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Section 8.3 Operations on Algebraic Fractions

Subsection 1. Use improper fractions

An improper fraction is one in which the numerator is larger than the denominator.

Subsubsection Examples

Example 8.31.

Write an improper fraction for the sum \(~2 + \dfrac{3}{8}\)

Solution

We write the whole number with the same denominator as the fraction. Thus,

\begin{equation*} \blert{2} + \dfrac{3}{8} = \blert{\dfrac{16}{8}}+\dfrac{3}{8} = \dfrac{16+3}{8} = \dfrac{19}{8} \end{equation*}
Example 8.32.

Write an improper fraction for the difference \(~3 - \dfrac{1}{5}\)

Solution

We write the whole number with the same denominator as the fraction. Thus,

\begin{equation*} \blert{3} - \dfrac{1}{5} = \blert{\dfrac{15}{5}}-\dfrac{1}{5} = \dfrac{15-1}{5} = \dfrac{14}{5} \end{equation*}

Subsubsection Exercises

Write an improper fraction for each sum or difference.

  1. \(\displaystyle 1 + \dfrac{3}{4}\)
  2. \(\displaystyle 2 - \dfrac{3}{10}\)
Answer
  1. \(\displaystyle \dfrac{7}{4}\)
  2. \(\displaystyle \dfrac{17}{10}\)

Write an improper fraction for each sum or difference.

  1. \(\displaystyle 1 - \dfrac{23}{100}\)
  2. \(\displaystyle 3 + \dfrac{3}{25}\)
Answer
  1. \(\displaystyle \dfrac{77}{100}\)
  2. \(\displaystyle \dfrac{78}{25}\)

Subsection 2. Find an LCD

The first step in adding unlike fractions is to find the lowest common denominator, or LCD.

Subsubsection Examples

Example 8.35.

Find the LCD for the fractions \(~\dfrac{2}{5} + \dfrac{3}{10}\)

Solution

We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.

\begin{align*} 15 \amp = \blert{3} ~~~\cdot~~~ 5\\ 10 \amp = ~~~~~~~~~~~~\blert{5}~~~\cdot~~~\blert{2} \end{align*}

The LCD is \(3 \cdot 5\cdot 2\text{,}\) or 30.

Example 8.36.

Find the LCD for the fractions \(~\dfrac{5}{6a} - \dfrac{2}{9a^2}\)

Solution

We factor each denominator and line up any common factors vertically. We use one factor from each column in the LCD.

\begin{align*} 6a \amp = \blert{2}~~~\cdot~~~ 3~~~\cdot~~~ a\\ 9a^2 \amp = ~~~~~~~~~~~~\blert{3}~~~\cdot~~~\blert{3}~~~\cdot~~~ \blert{a}~~~\cdot~~~ \blert{a} \end{align*}

The LCd is \(2 \cdot 3 \cdot 3 \cdot a \cdot a\text{,}\) or \(18a^2\text{.}\)

Subsubsection Exercises

Find the LCD for the fractions

  1. \(\displaystyle \dfrac{7}{8} - \dfrac{1}{6}\)
  2. \(\displaystyle \dfrac{52}{75} + \dfrac{13}{24}\)
Answer
  1. \(\displaystyle 24\)
  2. \(\displaystyle 600\)

Find the LCD for the fractions

  1. \(\displaystyle \dfrac{3}{4a^2b^2} + \dfrac{7}{10ab^3}\)
  2. \(\displaystyle \dfrac{3}{8t^4} - \dfrac{3}{5t^2}\)
Answer
  1. \(\displaystyle 20a^2b^3\)
  2. \(\displaystyle 40t^4\)

Subsection 3. Build fractions

Before we can add unlike fractions, we must build each fraction to an equivalent one with the LCD as denominator.

Subsubsection Examples

Example 8.39.

Write an equivalent fraction with the new denominator:

\begin{equation*} ~\dfrac{3}{s} = \dfrac{?}{2s^2} \end{equation*}
Solution

We first find the building factor for the fraction: what must we mulitply the old denominator by to get the new denominator? We factor the new denominator to see what factors are missing.

The new denominator is \(2 \cdot s \cdot 3\text{,}\) so we need to multiply the old denominator by \(2s\text{.}\) This is the building factor. We multiply top and bottom of the old fraction by the building factor:

\begin{equation*} \dfrac{3}{s} \cdot \blert{\dfrac{2s}{2s}} = \dfrac{6}{2s^2} \end{equation*}
Example 8.40.

Write an equivalent fraction with the new denominator:

\begin{equation*} ~\dfrac{2}{t+1} = \dfrac{?}{t^2+t} \end{equation*}
Solution

The new denominator factors as \(t(t+1)\text{,}\) so the building factor is \(t\text{.}\) We multiply top and bottom of the old fraction by \(t\) to obtain:

\begin{equation*} \dfrac{2}{t+1} \cdot \blert{\dfrac{t}{t}} = \dfrac{2t}{t^2+t} \end{equation*}

Subsubsection Exercises

Write an equivalent fraction with the new denominator.

  1. \(\displaystyle \dfrac{1}{n} = \dfrac{?}{n^2-n}\)
  2. \(\displaystyle \dfrac{4}{a-1} = \dfrac{?}{a^2-1}\)
Answer
  1. \(\displaystyle \dfrac{n-1}{n^2-n}\)
  2. \(\displaystyle \dfrac{4a+4}{a^2-1}\)

Write an equivalent fraction with the new denominator.

  1. \(\displaystyle \dfrac{b}{b-2} = \dfrac{?}{(b+3)(b-2)}\)
  2. \(\displaystyle \dfrac{x-1}{x^2+2x} = \dfrac{?}{(x^2-x)(x+2)}\)
Answer
  1. \(\displaystyle \dfrac{b^2+3b}{(b+3)(b-2)}\)
  2. \(\displaystyle \dfrac{x^2-2x+1}{(x^2-x)(x+2)}\)