<meta http-equiv="refresh" content="1; url=/nojavascript/"> Addition of Rational Numbers ( Read ) | Arithmetic | CK-12 Foundation

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Suppose you drank $\frac{7}{8}$ of a bottle of juice on Monday, $\frac{2}{5}$ of a bottle on Tuesday, and $\frac{3}{2}$ of a bottle on Wednesday. Can you identify which of the fractions of a bottle of juice are improper, and can you write the fractions with a common denominator? Also, can you add these fractions of a bottle of juice together? After completing this Concept, you'll be able to perform these tasks so that you can figure out how many bottles of juice you drank in total over the three days.

### Guidance

To add rational numbers, we must first remember how to rewrite mixed numbers as improper fractions. Begin by multiplying the denominator of the mixed number to the whole value. Add the numerator to this product. This value is the numerator of the improper fraction. The denominator is the original.

#### Example A

Write $11 \frac{2}{3}$ as an improper fraction:

Solution: $3 \times 11 = 33 + 2 = 35$ . This is the numerator of the improper fraction.

$11 \frac{2}{3} = \frac{35}{3}$

Now that we know how to rewrite a mixed number as an improper fraction, we can begin to add rational numbers. There is one thing to remember when finding the sum or difference of rational numbers: The denominators must be equivalent.

The Addition Property of Fractions: For all real numbers $a, \ b,$ and $c, \ \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}.$

#### Example B

Add the following sets of fractions (rational numbers):

a.) $\frac{1}{3}+\frac{2}{3}$

b.) $4\frac{1}{7}+\frac{2}{7}$

Solutions:

a.) Since the denominators are the same, we can go ahead and add the numerators:

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

b.) $4\frac{1}{7}$ is a mixed fraction, and reads "4 and one seventh." This means we can think of it as $4+\frac{1}{7}$ . Since the other fraction also has 7 as a denominator, we can add the two fractions:

$4\frac{1}{7}+\frac{2}{7}=4+\frac{1}{7}+\frac{2}{7}=4+\frac{1+2}{7}=4+\frac{3}{7}=4\frac{3}{7}.$

This mixed fraction can be turned into an improper fraction as follows:

$4\times 7=28$

$\frac{28+3}{7}=\frac{31}{7}$

Since these properties apply to all real numbers, they apply to fractions or rational numbers as well. Let's review them here:

The Commutative Property of Addition: For all real numbers $a$ ,and $b$ , $a + b = b + a$ .

To commute means to change locations, so the Commutative Property of Addition allows you to rearrange the objects in an addition problem.

The Associative Property of Addition: For all real numbers $a, \ b,$ and $c$ , $(a + b)+ c = a + (b + c).$

To associate means to group together, so the Associative Property of Addition allows you to regroup the objects in an addition problem.

The Identity Property of Addition: For any real number $a, \ a + 0 = a.$

Another way you sometimes see a rational number is as a decimal number, such as 2.5, 30.01, or 2.9999. We will practice some of the above properties on rational numbers in their different forms.

#### Example C

To convince ourselves that the algebraic properties are true, in this exercise we will check whether the following equations are equal:

a.) Commutative Property: $2.5 + 3.5 =3.5 + 2.5$

b.) Associate Property: $\frac{1}{9}+\left(\frac{2}{9}+\frac{5}{9}\right)=\left(\frac{1}{9}+\frac{2}{9}\right)+\frac{5}{9}$

Solutions:

a.) We will check each side separately to see if they equal the same thing.

$2.5 + 3.5 =6$

$3.5 + 2.5=6$

So we conclude that the equality is satisfied.

b.) We check each side of the equation here as well.

$\frac{1}{9}+\left(\frac{2}{9}+\frac{5}{9}\right)=\frac{1}{9}+\frac{7}{9}=\frac{8}{9}$

$\left(\frac{1}{9}+\frac{2}{9}\right)+\frac{5}{9}=\frac{3}{9}+\frac{5}{9}=\frac{8}{9}$

So we conclude that the equality is satisfied.

Common Denominators

In order to add two fractions, they must have a common denominator. This means that they must have the same number in the denominator. If two fractions to be added do not have common denominators, either one or both of the fractions can be changed so that they do have common denominators. In general, when two fractions have different denominators, use the pattern below.

