# 2.3: Addition of Rational Numbers

**At Grade**Created by: CK-12

**Practice**Addition of Rational Numbers

### Addition of Rational Numbers

In the last Concept, we learned how to represent numbers on a number line. To add numbers on a number line, we start at the position of the first number, and then move to the right by a number of units equal to the second number.

#### Representing Sums on a Number Line

1. Represent the sum \begin{align*}-2 + 3\end{align*} on a number line.

We start at the number -2, and then move 3 units to the right. We thus end at +1.

\begin{align*}-2 + 3 = 1\end{align*}

2. Represent the sum 2 - 3 on a number line.

Subtracting a number is basically just **adding a negative number**. Instead of moving to the right, we move to the left. Starting at the number 2, and then moving 3 to the left, means we end at -1.

\begin{align*}2 - 3 = -1\end{align*}

**Adding and Subtracting Rational** Numbers

When we add or subtract two fractions, the denominators must match before we can find the sum or difference. We have already seen how to find a common denominator for two rational numbers.

Simplify \begin{align*}\frac{3}{5} + \frac{1}{6}\end{align*}.

To combine these fractions, we need to rewrite them over a common denominator. We are looking for the **lowest common denominator** (LCD). We need to identify the **lowest common multiple** or **least common multiple** (LCM) of 5 and 6. That is the smallest number that both 5 and 6 divide into evenly (that is, without a remainder).

The lowest number that 5 and 6 both divide into evenly is 30. The LCM of 5 and 6 is 30, so the lowest common denominator for our fractions is also 30.

We need to rewrite our fractions as new **equivalent fractions** so that the denominator in each case is 30.

If you think back to our idea of a cake cut into a number of slices, \begin{align*}\frac{3}{5}\end{align*} means 3 slices of a cake that has been cut into 5 pieces. You can see that if we cut the same cake into 30 pieces (6 times as many) we would need 6 times as many slices to make up an equivalent fraction of the cake—in other words, 18 slices instead of 3.

\begin{align*}\frac{3}{5}\end{align*} is equivalent to \begin{align*}\frac{18}{30}\end{align*}.

By a similar argument, we can rewrite the fraction \begin{align*}\frac{1}{6}\end{align*} as a share of a cake that has been cut into 30 pieces. If we cut it into 5 times as many pieces, we need 5 times as many slices.

\begin{align*}\frac{1}{6}\end{align*} is equivalent to \begin{align*}\frac{5}{30}\end{align*}.

Now that both fractions have the same denominator, we can add them. If we add 18 pieces of cake to 5 pieces, we get a total of 23 pieces. 23 pieces of a cake that has been cut into 30 pieces means that our answer is \begin{align*}\frac{23}{30}\end{align*}.

\begin{align*}\frac{3}{5} + \frac{1}{6} = \frac{18}{30} + \frac{5}{30} = \frac{23}{30}\end{align*}

Notice that when we have fractions with a common denominator, we **add the numerators** but we **leave the denominators alone**. Here is this information in algebraic terms.

When adding fractions: \begin{align*} \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}\end{align*}

So far, we’ve only dealt with examples where it’s easy to find the least common multiple of the denominators. With larger numbers, it isn’t so easy to be sure that we have the LCD. We need a more systematic method. In the next example, we will use the method of **prime factors** to find the least common denominator.

#### Subtracting Rational Numbers

Simplify \begin{align*}\frac{29}{90} - \frac{13}{126}\end{align*}.

To find the lowest common multiple of 90 and 126, we first find the prime factors of 90 and 126. We do this by continually dividing the number by factors until we can’t divide any further. You may have seen a factor tree before.

The factor tree for 90 looks like this:

The factor tree for 126 looks like this:

The LCM for 90 and 126 is made from the **smallest possible collection of primes** that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree.

Prime |
Factors in 90 |
Factors in 126 |
We Need |
---|---|---|---|

2 | 1 | 1 | 1 |

3 | 2 | 2 | 2 |

5 | 1 | 0 | 1 |

7 | 0 | 1 | 1 |

So we need one 2, two 3’s, one 5 and one 7. That gives us \begin{align*}2 \cdot 3 \cdot 3 \cdot 5 \cdot 7 = 630\end{align*} as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation.

90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so \begin{align*}\frac{29}{90} = \frac{7 \cdot 29}{7 \cdot 90} = \frac{203}{630}\end{align*}.

126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so \begin{align*}\frac{13}{126} = \frac{5 \cdot 13}{5 \cdot 126} = \frac{65}{630}\end{align*}.

Now we complete the problem: \begin{align*}\frac{29}{90} - \frac{13}{126} = \frac{203}{630} - \frac{65}{630} = \frac{138}{630}\end{align*}.

