Add and Subtract Fractions with Different Denominators

Find the Least Common Denominator

In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators?

Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals $25$ cents and one dime equals $10$ cents, so the sum is $35$ cents. See Figure 1.

A quarter and a dime are shown. Below them, it reads 25 cents plus 10 cents. Below that, it reads 35 cents. — Together, a quarter and a dime are worth $35$ cents, or $\frac{35}{100}$ of a dollar.

Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is $100 .$ Since there are $100$ cents in one dollar, $25$ cents is $\frac{25}{100}$ and $10$ cents is $\frac{10}{100} .$ So we add $\frac{25}{100} + \frac{10}{100}$ to get $\frac{35}{100},$ which is $35$ cents.

You have practiced adding and subtracting fractions with common denominators. Now let’s see what you need to do with fractions that have different denominators.

First, we will use fraction tiles to model finding the common denominator of $\frac{1}{2}$ and $\frac{1}{3} .$

We’ll start with one $\frac{1}{2}$ tile and $\frac{1}{3}$ tile. We want to find a common fraction tile that we can use to match both $\frac{1}{2}$ and $\frac{1}{3}$ exactly.

If we try the $\frac{1}{4}$ pieces, $2$ of them exactly match the $\frac{1}{2}$ piece, but they do not exactly match the $\frac{1}{3}$ piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into two pieces, each labeled 1 fourth. Underneath the second rectangle are two pieces, each labeled 1 fourth. These rectangles together are longer than the rectangle labeled as 1 third.

If we try the $\frac{1}{5}$ pieces, they do not exactly cover the $\frac{1}{2}$ piece or the $\frac{1}{3}$ piece.

If we try the $\frac{1}{6}$ pieces, we see that exactly $3$ of them cover the $\frac{1}{2}$ piece, and exactly $2$ of them cover the $\frac{1}{3}$ piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle are three smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 half rectangle. Below the 1 third rectangle are two smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 third rectangle.

If we were to try the $\frac{1}{12}$ pieces, they would also work.

Even smaller tiles, such as $\frac{1}{24}$ and $\frac{1}{48},$ would also exactly cover the $\frac{1}{2}$ piece and the $\frac{1}{3}$ piece.

The denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of $\frac{1}{2}$ and $\frac{1}{3}$ is $6 .$

Notice that all of the tiles that cover $\frac{1}{2}$ and $\frac{1}{3}$ have something in common: Their denominators are common multiples of $2$ and $3,$ the denominators of $\frac{1}{2}$ and $\frac{1}{3} .$ The least common multiple (LCM) of the denominators is $6,$ and so we say that $6$ is the least common denominator (LCD) of the fractions $\frac{1}{2}$ and $\frac{1}{3} .$

To find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD.

Find the LCD for the fractions $\frac{7}{12}$ and $\frac{5}{18} .$

Solution

Factor each denominator into its primes.
List the primes of 12 and the primes of 18 lining them up in columns when possible.
Bring down the columns.
Multiply the factors. The product is the LCM.	LCM $= 36$
The LCM of 12 and 18 is 36, so the LCD of $\frac{7}{12}$ and $\frac{5}{18}$ is 36.	LCD of $\frac{7}{12}$ and $\frac{5}{18}$ is 36.

To find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM.

Find the least common denominator for the fractions $\frac{8}{15}$ and $\frac{11}{24} .$

Solution

To find the LCD, we find the LCM of the denominators.

Find the LCM of $15$ and $24 .$

The top line shows 15 equals 3 times 5. The next line shows 24 equals 2 times 2 times 2 times 3. The 3s are lined up vertically. The next line shows LCM equals 2 times 2 times 2 times 3 times 5. The last line shows LCM equals 120.

The LCM of $15$ and $24$ is $120 .$ So, the LCD of $\frac{8}{15}$ and $\frac{11}{24}$ is $120 .$

Convert Fractions to Equivalent Fractions with the LCD

Earlier, we used fraction tiles to see that the LCD of $\frac{1}{4}$ when $\frac{1}{6}$ is $12 .$ We saw that three $\frac{1}{12}$ pieces exactly covered $\frac{1}{4}$ and two $\frac{1}{12}$ pieces exactly covered $\frac{1}{6},$ so

\frac{1}{4} = \frac{3}{12} and \frac{1}{6} = \frac{2}{12} .

