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Section 8.3 Lowest Common Denominator

Subsection Building Fractions

Let's begin with a quick review of the steps for adding fractions in arithmetic. Recall that we can only simplify a sum of like fractions. To add or subtract fractions with unlike denominators, we must convert the fractions to equivalent forms with the same denominator.

As an example, we'll compute the sum

\begin{equation*} \dfrac{2}{3} + \dfrac{3}{4} \end{equation*}

  1. We first find the lowest common denominator for the two fractions. The LCD is the smallest number that is a multiple of both 3 and 4, so in this example the LCD is 12.

  2. Next, we build each fraction to an equivalent fraction with denominator 12. First, we multiply \(\dfrac{2}{3}\) by \(\dfrac{4}{4}\text{.}\) Because \(\dfrac{4}{4}\) is equal to 1, we have not changed the value of the fraction. 4 is called the building factor for \(\dfrac{2}{3}\text{.}\) Similarly, we multiply \(\dfrac{3}{4}\) by \(\dfrac{3}{3}\text{.}\)

    \begin{align*} \dfrac{2}{3} \amp = \dfrac{2 \cdot \blert{4}}{3 \cdot \blert{4}} = \dfrac{8}{12} \amp \amp \blert{\text{The building factor is 4.}}\\ \dfrac{3}{4} \amp = \dfrac{3 \cdot \blert{3}}{4 \cdot \blert{3}} = \dfrac{9}{12} \amp \amp \blert{\text{The building factor is 3.}} \end{align*}

  3. The new fractions, \(\dfrac{8}{12}\) and \(\dfrac{9}{12}\text{,}\) are like fractions, so we add them by combining their numerators.

    \begin{equation*} \dfrac{2}{3} + \dfrac{3}{4} = \dfrac{8}{12} + \dfrac{9}{12} = \dfrac{8+9}{12} = \dfrac{17}{12} \end{equation*}
Look Closer.

Building is an application of the fundamental principle of fractions,

\begin{equation*} \blert{ \dfrac{a}{b} = \dfrac{a \cdot c}{b \cdot c}},~~~~~\text{if} ~~~c \not= 0 \end{equation*}

In words, this says that we can multiply the numerator and denominator of a fraction by the same (nonzero) quantity without changing the value of the fraction.

Reading Questions Reading Questions

1.

What does it mean to build a fraction?

Answer.

To convert the fraction to an equivalent form with a larger denominator.

We can visualize the building process with diagrams. Here are pictures of the original fractions in our example, \(\dfrac{2}{3}\) and \(\dfrac{3}{4}\text{.}\)

rectangle for 2/3
rectangle for 3/4

Thirds and fourths are not like fractions -- they are not pieces of the same size. But if we can break up one-third and one-fourth into smaller pieces that are the same size, then we can combine all the pieces. This is what the LCD is for. We can break up both thirds and fourths into twelfths, as shown below.

rectangle for 8/12
rectangle for 9/12

We can see from the figure that \(\dfrac{2}{3} = \dfrac{8}{12}\) and \(\dfrac{3}{4} = \dfrac{9}{12}\text{.}\) Once we have written the fractions with the same denominator, they are like fractions and we can add them. This is what building fractions is all about.

Subsection Adding and Subtracting Unlike Fractions

We add or subtract algebraic fractions in the same way.

To Add or Subtract Algebraic Fractions.

  1. Find the lowest common denominator (LCD) for the fractions.

  2. Build each fraction to an equivalent one with the LCD as denominator.

  3. Add or subtract the resulting like fractions: Add or subtract their numerators, and keep the same denominator.

  4. Reduce the sum or difference if necessary.

Example 8.22.

Subtract: \(~~\dfrac{x+2}{6} - \dfrac{x-1}{15}\)

Solution.

