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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

In the previous two concepts we have eased our way up to this one. 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.

Add or Subtract the Rational Expressions

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.

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*}

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 what is 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*}

Answers for Review Problems

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

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