On the left is a rectangle labeled 1 fourth. Below it is an identical rectangle split vertically into 3 equal pieces, each labeled 1 twelfth. On the right is a rectangle labeled 1 sixth. Below it is an identical rectangle split vertically into 2 equal pieces, each labeled 1 twelfth.

We say that $\frac{1}{4}$ and $\frac{3}{12}$ are equivalent fractions and also that $\frac{1}{6}$ and $\frac{2}{12}$ are equivalent fractions.

We can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The Equivalent Fractions Property is repeated below for reference.

To add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let’s see how to change $\frac{1}{4}$ and $\frac{1}{6}$ to equivalent fractions with denominator $12$ without using models.

Convert $\frac{1}{4}$ and $\frac{1}{6}$ to equivalent fractions with denominator $12,$ their LCD.

Solution

Find the LCD.	The LCD of $\frac{1}{4}$ and $\frac{1}{6}$ is 12.
Find the number to multiply 4 to get 12.
Find the number to multiply 6 to get 12.
Use the Equivalent Fractions Property to convert each fraction to an equivalent fraction with the LCD, multiplying both the numerator and denominator of each fraction by the same number.	$Fractions 1/4 and 1/6 are shown being converted to equivalent fractions by multiplying their numerators and denominators by 3 and 2, respectively, to find a common denominator.$
Simplify the numerators and denominators.	$Two fractions, 3/12 and 2/12, are displayed on a white background.$

We do not reduce the resulting fractions. If we did, we would get back to our original fractions and lose the common denominator.

Convert $\frac{8}{15}$ and $\frac{11}{24}$ to equivalent fractions with denominator $120,$ their LCD.

Solution

	The LCD is 120. We will start at Step 2.
Find the number that must multiply 15 to get 120.
Find the number that must multiply 24 to get 120.
Use the Equivalent Fractions Property.	$Fractions illustrating that multiplying both the numerator and denominator by the same non-zero number (highlighted in red) results in an equivalent fraction.$
Simplify the numerators and denominators.	$Two fractions, 64 over 120 and 55 over 120, are displayed on a white background.$

Add and Subtract Fractions with Different Denominators

Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators.

Add: $\frac{1}{2} + \frac{1}{3} .$

Solution

	$\frac{1}{2} + \frac{1}{3}$
Find the LCD of 2, 3.
Change into equivalent fractions with the LCD 6.
Simplify the numerators and denominators.	$\frac{3}{6} + \frac{2}{6}$
Add.	$\frac{5}{6}$

Remember, always check to see if the answer can be simplified. Since $5$ and $6$ have no common factors, the fraction $\frac{5}{6}$ cannot be reduced.

Subtract: $\frac{1}{2} - (- \frac{1}{4}) .$

Solution

	$\frac{1}{2} - (- \frac{1}{4})$
Find the LCD of 2 and 4.
Rewrite as equivalent fractions using the LCD 4.	$A mathematical expression featuring a fraction (1 times 2, over 2 times 2, with the '2's highlighted in red) subtracted by a negative fraction (-1/4).$
Simplify the first fraction.	$\frac{2}{4} - (- \frac{1}{4})$
Subtract.	$\frac{2 - (−1)}{4}$
Simplify.	$\frac{3}{4}$

One of the fractions already had the least common denominator, so we only had to convert the other fraction.

Add: $\frac{7}{12} + \frac{5}{18} .$

Solution

	$\frac{7}{12} + \frac{5}{18}$
Find the LCD of 12 and 18.
Rewrite as equivalent fractions with the LCD.	$An arithmetic expression showing the addition of two fractions, 7/12 and 5/18, where each fraction is multiplied by a factor (3/3 and 2/2 respectively) to achieve a common denominator for summation.$
Simplify the numerators and denominators.	$\frac{21}{36} + \frac{10}{36}$
Add.	$\frac{31}{36}$

Because $31$ is a prime number, it has no factors in common with $36 .$ The answer is simplified.

When we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The “missing” factors of each denominator are the numbers you need.

The first line says 12 equals 2 times 2 times 3. There is a blank space next to the 3. The next line says 18 equals 2 times 3 times 3. There is a blank space between the 2 and the first 3. There are red lines drawn from the blank spaces. This is labeled as missing factors. There is a horizontal line. Below the line, it says LCD equals 2 times 2 times 3 times 3. Below this, it says LCD equals 36.