Step 1 We find the LCD: The smallest common multiple of 6 and 15 is 30. Step 2 We build each fraction to an equivalent one with denominator 30. The building factor for the first fraction is 5, and for the second fraction the building factor is 2.

\begin{gather*} \dfrac{x+2}{6} = \dfrac{(x+2) \cdot \blert{5}}{6 \cdot \blert{5}} = \dfrac{5x+10}{30}\\ \dfrac{x-1}{15} = \dfrac{(x-1) \cdot \blert{2}}{15 \cdot \blert{2}} = \dfrac{2x-2}{30} \end{gather*}

Step 3 We subtract the resulting like fractions to obtain

\begin{align*} \dfrac{x+2}{6} - \dfrac{x-1}{15} =~ \amp \dfrac{5x+10}{30} - \dfrac{2x-2}{30}\\ \amp \blert{\text{Combine the numerators; keep the same denominator.}}\\ =~\amp \dfrac{(5x+10)-(2x-2)}{30}~~~~~~\blert{\text{Simplify the numerator.}}\\ = ~\amp \dfrac{5x+10-2x+2}{30} = \dfrac{3x+12}{30} \end{align*}

Step 4 We reduce the fraction.

\begin{equation*} \dfrac{3x+12}{30} = \dfrac{\cancel{3}(x+4)}{\cancel{3} \cdot 10} = \dfrac{x+4}{10} \end{equation*}

Reading Questions Reading Questions

2.

What is the first step in adding or subtracting unlike fractions?

Answer.

Find the lowest common denominator (LCD) for the fractions.

Caution 8.23.

In Example 8.22, be careful to subtract each term of the numerator in the second fraction.

\begin{equation*} \dfrac{5x+10}{30} - \dfrac{2x-2}{30} = \dfrac{3x\alert{+12}}{30} \end{equation*}

because \(~~(5x+10) - (2x-2) = 5x+10 -2x \alert{+2} = 3x+12\)

Subsection Finding the Lowest Common Denominator

How can we find an LCD for fractions whose denominators are algebraic expressions?

The lowest common denominator for two or more algebraic fractions is the simplest algebraic expression that is a multiple of each denominator.

The easiest case occurs when neither denominator can be factored. In that case, their LCD is just the product of the two expressions. For example, the LCD for the fractions \(\dfrac{6}{x}\) and \(\dfrac{x}{x-2}\) is \(x(x-2)\text{.}\)

Example 8.24.

Add: \(~~\dfrac{6}{x} + \dfrac{x}{x-2}\)

Solution.

Step 1 Find the LCD: The LCD is \(x(x-2)\text{.}\)

Step 2 Build: We build each fraction to an equivalent one with denominator \(x(x-2)\text{.}\) To find the building factors, we compare the given denominator with the desired LCD. The building factor for each fraction is the missing factor.

Thus, the building factor for \(\dfrac{6}{x}\) is \(x-2,\) and the building factor for \(\dfrac{x}{x-2}\) is \(x\text{.}\)

\begin{gather*} \dfrac{6}{x} = \dfrac{6 \blert{(x-2)}}{x\blert{(x-2)}} = \dfrac{6x-12}{x(x-2)}\\ \dfrac{x}{x-2} = \dfrac{x\blert{(x)}}{(x-2)\blert{(x)}} = \dfrac{x^2}{x(x-2)} \end{gather*}

Step 3 Combine: We combine the resulting like fractions.

\begin{align*} \dfrac{6}{x} + \dfrac{x}{x-2} =~ \amp \dfrac{6x-12}{x(x-2)} + \dfrac{x^2}{x(x-2)}\\ \amp \blert{\text{Combine the numerators; keep the same denominator.}}\\ =~ \amp \dfrac{x^2+6x+12}{x(x-2)} \end{align*}

Step 4 Reduce: The numerator of this fraction cannot be factored, so the sum cannot be reduced.

Reading Questions Reading Questions

3.

When can we multiply two denominators together to find their LCD?

Answer.

When neither denominator factors.

4.

How do we find the building factor for each fraction?

Answer.

Compare the given denominator with the desired LCD.

However, if the denominators contain any common factors, the simplest common denominator is not just the product of the two denominators. In this case we must first factor each denominator to find their LCD.

Example 8.25.

Find the LCD for \(~~\dfrac{5}{12a^3} + \dfrac{7}{18a}\)

Solution.