$\frac{a}{b}+\frac{c}{d}=\frac{a}{b}\cdot \frac{d}{d}+\frac{c}{d}\cdot \frac{b}{b}=\frac{ad}{bd}+\frac{bc}{bd}=\frac{ad+bc}{bd}$

To understand how this works, see the next example.

#### Example D

1. f
2. $\frac{1}{5}+\frac{3}{10}$

Solutions:

1. Follow the pattern; that is, multiply the top and bottom of each fraction by the denominator of the other fraction:

$\frac{2}{11}+\frac{1}{3}=\frac{2}{11}\cdot \frac{3}{3}+\frac{1}{3}\cdot \frac{11}{11}=\frac{6}{33}+\frac{11}{33}=\frac{6+11}{33}=\frac{17}{33}$

2. In this example, our denominator, 10, is a multiple of the other denominator, 5. There is no need to change the fraction with a denominator of 10. Simply multiply the top and bottom of the first fraction in order to make its denominator 10:

$\frac{1}{5}+\frac{3}{10}=\frac{1}{5}\cdot \frac{2}{2}+\frac{3}{10}=\frac{2}{10}+\frac{3}{10}=\frac{2+3}{10}=\frac{5}{10}=\frac{1}{2}$

### Guided Practice

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

Solution:

We can break up the mixed fractions:

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

Using the Commutative Property we can rearrange and simplify by adding integers:

$5+\frac{1}{3}+2+\frac{3}{4}=5+2+\frac{1}{3}+\frac{3}{4}=7+\frac{1}{3}+\frac{3}{4}.$

Now we just need to add the fractions. Since they do not have common denominators, we have to give them common denominators. The denominators do not share any factors, so we need to multiply them by each other:

$\frac{1}{3}+\frac{3}{4}=\frac{1\times 4}{3\times 4}+\frac{3\times 3}{4\times 3}=\frac{4}{12}+\frac{9}{12}=\frac{4+9}{12}=\frac{13}{12}.$

Now we know what the sum of the fractions is:

$7+\frac{1}{3}+\frac{3}{4}=7+\frac{13}{12}.$

Since our answer needs to be a mixed fraction, we will turn the improper fraction into a mixed fraction. Since 12 goes into 13 one time with a remainder of 1, we get:

$7+\frac{13}{12}=7+1\frac{1}{12}=8\frac{1}{12}.$

$5\frac{1}{3}+2\frac{3}{4}=8\frac{1}{12}.$

### Explore More

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Addition of Rational Numbers (7:40)

Find the sum. Write the answer in its simplest form.

1. $\frac{3}{7} + \frac{2}{7}$
2. $\frac{3}{10} + \frac{1}{5}$
3. $\frac{5}{16} + \frac{5}{12}$
4. $\frac{3}{8} + \frac{9}{16}$
5. $\frac{8}{25} + \frac{7}{10}$
6. $\frac{1}{6} + \frac{1}{4}$
7. $\frac{7}{15} + \frac{2}{9}$
8. $\frac{5}{19} + \frac{2}{27}$
9. $-2.6 + 11.19$
10. $-8 + 13$
11. $-7.1 + (-5.63)$
12. $9.99 + (-0.01)$
13. $4 \frac{7}{8} + 1\frac{1}{2}$
14. $-3 \frac{1}{3} + \left (-2 \frac{3}{4} \right )$

In 15–17, which property of addition does each situation involve?

1. Whichever order your groceries are scanned at the store, the total will be the same.
2. Suppose you go buy a DVD for $8.00, another for$29.99, and a third for \$14.99. You can add $(8 + 29.99) + 14.99$ or you can add $8 + (29.99 + 14.99)$ to obtain the total.
3. Nadia, Peter, and Ian are pooling their money to buy a gallon of ice cream. Nadia is the oldest and gets the greatest allowance. She contributes half of the cost. Ian is next oldest and contributes one third of the cost. Peter, the youngest, gets the smallest allowance and contributes one fourth of the cost. They figure that this will be enough money. When they get to the check-out, they realize that they forgot about sales tax and worry there will not be enough money. Amazingly, they have exactly the right amount of money. What fraction of the cost of the ice cream was added as tax?

In 18–24, evaluate each expression for $v = 5.8.$

1. $9.1 + v$
2. $v+(-v)$
3. $-v+4.12$
4. $-23.14+ -v$
5. $7.86+(-v)$
6. $-v+3.5$
7. $-v+v$

### Vocabulary Language: English Spanish

For all real numbers $a, \ b,$ and $c, \ \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}.$