The LCD, $36,$ has $2$ factors of $2$ and $2$ factors of $3 .$

Twelve has two factors of $2,$ but only one of $3$ —so it is ‘missing‘ one $3 .$ We multiplied the numerator and denominator of $\frac{7}{12}$ by $3$ to get an equivalent fraction with denominator $36 .$

Eighteen is missing one factor of $2$ —so you multiply the numerator and denominator $\frac{5}{18}$ by $2$ to get an equivalent fraction with denominator $36 .$ We will apply this method as we subtract the fractions in the next example.

Subtract: $\frac{7}{15} - \frac{19}{24} .$

Solution

	$\frac{7}{15} - \frac{19}{24}$
Find the LCD. 15 is 'missing' three factors of 2 24 is 'missing' a factor of 5
Rewrite as equivalent fractions with the LCD.	$A math problem displaying the subtraction of two fractions, (7 * 8) / (15 * 8) and (19 * 5) / (24 * 5). Common factors 8 and 5 in the numerators and denominators are highlighted in red, suggesting simplification.$
Simplify each numerator and denominator.	$\frac{56}{120} - \frac{95}{120}$
Subtract.	$- \frac{39}{120}$
Rewrite showing the common factor of 3.	$- \frac{13 \cdot 3}{40 \cdot 3}$
Remove the common factor to simplify.	$- \frac{13}{40}$

Add: $- \frac{11}{30} + \frac{23}{42} .$

Solution

	$- \frac{11}{30} + \frac{23}{42}$
Find the LCD.
Rewrite as equivalent fractions with the LCD.	$Math expression for adding fractions: -11/30 and 23/42. Each fraction's numerator and denominator are multiplied by a distinct factor (7 or 5, in red) to find a common denominator.$
Simplify each numerator and denominator.	$- \frac{77}{210} + \frac{115}{210}$
Add.	$\frac{38}{210}$
Rewrite showing the common factor of 2.	$\frac{19 \cdot 2}{105 \cdot 2}$
Remove the common factor to simplify.	$\frac{19}{105}$

In the next example, one of the fractions has a variable in its numerator. We follow the same steps as when both numerators are numbers.

Add: $\frac{3}{5} + \frac{x}{8} .$

Solution

The fractions have different denominators.

	$\frac{3}{5} + \frac{x}{8}$
Find the LCD.
Rewrite as equivalent fractions with the LCD.	$A mathematical expression featuring two fractions being added. The first fraction is (3 times 8) divided by (5 times 8). The second fraction is (x times 5) divided by (8 times 5).$
Simplify the numerators and denominators.	$\frac{24}{40} + \frac{5 x}{40}$
Add.	$\frac{24 + 5 x}{40}$

We cannot add $24$ and $5 x$ since they are not like terms, so we cannot simplify the expression any further.

Identify and Use Fraction Operations

By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions

Simplify:

ⓐ $- \frac{1}{4} + \frac{1}{6}$
ⓑ $- \frac{1}{4} \div \frac{1}{6}$

Solution

First we ask ourselves, “What is the operation?”

ⓐ The operation is addition.

Do the fractions have a common denominator? No.

	$- \frac{1}{4} + \frac{1}{6}$
Find the LCD.
Rewrite each fraction as an equivalent fraction with the LCD.	$A mathematical expression showing the addition of two fractions: - (1 * 3) / (4 * 3) + (1 * 2) / (6 * 2). The numbers 3 and 2, used to adjust fractions to a common denominator, are highlighted in red.$
Simplify the numerators and denominators.	$- \frac{3}{12} + \frac{2}{12}$
Add the numerators and place the sum over the common denominator.	$- \frac{1}{12}$
Check to see if the answer can be simplified. It cannot.

ⓑ The operation is division. We do not need a common denominator.

Step-by-step process for dividing fractions, showing the conversion to multiplication by the reciprocal and subsequent simplification.
	$- \frac{1}{4} \div \frac{1}{6}$
To divide fractions, multiply the first fraction by the reciprocal of the second.	$- \frac{1}{4} \cdot \frac{6}{1}$
Multiply.	$- \frac{6}{4}$
Simplify.	$- \frac{3}{2}$

Simplify:

ⓐ $\frac{5 x}{6} - \frac{3}{10}$
ⓑ $\frac{5 x}{6} \cdot \frac{3}{10}$

Solution

ⓐ The operation is subtraction. The fractions do not have a common denominator.