The LCD is not \(12a^3(18a) = 216a^4\text{.}\) It is true that 216 is a multiple of both 12 and 18, but it is not the smallest one! We can find a smaller common denominator by factoring each denominator.

\begin{gather*} 12a^3 = \alert{2} \cdot \alert{2} \cdot 3 \cdot \alert{a} \cdot \alert{a} \cdot \alert{a}\\ 18a = 2 \cdot \alert{3} \cdot \alert{3} \cdot a \end{gather*}

To find an expression that both \(12a^3\) and \(18a\) divide into evenly, we need only enough factors to "cover" each of them separately. In this case two 2's, two 3's, and three \(a\)'s are sufficient, so the LCD is

\begin{equation*} \text{LCD} = \alert{2 \cdot 2 \cdot 3 \cdot 3 \cdot a \cdot a \cdot a} = 36a^3 \end{equation*}

You can check that \(36a^3\) is a multiple of both \(12a^3\) and \(18a\text{.}\)

Look Closer.

To find the LCD in Example 8.25, notice that for each factor, we chose the denominator with the most copies of that factor, and included them in the LCD. For example, \(12a^3\) has only one factor of 3, but \(18a\) has two factors of 3, so we used two factors of 3 in the LCD.

To Find the LCD.

  1. Factor each denominator completely, and arrange the factors in order.

  2. For each factor,

    1. Which denominator has the most copies of that factor? Circle them. (If there is a tie, either denominator will do.)

    2. Include all the circled factors in the LCD.

  3. Multiply together the factors of the LCD.

Reading Questions Reading Questions

5.

We would like to add two unlike fractions. If each denominator contains one factor of \(x\text{,}\) how many factors of do we include in the LCD?

Answer.

One

Now let us see how to incorporate this new skill into the steps for adding fractions.

Example 8.26.

Add: \(~~\dfrac{x-4}{x^2-2x} + \dfrac{4}{x^2-4}\)

Solution.

Step 1 To find the LCD, we first factor each denominator completely.

\begin{gather*} x^2-2x = x(x-2)\\ x^2-4 = (x-2)(x+2) \end{gather*}

The LCD is \(x(x-2)(x+2)\text{.}\)

Step 2 We build each fraction to an equivalent one with the LCD as its denominator.

\begin{align*} \dfrac{x-4}{x(x-2)} \amp = \dfrac{(x-4)\blert{(x+2)}}{x(x-2)\blert{(x+2)}} = \dfrac{x^2-2x-8}{x(x-2)(x+2)}\\ \dfrac{4}{(x-2)(x+2)} \amp = \dfrac{4\blert{x}}{(x-2)(x+2)\blert{x}} \end{align*}

Step 3 Because the fractions are now like fractions, we can combine their numerators.

\begin{align*} \dfrac{x-4}{x^2-2x} + \dfrac{4}{x^2-4} \amp = \dfrac{x^2-2x-8}{x(x-2)(x+2)} + \dfrac{4x}{x(x-2)(x+2)}\\ \amp = \dfrac{x^2+2x-8}{x(x-2)(x+2)} \end{align*}

Step 4 We reduce the fraction by first factoring numerator and denominator, then dividing out any common factors.

\begin{equation*} \dfrac{x^2+2x-8}{x(x-2)(x+2)} = \dfrac{(x+4)\cancel{\blert{(x-2)}}}{x\cancel{\blert{(x-2)}}(x+2)} = \dfrac{x+4}{x(x+2)} \end{equation*}

Reading Questions Reading Questions

6.

State the four steps for adding or subtracting fractions.

Answer.

Find the LCD, build, combine, reduce.

Subsection Skills Warm-Up

Exercises Exercises

Exercise Group.

Simplify.

1.
\(2x+5-(3x+2)\)
2.
\(4(x-3)-3(2x-5)\)
3.
\(2(3x-x^2)+x(2x-4)\)
4.
\((x-3)2x+(x+2)4\)
5.
\(x^2(x-1)-2x(x+2)\)
6.
\(3x(x-4)-2(x^2-16)-4x\)

Subsubsection Answers to Skills Warm-Up

  1. \(\displaystyle -x+3\)

  2. \(\displaystyle -2x+3\)

  3. \(\displaystyle 2x\)

  4. \(\displaystyle 2x^2-2x+8\)

  5. \(\displaystyle x^3-3x^2-4x\)

  6. \(\displaystyle x^2+8x+32\)

Subsection

Subsubsection Activity 1: Unlike Fractions

Exercises Exercises
1.

