frac{25}{90}\\ &= \frac{49}{90}\end{align*}

We multiplied the first fraction by \begin{align*}\frac{6}{6}\end{align*} to obtain 90 in the denominator. Recall that a number over itself is \begin{align*}6 \div 6=1\end{align*}. Therefore, we haven’t changed the value of the fraction. We multiplied the second fraction by \begin{align*}\frac{5}{5}\end{align*}. We will now apply this idea to rational expressions.

Let's add or subtract the following rational expressions.

1. Add \begin{align*}\frac{x+5}{x^2-3x} + \frac{3}{x^2+2x}\end{align*}.

First factor each denominator to find the LCD. The first denominator, factored, is \begin{align*}x^2-3x=x(x-3)\end{align*}. The second denominator is \begin{align*}x^2+2x=x(x+2)\end{align*}. Both denominators have and \begin{align*}x\end{align*}, so we only need to list it once. The LCD is \begin{align*}x(x-3)(x+2)\end{align*}.

\begin{align*}\frac{x+5}{x^2-3x} + \frac{3}{x^2+2x} = \frac{x+5}{x(x-3)} + \frac{3}{x(x+2)}\end{align*}

Looking at the two denominators factored, we see that the first fraction needs to be multiplied by \begin{align*}\frac{x+2}{x+2}\end{align*} and the second fraction needs to be multiplied by \begin{align*}\frac{x+3}{x+3}\end{align*}.

\begin{align*}&= {\color{red}\frac{x+2}{x+2}} \cdot \frac{x+5}{x{\color{blue}(x-3)}} + \frac{3}{x{\color{red}(x+2)}} \cdot {\color{blue}\frac{x-3}{x-3}} \\ &= \frac{(x+2)(x+5)+3(x-3)}{x{\color{red}(x+2)} {\color{blue}(x-3)}}\end{align*}

At this point, we need to FOIL the first expression and distribute the 3 to the second. Lastly we need to combine like terms.

\begin{align*}&= \frac{x^2+7x+10+3x-9}{x(x+2)(x-3)} \\ &= \frac{x^2+10x+1}{x(x+2)(x-3)}\end{align*}

The quadratic in the numerator is not factorable, so we are done.

1. Add \begin{align*}\frac{4}{x+6} + \frac{x-2}{3x+1}\end{align*}.

The denominators have no common factors, so the LCD will be \begin{align*}(x+6)(3x+1)\end{align*}.

\begin{align*}\frac{4}{x+6} + \frac{x-2}{3x+1} &= {\color{red}\frac{3x+1}{3x+1}} \cdot \frac{4}{{\color{blue}x+6}} + \frac{x-2}{{\color{red}3x+1}} \cdot {\color{blue}\frac{x+6}{x+6}} \\ &= \frac{4(3x+1)}{{\color{red}(3x+1)}{\color{blue}(x+6)}} + \frac{(x-2)(x+6)}{{\color{red}(3x+1)} {\color{blue}(x+6)}} \\ &= \frac{12x+4+x^2+4x-12}{{\color{red}(3x+1)} {\color{blue}(x+6)}} \\ &= \frac{x^2+16x-8}{(3x+1)(x+6)}\end{align*}

1. Subtract \begin{align*}\frac{x-1}{x^2+5x+4} - \frac{x+2}{2x^2+13x+20}\end{align*}.

To find the LCD, we need to factor the denominators.

\begin{align*}x^2+5x+4 &= {\color{red}(x+1)} {\color{green}(x+4)} \\ 2x^2+13x+20 &= {\color{blue}(2x+5)} {\color{green}(x+4)} \\ LCD &= {\color{red}(x+1)} {\color{blue}(2x+5)} {\color{green}(x+4)}\end{align*}

\begin{align*}\frac{x-1}{x^2+5x+4} - \frac{x+2}{2x^2+13x+20} &= \frac{x-1}{{\color{red}(x+1)} {\color{green}(x+4)}} - \frac{x+2}{{\color{blue}(2x+5)}(x+4)} \\ &= {\color{blue}\frac{2x+5}{2x+5}} \cdot \frac{x-1}{{\color{red}(x+1)}{\color{green}(x+4)}} - \frac{x+2}{{\color{blue}(2x+5)}{\color{green}(x+4)}} \cdot {\color{red}\frac{x+1}{x+1}} \\ &= \frac{(2x+5)(x-1)-(x+2)(x+1)}{{\color{red}(x+1)}{\color{blue}(2x+5)}{\color{green}(x+4)}} \\ &= \frac{2x^2+3x-5-(x^2+3x+2)}{(x+1)(2x+5)(x+4)} \\ &= \frac{2x^2+3x-5-x^2-3x-2}{(x+1)(2x+5)(x+4)} \\ &= \frac{x^2-7}{(x+1)(2x+5)(x+4)}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the total length of the line segment.