This table illustrates the step-by-step process of subtracting algebraic fractions by finding a common denominator.
	$\frac{5 x}{6} - \frac{3}{10}$
Rewrite each fraction as an equivalent fraction with the LCD, 30.	$\frac{5 x \cdot 5}{6 \cdot 5} - \frac{3 \cdot 3}{10 \cdot 3}$
	$\frac{25 x}{30} - \frac{9}{30}$
Subtract the numerators and place the difference over the common denominator.	$\frac{25 x - 9}{30}$

ⓑ The operation is multiplication; no need for a common denominator.

Step-by-step example demonstrating the multiplication and simplification of algebraic fractions.
	$\frac{5 x}{6} \cdot \frac{3}{10}$
To multiply fractions, multiply the numerators and multiply the denominators.	$\frac{5 x \cdot 3}{6 \cdot 10}$
Rewrite, showing common factors.	$\frac{5 \cdot x \cdot 3}{2 \cdot 3 \cdot 2 \cdot 5}$
Remove common factors to simplify.	$\frac{x}{4}$

Use the Order of Operations to Simplify Complex Fractions

In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example,

\frac{\frac{3}{4}}{\frac{5}{8}} = \frac{3}{4} \div \frac{5}{8}

Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator.

Simplify: $\frac{{(\frac{1}{2})}^{2}}{4 + 3^{2}} .$

Solution

Step-by-step simplification of the expression (1/2)^2 / (4 + 3^2), detailing each mathematical transformation to reach the final value of 1/52.
	$\frac{{(\frac{1}{2})}^{2}}{4 + 3^{2}}$
Simplify the numerator.	$\frac{\frac{1}{4}}{4 + 3^{2}}$
Simplify the term with the exponent in the denominator.	$\frac{\frac{1}{4}}{4 + 9}$
Add the terms in the denominator.	$\frac{\frac{1}{4}}{13}$
Divide the numerator by the denominator.	$\frac{1}{4} \div 13$
Rewrite as multiplication by the reciprocal.	$\frac{1}{4} \cdot \frac{1}{13}$
Multiply.	$\frac{1}{52}$

Simplify: $\frac{\frac{1}{2} + \frac{2}{3}}{\frac{3}{4} - \frac{1}{6}} .$

Solution

Step-by-step solution demonstrating the simplification of a complex fraction, detailing each operation and the resulting mathematical expression.
	$\frac{\frac{1}{2} + \frac{2}{3}}{\frac{3}{4} - \frac{1}{6}}$
Rewrite numerator with the LCD of 6 and denominator with LCD of 12.	$\frac{\frac{3}{6} + \frac{4}{6}}{\frac{9}{12} - \frac{2}{12}}$
Add in the numerator. Subtract in the denominator.	$\frac{\frac{7}{6}}{\frac{7}{12}}$
Divide the numerator by the denominator.	$\frac{7}{6} \div \frac{7}{12}$
Rewrite as multiplication by the reciprocal.	$\frac{7}{6} \cdot \frac{12}{7}$
Rewrite, showing common factors.	$\frac{7 \cdot 6 \cdot 2}{6 \cdot 7 \cdot 1}$
Simplify.	2

Evaluate Variable Expressions with Fractions

We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

Evaluate

x + \frac{1}{3}

when

ⓐ $x = - \frac{1}{3}$
ⓑ $x = - \frac{3}{4} .$

Solution

ⓐ To evaluate $x + \frac{1}{3}$ when $x = - \frac{1}{3},$ substitute $- \frac{1}{3}$ for $x$ in the expression.

	$x + \frac{1}{3}$
$The image displays the instruction 'Substitute -1/3 for x.' in a dark teal font, with the fraction '-1/3' highlighted in red, against a white background.$
Simplify.	$0$

ⓑ To evaluate $x + \frac{1}{3}$ when $x = - \frac{3}{4},$ we substitute $- \frac{3}{4}$ for $x$ in the expression.

	$x + \frac{1}{3}$
$The text 'Substitute -3/4 for x.' is shown in a math problem, indicating an instruction to replace the variable 'x' with the fractional value -3/4.$
Rewrite as equivalent fractions with the LCD, 12.	$- \frac{3 \cdot 3}{4 \cdot 3} + \frac{1 \cdot 4}{3 \cdot 4}$
Simplify the numerators and denominators.	$- \frac{9}{12} + \frac{4}{12}$
Add.	$- \frac{5}{12}$

Evaluate $y - \frac{5}{6}$ when $y = - \frac{2}{3} .$

Solution

We substitute $- \frac{2}{3}$ for $y$ in the expression.