Add: \(~~~\dfrac{5x}{6}+\dfrac{3-2x}{4}\)

\(\blert{\text{Step 1 Find the LCD}}\)

\(\blert{\text{Step 2 Build each fraction.}}\)

\(\blert{\text{Step 3 Add like fractions.}}\)

\(\blert{\text{Step 4 Reduce if possible.}}\)

2.

Follow the steps below to subtract \(~~~\dfrac{2}{x+2}-\dfrac{3}{x-2}\)

\(\blert{\text{Step 1 Find the LCD}}\)

\(\blert{\text{Step 2 Build each fraction.}}\)

\(\blert{\text{Step 3 Subtract like fractions.}}\)

\(\blert{\text{Step 4 Reduce if possible.}}\)

Subsubsection Activity 2: Finding the LCD

Exercises Exercises
1.

Find the LCD for \(~\dfrac{3}{4x^2}+\dfrac{5}{6xy}\)

\(\blert{\text{Factor each denominator.}}\)

\(\blert{\text{Choose the correct factors.}}\)

\(\blert{\text{LCD =}}\)

2.

Subtract \(~\dfrac{b}{3b+9}-\dfrac{b-1}{2b^2+6b}\)

\(\blert{\text{Step 1 Find the LCD}}\)

\(~~~~~~~\blert{\text{Factor each denominator.}}\)

\(~~~~~~~\blert{\text{Choose the correct factors.}}\)

\(~~~~~~~\blert{\text{LCD =}}\)

\(\blert{\text{Step 2 Build each fraction.}}\)

\(\blert{\text{Step 3 Subtract like fractions.}}\)

\(\blert{\text{Step 4 Reduce if possible.}}\)

Subsubsection Wrap-Up

Objectives.

In this Lesson we practiced the following skills:

  • Finding an LCD

  • Adding and subtracting unlike fractions

Questions.

  1. In Activity 1, Problem 2, what was the LCD?

  2. In Activity 1, Problem 2, what was the building factor for the first fraction?

  3. In Activity 2, Problem 2, explain how to find the building factor for each fraction.

Subsection Homework Preview

Exercises Exercises

1.

Add: \(~~~\dfrac{x+1}{x} + \dfrac{2x+1}{x+2}\)

2.

Find the LCD for \(~~~\dfrac{1}{3a^2-6a}\) and \(\dfrac{1}{3a^2+3a-6}\)

3.

Subtract: \(~~~\dfrac{x+1}{2x-4} + \dfrac{2x}{x^2-4}\)

Subsubsection Answers to Homework Preview

  1. \(\displaystyle \dfrac{3x^2+4x+2}{x(x+2)}\)

  2. \(\displaystyle 3a(a-1)(a+2)(a-2)\)

  3. \(\displaystyle \dfrac{x^2-x+2}{2(x^2-4)}\)

Exercises Homework 8.3

Exercise Group.

For Problems 1–12, find the LCD, then add or subtract.

1.

\(\dfrac{3}{x} - \dfrac{4}{y}\)

2.

\(\dfrac{-3}{a} - \dfrac{2a}{b}\)

3.

\(1 - \dfrac{3}{a}\)

4.

\(\dfrac{1}{x} + \dfrac{x}{y} + \dfrac{x}{z}\)

5.

\(\dfrac{5}{xy} - \dfrac{x}{y}\)

6.

\(\dfrac{2}{z} - \dfrac{1}{2z}\)

7.

\(\dfrac{u-4}{3} + \dfrac{6}{v}\)

8.

\(\dfrac{3}{x} - \dfrac{2}{x+1}\)

9.

\(\dfrac{3}{n+3} + \dfrac{4n}{n-3}\)

10.

\(h - \dfrac{3}{h+2}\)

11.

\(\dfrac{v+1}{v} + \dfrac{1}{v-1}\)

12.

\(\dfrac{2}{x} + \dfrac{x}{x-2}-2\)

Exercise Group.

For Problems 13–24,

  1. Find the lowest common denominator for the fractions.

  2. Add or subtract.

13.

\(\dfrac{-2}{5n} + \dfrac{q}{4n}\)

14.

\(\dfrac{-6}{s} + \dfrac{3}{s^2}\)

15.

\(\dfrac{z}{4x^2} - \dfrac{1}{6xz}\)

16.