We need to add the two parts of the segment to get the whole.

\begin{align*}\frac{3}{x-2} + \frac{2}{x+1}\end{align*}

The denominators have no common factors, so the LCD will be \begin{align*}(x-2)(x+1)\end{align*}.

\begin{align*}\frac{3}{x-2} + \frac{2}{x+1} &= {\color{red}\frac{x+1}{x+1}} \cdot \frac{3}{{\color{blue}x-2}} + \frac{2}{{\color{red}x+1}} \cdot {\color{blue}\frac{x-2}{x-2}} \\ &= \frac{3(x+1)}{{\color{red}(x+1)}{\color{blue}(x-2)}} + \frac{(2)(x-2)}{{\color{red}(x+1)} {\color{blue}(x-2)}} \\ &= \frac{3x+3+2x-4}{{\color{red}(x+1)} {\color{blue}(x-2)}} \\ &= \frac{5x-1}{(x+1)(x-2)}\end{align*}

Therefore, the total length of the line segment is \begin{align*}\frac{5x-1}{(x+1)(x-2)}\end{align*}.

Perform the indicated operation.

#### Example 2

\begin{align*}\frac{3}{x^2-6x} + \frac{5-x}{2x-12}\end{align*}

The LCD is \begin{align*}3x(x-6)\end{align*}.

\begin{align*}\frac{3}{x^2-6x} + \frac{5-x}{2x-12} &= \frac{2}{2} \cdot \frac{3}{x(x-6)} + \frac{5-x}{2(x-6)} \cdot \frac{x}{x} \\ &= \frac{6+x(5-x)}{2x(x-6)} \\ &= \frac{6+5x-x^2}{2x(x-6)} \\ &= \frac{-1(x^2-5x-6)}{2x(x-6)}\end{align*}

We pulled a -1 out of the numerator so we can factor it.

\begin{align*}&= \frac{-1 \bcancel{(x-6)}(x+1)}{2x \bcancel{(x-6)}} \\ &= \frac{-x-1}{2x}\end{align*}

#### Example 3

\begin{align*}\frac{x}{x^2+4x+4} - \frac{x-5}{x^2+5x+6}\end{align*}

The LCD is \begin{align*}(x+2)(x+2)(x+3)\end{align*}.

\begin{align*}\frac{x}{x^2+4x+4} - \frac{x-5}{x^2+5x+6} &= \frac{x+3}{x+3} \cdot \frac{x}{(x+2)(x+2)} - \frac{x-5}{(x+2)(x+3)} \cdot \frac{x+2}{x+2} \\ &= \frac{x(x+3)-(x-5)(x+2)}{(x+2)(x+2)(x+3)} \\ &= \frac{x^2+3x-(x^2-3x-10)}{(x+2)^2(x+3)} \\ &= \frac{x^2+3x-x^2+3x+10}{(x+2)^2(x+3)} \\ &= \frac{6x+10}{(x+2)^2(x+3)} \\ &= \frac{2(3x+5)}{(x+2)^2(x+3)}\end{align*}

#### Example 4

\begin{align*}\frac{2x}{x^2-x-20} + \frac{x^2-9}{x^2-1}\end{align*}

The LCD is \begin{align*}(x-5)(x+4)(x+1)(x-1)\end{align*}.

\begin{align*}\frac{2x}{x^2-x-20} + \frac{x^2-9}{x^2-1} &= \frac{(x+1)(x-1)}{(x+1)(x-1)} \cdot \frac{2x}{(x-5)(x+4)} + \frac{x^2-9}{(x+1)(x-1)} \cdot \frac{(x-5)(x+4)}{(x-5)(x+4)} \\ &= \frac{2x(x+1)(x-1)+(x^2-9)(x-5)(x+4)}{(x-5)(x+4)(x+1)(x-1)} \\ &= \frac{2x^3-2x+x^4-x^3-29x^2+9x+180}{(x-5)(x+4)(x+1)(x-1)} \\ &= \frac{x^4+x^3-29x^2+7x+180}{(x-5)(x+4)(x+1)(x-1)}\end{align*}

### Review

Find the LCD.

1. \begin{align*}3x, \ 7x\end{align*}
2. \begin{align*}x-2, \ 2x-1\end{align*}
3. \begin{align*}x^2 - 9, \ x^2-x-6\end{align*}
4. \begin{align*}4x, \ x^2-6x\end{align*}
5. \begin{align*}x^2-4, \ x^2+4x+4, \ x^2+3x-10\end{align*}

Perform the indicated operation.

1. \begin{align*}\frac{5}{3x} + \frac{x}{2}\end{align*}
2. \begin{align*}\frac{x+1}{x^2} - \frac{5}{7x}\end{align*}
3. \begin{align*}\frac{x-5}{4x} + \frac{3}{x+2}\end{align*}
4. \begin{align*}\frac{5}{2x+6} + \frac{x-2}{x^2+2x-3}\end{align*}
5. \begin{align*}\frac{4x+3}{2x^2+11x-6} - \frac{3x-1}{2x^2-x}\end{align*}
6. \begin{align*}\frac{x}{3x^2+x-2} - \frac{2}{15x-10}\end{align*}
7. \begin{align*}\frac{3x}{x^2-3x-10} + \frac{x+1}{x^2-2x-15} - \frac{2}{x^2+5x+6}\end{align*}
8. \begin{align*}\frac{7+x}{x^2-2x} - \frac{x-6}{3x^2+5x} - \frac{x+4}{3x^2-x-10}\end{align*}
9. \begin{align*}\frac{3x+2}{x^2-1} - \frac{10x-7}{5x^2+5x} + \frac{3}{x-1}\end{align*}
10. \begin{align*}\frac{x+6}{2x-1} + \frac{2x}{3x+2} - \frac{5}{x}\end{align*}

To see the Review answers, open this PDF file and look for section 9.12.

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