	$y - \frac{5}{6}$
$The text 'Substitute -2/3 for y.' is displayed, instructing to replace the variable y with the fraction -2/3.$	$A mathematical expression showing the subtraction of two fractions: negative two-thirds minus five-sixths. The first fraction's numerator and denominator are colored red.$
Rewrite as equivalent fractions with the LCD, 6.	$- \frac{4}{6} - \frac{5}{6}$
Subtract.	$- \frac{9}{6}$
Simplify.	$- \frac{3}{2}$

Evaluate $2 x^{2} y$ when $x = \frac{1}{4}$ and $y = - \frac{2}{3} .$

Solution

Substitute the values into the expression. In $2 x^{2} y,$ the exponent applies only to $x .$


	$The image displays the mathematical expression 2(1/4)^2(-2/3), showing a multiplication of a whole number, a fraction squared, and a negative fraction. The numbers 1/4 are red, and -2/3 are blue.$
Simplify exponents first.	$A mathematical expression showing the product of three terms: 2, the fraction 1/16, and the negative fraction -2/3. The terms are enclosed in parentheses, indicating multiplication.$
Multiply. The product will be negative.	$A mathematical expression showing the multiplication of three fractions: negative two over one, one over sixteen, and two over three.$
Simplify.	$A mathematical expression showing the fraction -4/48, presented in black text on a white background. The negative sign precedes the fraction bar, with '4' as the numerator and '48' as the denominator.$
Remove the common factors.	$A mathematical expression showing a fraction with a negative sign, where the common factor '4' in the numerator and denominator has been crossed out as part of simplification, resulting in -(14)/(412).$
Simplify.	$A mathematical expression displaying the fraction minus 1 over 12, written as -1/12.$

Evaluate $\frac{p + q}{r}$ when $p = −4, q = −2,$ and $r = 8 .$

Solution

We substitute the values into the expression and simplify.

	$\frac{p + q}{r}$

Add in the numerator first.	$- \frac{6}{8}$
Simplify.	$- \frac{3}{4}$

Key Concepts

Find the least common denominator (LCD) of two fractions.
1. Factor each denominator into its primes.
2. List the primes, matching primes in columns when possible.
3. Bring down the columns.
4. Multiply the factors. The product is the LCM of the denominators.
5. The LCM of the denominators is the LCD of the fractions.
Equivalent Fractions Property
- If $a, b$ , and $c$ are whole numbers where $b \neq 0$ , $c \neq 0$ then
  $\frac{a}{b} = \frac{a \cdot c}{b \cdot c}$ and $\frac{a \cdot c}{b \cdot c} = \frac{a}{b}$
Convert two fractions to equivalent fractions with their LCD as the common denominator.
1. Find the LCD.
2. For each fraction, determine the number needed to multiply the denominator to get the LCD.
3. Use the Equivalent Fractions Property to multiply the numerator and denominator by the number from Step 2.
4. Simplify the numerator and denominator.
Add or subtract fractions with different denominators.
1. Find the LCD.
2. Convert each fraction to an equivalent form with the LCD as the denominator.
3. Add or subtract the fractions.
4. Write the result in simplified form.
Summary of Fraction Operations
- Fraction multiplication: Multiply the numerators and multiply the denominators.
  $\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d}$
- Fraction division: Multiply the first fraction by the reciprocal of the second.
  $\frac{a}{b} + \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$
- Fraction addition: Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.
  $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$
- Fraction subtraction: Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.
  $\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$
Simplify complex fractions.
1. Simplify the numerator.
2. Simplify the denominator.
3. Divide the numerator by the denominator.
4. Simplify if possible.

Practice Makes Perfect

Find the Least Common Denominator (LCD)

In the following exercises, find the least common denominator (LCD) for each set of fractions.