\(\dfrac{5}{2x} + \dfrac{3}{4x^2}\)

17.

\(\dfrac{2z-3}{8z} + \dfrac{z-2}{6z}\)

18.

\(\dfrac{3}{2a-b} + \dfrac{1}{8a-4b}\)

19.

\(\dfrac{1}{x} + \dfrac{1}{d-x}\)

20.

\(a + \dfrac{N-a^2}{2a}\)

21.

\(\dfrac{1}{2} \cdot \dfrac{a}{t} - \dfrac{m}{a}\)

22.

\(\dfrac{q}{4 \pi r} + \dfrac{qa}{2 \pi r^2}\)

23.

\(\dfrac{L}{c-V} + \dfrac{L}{c+V}\)

24.

\(\dfrac{q}{r-a} - \dfrac{2q}{r} + \dfrac{q}{r+a}\)

Exercise Group.

For Problems 25–30,

  1. Find the lowest common denominator for the fractions.

  2. Add or subtract.

25.

\(\dfrac{2x+3}{x-1} + \dfrac{2x-5}{1-x}\)

26.

\(\dfrac{5}{2p-4} - \dfrac{2}{6-3p}\)

27.

\(\dfrac{-2}{m^2+3m} + \dfrac{1}{m^2-9}\)

28.

\(\dfrac{4}{k^2-3k} + \dfrac{1}{k^2+k}\)

29.

\(\dfrac{x-1}{x^2+3x} + \dfrac{x}{x^2+6x+9}\)

30.

\(\dfrac{n+2}{2n^2-3n} - \dfrac{7}{6n-9}\)

Exercise Group.

For Problems 31–33, combine the fractions.

31.

  1. \(\displaystyle \dfrac{2b}{a} \cdot \dfrac{4}{b}\)

  2. \(\displaystyle \dfrac{2b}{a} - \dfrac{4}{b}\)

  3. \(\displaystyle \dfrac{2b}{a} \div \dfrac{4}{b}\)

  4. \(\displaystyle \dfrac{2b}{a} + \dfrac{4}{b}\)

32.

  1. \(\displaystyle \dfrac{2y}{3x^2} \div \dfrac{6}{xy}\)

  2. \(\displaystyle \dfrac{2y}{3x^2} + \dfrac{6}{xy}\)

  3. \(\displaystyle \dfrac{2y}{3x^2} \cdot \dfrac{6}{xy}\)

  4. \(\displaystyle \dfrac{2y}{3x^2} - \dfrac{6}{xy}\)

33.

  1. \(\displaystyle \dfrac{3}{2+a} - \dfrac{2}{a^2-4}\)

  2. \(\displaystyle \dfrac{3}{2+a} \cdot \dfrac{2}{a^2-4}\)

  3. \(\displaystyle \dfrac{3}{2+a} + \dfrac{2}{a^2-4}\)

  4. \(\displaystyle \dfrac{3}{2+a} \div \dfrac{2}{a^2-4}\)

Exercise Group.

For Problems 34–36, write algebraic fractions in simplest form.

34.

The dimensions of a rectangular rug are \(\dfrac{12}{x}\) feet and \(\dfrac{12}{x-2}\) feet.

  1. Write an expression for the area of the rug.

  2. Write an expression for the perimeter of the rug.

35.

Colonial Airline has a commuter flight between Richmond and Washington, a distance of 100 miles. The plane flies at \(x\) miles per hour in still air. Today there is a steady wind from the north at 10 miles per hour.

  1. How long will the flight from Richmond to Washington take?

  2. How long will the flight from Washington to Richmond take?

  3. How long will a round trip take?

  4. Evaluate your fractions in parts (a)-(c) for \(x=150\text{.}\)

36.

Francine's cocker spaniel eats a large bag of dog food in \(d\) days, and Delbert's sheep dog takes 5 days less to eat the same size bag.

  1. What fraction of a bag of dog food does Francine's cocker spaniel eat in one day?

  2. What fraction of a bag of dog food does Delbert's sheep dog eat in one day?

  3. If Delbert and Francine get married, what fraction of a bag of dog food will their dogs eat in one day?

  4. If \(d=25\text{,}\) how soon will Delbert and Francine have to buy more dog food?