This fraction **simplifies**. To be sure of finding the **simplest form** for \begin{align*}\frac{138}{630}\end{align*}, we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23.

\begin{align*}\frac{138}{630} = \frac{2 \cdot 3 \cdot 23}{2 \cdot 3 \cdot 3 \cdot 5 \cdot 7}\end{align*}; one factor of 2 and one factor of 3 cancels out, leaving \begin{align*}\frac{23}{3 \cdot 5 \cdot 7}\end{align*} or \begin{align*}\frac{23}{105}\end{align*} as our answer.

**Identify and Apply Properties of Addition**

Three mathematical properties which involve addition are the **commutative, associative,** and the **additive identity properties**.

**Commutative property:** When two numbers are added, the sum is the same even if the order of the items being added changes.

**Example:** \begin{align*}3 + 2 = 2 + 3\end{align*}

**Associative Property:** When three or more numbers are added, the sum is the same regardless of how they are grouped.

**Example:** \begin{align*}(2 + 3) + 4 = 2 + (3 + 4) \end{align*}

**Additive Identity Property:** The sum of any number and zero is the original number.

**Example:** \begin{align*}5 + 0 = 5\end{align*}

### Example

#### Example 1

Simplify \begin{align*}\frac{1}{3} - \frac{1}{9}\end{align*}.

The lowest common multiple of 9 and 3 is 9, so 9 is our common denominator. That means we don’t have to alter the second fraction at all.

3 divides into 9 three times, so \begin{align*}\frac{1}{3} = \frac{3 \cdot 1}{3 \cdot 3} = \frac{3}{9}\end{align*}. Our sum becomes \begin{align*}\frac{3}{9} - \frac{1}{9}\end{align*}. We can subtract fractions with a common denominator by subtracting their numerators, just like adding. In other words:

When subtracting fractions: \begin{align*}\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}\end{align*}

\begin{align*}\frac{1}{3} - \frac{1}{9} = \frac{2}{9}\end{align*}

### Review

- Write the sum that the following moves on a number line represent.

For 2-7, add the following rational numbers. Write each answer in its **simplest form**.

- \begin{align*}\frac{3}{7} + \frac{2}{7}\end{align*}
- \begin{align*}\frac{3}{10} + \frac{1}{5}\end{align*}
- \begin{align*}\frac{5}{16} + \frac{5}{12}\end{align*}
- \begin{align*}\frac{3}{8} + \frac{9}{16}\end{align*}
- \begin{align*}\frac{8}{25} + \frac{7}{10}\end{align*}
- \begin{align*}\frac{1}{6} + \frac{1}{4}\end{align*}

For 8-14, subtract the following rational numbers. Be sure that your answer is in the **simplest form**.

- \begin{align*}\frac{3}{4} - \frac{1}{3}\end{align*}
- \begin{align*}\frac{15}{11} - \frac{9}{7}\end{align*}
- \begin{align*}\frac{2}{13} - \frac{1}{11}\end{align*}
- \begin{align*}\frac{7}{27} - \frac{9}{39}\end{align*}
- \begin{align*}\frac{6}{11} - \frac{3}{22}\end{align*}
- \begin{align*}\frac{13}{64} - \frac{7}{40}\end{align*}
- \begin{align*}\frac{11}{70} - \frac{11}{30}\end{align*}
- Consider the equation \begin{align*}y = 3x + 2\end{align*}. Determine the change in \begin{align*}y\end{align*} between \begin{align*}x = 3\end{align*} and \begin{align*}x = 7\end{align*}.
- Consider the equation \begin{align*}y = \frac{2}{3} x + \frac{1}{2}\end{align*}. Determine the change in \begin{align*}y\end{align*} between \begin{align*}x = 1\end{align*} and \begin{align*}x = 2\end{align*}.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 2.3.

### Notes/Highlights Having trouble? Report an issue.

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Additive Identity Property

The sum of any number and zero is the number itself.Associative Property

The associative property states that you can change the groupings of numbers being added or multiplied without changing the sum. For example: (2+3) + 4 = 2 + (3+4), and (2 X 3) X 4 = 2 X (3 X 4).Commutative Property

The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example .Equivalent Fractions

Equivalent fractions are fractions that can each be simplified to the same fraction. An equivalent fraction is created by multiplying both the numerator and denominator of the original fraction by the same number.multiplicative identity property

The product of any number and one is the number itself.simplest form

The simplest form of a fraction has no common factors in the numerator and the denominator. The simplest form of 3/6 is 1/2.### Image Attributions

Here you'll learn how to find the sum and difference of rational numbers by applying the properties of addition and subtraction.

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