$\frac{2}{3}$ and $\frac{3}{4}$

$\frac{3}{4}$ and $\frac{2}{5}$

Solution

$\frac{7}{12}$ and $\frac{5}{8}$

$\frac{9}{16}$ and $\frac{7}{12}$

Solution

$\frac{13}{30}$ and $\frac{25}{42}$

$\frac{23}{30}$ and $\frac{5}{48}$

Solution

240

$\frac{21}{35}$ and $\frac{39}{56}$

$\frac{18}{35}$ and $\frac{33}{49}$

Solution

245

$\frac{2}{3}, \frac{1}{6},$ and $\frac{3}{4}$

$\frac{2}{3}, \frac{1}{4},$ and $\frac{3}{5}$

Solution

Convert Fractions to Equivalent Fractions with the LCD

In the following exercises, convert to equivalent fractions using the LCD.

$\frac{1}{3}$ and $\frac{1}{4},$ LCD $= 12$

$\frac{1}{4}$ and $\frac{1}{5},$ LCD $= 20$

Solution

$\frac{5}{20}, \frac{4}{20}$

$\frac{5}{12}$ and $\frac{7}{8},$ LCD $= 24$

$\frac{7}{12}$ and $\frac{5}{8},$ LCD $= 24$

Solution

$\frac{14}{24}, \frac{15}{24}$

$\frac{13}{16}$ and $- \frac{11}{12},$ LCD $= 48$

$\frac{11}{16}$ and $- \frac{5}{12},$ LCD $= 48$

Solution

$\frac{33}{48}, - \frac{20}{48}$

$\frac{1}{3}, \frac{5}{6},$ and $\frac{3}{4},$ LCD $= 12$

$\frac{1}{3}, \frac{3}{4},$ and $\frac{3}{5},$ LCD $= 60$

Solution

$\frac{20}{60}, \frac{45}{60}, \frac{36}{60}$

Add and Subtract Fractions with Different Denominators

In the following exercises, add or subtract. Write the result in simplified form.

$\frac{1}{3} + \frac{1}{5}$

$\frac{1}{4} + \frac{1}{5}$

Solution

$\frac{9}{20}$

$\frac{1}{2} + \frac{1}{7}$

$\frac{1}{3} + \frac{1}{8}$

Solution

$\frac{11}{24}$

$\frac{1}{3} - (- \frac{1}{9})$

$\frac{1}{4} - (- \frac{1}{8})$

Solution

$\frac{3}{8}$

$\frac{1}{5} - (- \frac{1}{10})$

$\frac{1}{2} - (- \frac{1}{6})$

Solution

$\frac{2}{3}$

$\frac{2}{3} + \frac{3}{4}$

$\frac{3}{4} + \frac{2}{5}$

Solution

$\frac{23}{20}$

$\frac{7}{12} + \frac{5}{8}$

$\frac{5}{12} + \frac{3}{8}$

Solution

$\frac{19}{24}$

$\frac{7}{12} - \frac{9}{16}$

$\frac{7}{16} - \frac{5}{12}$

Solution

$\frac{1}{48}$

$\frac{11}{12} - \frac{3}{8}$

$\frac{5}{8} - \frac{7}{12}$

Solution

$\frac{1}{24}$

$\frac{2}{3} - \frac{3}{8}$

$\frac{5}{6} - \frac{3}{4}$

Solution

$\frac{1}{12}$

$- \frac{11}{30} + \frac{27}{40}$

$- \frac{9}{20} + \frac{17}{30}$

Solution

$\frac{7}{60}$

$- \frac{13}{30} + \frac{25}{42}$

$- \frac{23}{30} + \frac{5}{48}$

Solution

$- \frac{53}{80}$

$- \frac{39}{56} - \frac{22}{35}$

$- \frac{33}{49} - \frac{18}{35}$

Solution

$- \frac{291}{245}$

$- \frac{2}{3} - (- \frac{3}{4})$

$- \frac{3}{4} - (- \frac{4}{5})$

Solution

$\frac{1}{20}$

$- \frac{9}{16} - (- \frac{4}{5})$

$- \frac{7}{20} - (- \frac{5}{8})$

Solution

$\frac{11}{40}$

$1 + \frac{7}{8}$

$1 + \frac{5}{6}$

Solution

$\frac{11}{6}$

$1 - \frac{5}{9}$

$1 - \frac{3}{10}$

Solution

$\frac{7}{10}$

$\frac{x}{3} + \frac{1}{4}$

$\frac{y}{2} + \frac{2}{3}$

Solution

$\frac{3 y + 4}{6}$

$\frac{y}{4} - \frac{3}{5}$

$\frac{x}{5} - \frac{1}{4}$

Solution

$\frac{4 x - 5}{20}$

Identify and Use Fraction Operations

In the following exercises, perform the indicated operations. Write your answers in simplified form.

ⓐ $\frac{3}{4} + \frac{1}{6}$
ⓑ $\frac{3}{4} \div \frac{1}{6}$

ⓐ $\frac{2}{3} + \frac{1}{6}$
ⓑ $\frac{2}{3} \div \frac{1}{6}$

Solution

ⓐ $\frac{5}{6}$
ⓑ $4$

ⓐ $- \frac{2}{5} - \frac{1}{8}$
ⓑ $- \frac{2}{5} \cdot \frac{1}{8}$

ⓐ $- \frac{4}{5} - \frac{1}{8}$
ⓑ $- \frac{4}{5} \cdot \frac{1}{8}$

Solution

ⓐ $- \frac{37}{40}$
ⓑ $- \frac{1}{10}$

ⓐ $\frac{5 n}{6} \div \frac{8}{15}$
ⓑ $\frac{5 n}{6} - \frac{8}{15}$

ⓐ $\frac{3 a}{8} \div \frac{7}{12}$
ⓑ $\frac{3 a}{8} - \frac{7}{12}$

Solution

ⓐ $\frac{9 a}{14}$
ⓑ $\frac{9 a - 14}{24}$

ⓐ $\frac{9}{10} \cdot (- \frac{11 d}{12})$
ⓑ $\frac{9}{10} + (- \frac{11 d}{12})$

ⓐ $\frac{4}{15} \cdot (- \frac{5 q}{9})$
ⓑ $\frac{4}{15} + (- \frac{5 q}{9})$

Solution

ⓐ $- \frac{4 q}{27}$
ⓑ $\frac{12 - 25 q}{45}$

$- \frac{3}{8} \div (- \frac{3}{10})$

$- \frac{5}{12} \div (- \frac{5}{9})$

Solution

$\frac{3}{4}$

$- \frac{3}{8} + \frac{5}{12}$

$- \frac{1}{8} + \frac{7}{12}$

Solution

$\frac{11}{24}$

$\frac{5}{6} - \frac{1}{9}$

$\frac{5}{9} - \frac{1}{6}$

Solution

$\frac{7}{18}$

$\frac{3}{8} \cdot (- \frac{10}{21})$

$\frac{7}{12} \cdot (- \frac{8}{35})$

Solution

$- \frac{2}{15}$

$- \frac{7}{15} - \frac{y}{4}$

$- \frac{3}{8} - \frac{x}{11}$

Solution

$\frac{−33 - 8 x}{88}$

$\frac{11}{12 a} \cdot \frac{9 a}{16}$

$\frac{10 y}{13} \cdot \frac{8}{15 y}$

Solution

$\frac{16}{39}$

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

$\frac{{(\frac{1}{5})}^{2}}{2 + 3^{2}}$

$\frac{{(\frac{1}{3})}^{2}}{5 + 2^{2}}$

Solution

$\frac{1}{81}$

$\frac{2^{3} + 4^{2}}{{(\frac{2}{3})}^{2}}$

$\frac{3^{3} - 3^{2}}{{(\frac{3}{4})}^{2}}$

Solution

$\frac{{(\frac{3}{5})}^{2}}{{(\frac{3}{7})}^{2}}$

$\frac{{(\frac{3}{4})}^{2}}{{(\frac{5}{8})}^{2}}$

Solution

$\frac{36}{25}$

$\frac{2}{\frac{1}{3} + \frac{1}{5}}$

$\frac{5}{\frac{1}{4} + \frac{1}{3}}$

Solution

$\frac{60}{7}$

$\frac{\frac{2}{3} + \frac{1}{2}}{\frac{3}{4} - \frac{2}{3}}$

$\frac{\frac{3}{4} + \frac{1}{2}}{\frac{5}{6} - \frac{2}{3}}$

Solution

$\frac{15}{2}$

$\frac{\frac{7}{8} - \frac{2}{3}}{\frac{1}{2} + \frac{3}{8}}$

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

Solution

$\frac{3}{13}$

Mixed Practice

In the following exercises, simplify.

$\frac{1}{2} + \frac{2}{3} \cdot \frac{5}{12}$

$\frac{1}{3} + \frac{2}{5} \cdot \frac{3}{4}$

Solution

$\frac{19}{30}$

$1 - \frac{3}{5} \div \frac{1}{10}$

$1 - \frac{5}{6} \div \frac{1}{12}$

Solution

−9

$\frac{2}{3} + \frac{1}{6} + \frac{3}{4}$

$\frac{2}{3} + \frac{1}{4} + \frac{3}{5}$

Solution

$\frac{91}{60}$

$\frac{3}{8} - \frac{1}{6} + \frac{3}{4}$

$\frac{2}{5} + \frac{5}{8} - \frac{3}{4}$

Solution

$\frac{11}{40}$

$12 (\frac{9}{20} - \frac{4}{15})$

$8 (\frac{15}{16} - \frac{5}{6})$

Solution

$\frac{5}{6}$

$\frac{\frac{5}{8} + \frac{1}{6}}{\frac{19}{24}}$

$\frac{\frac{1}{6} + \frac{3}{10}}{\frac{14}{30}}$

Solution

$(\frac{5}{9} + \frac{1}{6}) \div (\frac{2}{3} - \frac{1}{2})$

$(\frac{3}{4} + \frac{1}{6}) \div (\frac{5}{8} - \frac{1}{3})$

Solution

$\frac{22}{7}$

In the following exercises, evaluate the given expression. Express your answers in simplified form, using improper fractions if necessary.

x + \frac{1}{2}

when

ⓐ $x = - \frac{1}{8}$
ⓑ $x = - \frac{1}{2}$

x + \frac{2}{3}

when

ⓐ $x = - \frac{1}{6}$
ⓑ $x = - \frac{5}{3}$

Solution

ⓐ $\frac{1}{2}$
ⓑ $−1$

x + (- \frac{5}{6})

when

ⓐ $x = \frac{1}{3}$
ⓑ $x = - \frac{1}{6}$

x + (- \frac{11}{12})

when

ⓐ $x = \frac{11}{12}$
ⓑ $x = \frac{3}{4}$

Solution

ⓐ $0$
ⓑ $- \frac{1}{6}$

x - \frac{2}{5}

when

ⓐ $x = \frac{3}{5}$
ⓑ $x = - \frac{3}{5}$

x - \frac{1}{3}

when

ⓐ $x = \frac{2}{3}$
ⓑ $x = - \frac{2}{3}$

Solution

ⓐ $\frac{1}{3}$
ⓑ $−1$

\frac{7}{10} - w

when

ⓐ $w = \frac{1}{2}$
ⓑ $w = - \frac{1}{2}$

\frac{5}{12} - w

when

ⓐ $w = \frac{1}{4}$
ⓑ $w = - \frac{1}{4}$

Solution

ⓐ $\frac{1}{6}$
ⓑ $\frac{2}{3}$

$4 p^{2} q$ when $p = - \frac{1}{2}$ and $q = \frac{5}{9}$

$5 m^{2} n$ when $m = - \frac{2}{5}$ and $n = \frac{1}{3}$

Solution

$\frac{4}{15}$

$2 x^{2} y^{3}$ when $x = - \frac{2}{3}$ and $y = - \frac{1}{2}$

$8 u^{2} v^{3}$ when $u = - \frac{3}{4}$ and $v = - \frac{1}{2}$

Solution

$- \frac{9}{16}$

$\frac{u + v}{w}$ when $u = −4, v = −8, w = 2$

$\frac{m + n}{p}$ when $m = −6, n = −2, p = 4$

Solution

−2

$\frac{a + b}{a - b}$ when $a = −3, b = 8$

$\frac{r - s}{r + s}$ when $r = 10, s = −5$

Solution

Everyday Math

Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs $\frac{3}{16}$ yard of print fabric and $\frac{3}{8}$ yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?

Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs $1 \frac{1}{4}$ cups of sugar for the chocolate chip cookies, and $1 \frac{1}{8}$ cups for the oatmeal cookies How much sugar does she need altogether?

Solution

She needs $2 \frac{3}{8}$ cups

Writing Exercises

Explain why it is necessary to have a common denominator to add or subtract fractions.

Explain how to find the LCD of two fractions.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

An empty self-evaluation chart for math students to gauge their understanding of fraction concepts, from basic addition to complex expressions, marked as 'Confidently', 'With some help', or 'No-I don't get it!'.

ⓑ After looking at the checklist, do you think you are well prepared for the next section? Why or why not?

least common denominator (LCD)

The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.