Problem 62 - Prealgebra

Question

There are two ways to work the problems below. You can combine the fractions inside the parentheses first and then multiply, or you can apply the distributive property first, then add. $$6\left(\frac{1}{3}+\frac{1}{2}\right)$$

Step-by-Step Solution

Answer

The result is 5.

Step 1: Combine the Fractions

First, let's find a common denominator to add the fractions \( \frac{1}{3} \) and \( \frac{1}{2} \). The least common denominator for 3 and 2 is 6. Convert each fraction to an equivalent fraction with a denominator of 6: \( \frac{1}{3} = \frac{2}{6} \) and \( \frac{1}{2} = \frac{3}{6} \). Now add them: \( \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \).

Step 2: Multiply by 6

Now multiply the sum of the fractions by 6: \( 6 \times \frac{5}{6} = 5 \). Thus, the result after combining the fractions and multiplying is 5.

Key Concepts

Least Common Denominator

The least common denominator (LCD) is a concept used in fraction operations to make adding and subtracting fractions possible. When dealing with fractions like \( \frac{1}{3} \) and \( \frac{1}{2} \), their denominators differ, which means we cannot simply add them together as they are. We need a common denominator that both denominators can divide into evenly. This is where the LCD comes in.
The least common denominator of any fractions is the smallest number that is a multiple of each of the denominators. For the fractions in our problem, the denominators are 3 and 2. The multiples of 3 are 3, 6, 9, and so on, while the multiples of 2 are 2, 4, 6, and so forth. The smallest multiple both denominators have in common is 6.
To add \( \frac{1}{3} \) and \( \frac{1}{2} \), we convert them into equivalent fractions with the denominator of 6. From step 1, we convert \( \frac{1}{3} \) to \( \frac{2}{6} \) and \( \frac{1}{2} \) to \( \frac{3}{6} \). Now it's simple to add them, resulting in \( \frac{5}{6} \). Finding the LCD helps in creating equivalent fractions that can be directly added or subtracted.

Distributive Property

The distributive property is a fundamental concept in algebra that allows you to simplify expressions that involve both addition (or subtraction) and multiplication. This property states that for any numbers \(a\), \(b\), and \(c\), the equation \(a(b + c) = ab + ac\) holds true. In simpler terms, multiplication distributes over addition.
Let's apply this to the original exercise: \(6(\frac{1}{3} + \frac{1}{2})\). Using the distributive property, you can first distribute the 6 to each fraction inside the parentheses, simplifying each separately.
We compute:

• \(6 \times \frac{1}{3} = 2\)
• \(6 \times \frac{1}{2} = 3\)

Then we add these results: \(2 + 3 = 5\). This method, applying the distributive property first, yields the same final outcome, demonstrating the property’s utility in solving complex expressions efficiently. It is a versatile tool that simplifies the calculation by breaking it down into more manageable parts.

Equivalent Fractions

Equivalent fractions are fractions that have different numerators and denominators but represent the same value or proportion. They are essential when performing any operation involving fractions. To create equivalent fractions, you multiply or divide both the numerator and the denominator of a fraction by the same non-zero number.
In our exercise, we made use of equivalent fractions to add \( \frac{1}{3} \) and \( \frac{1}{2} \). Given their different denominators, we found the least common denominator (LCD), which was 6. Then, we converted \( \frac{1}{3} \) into an equivalent fraction \( \frac{2}{6} \), because when you multiply the numerator and the denominator by 2, the fraction remains equivalent. Similarly, we transformed \( \frac{1}{2} \) into \( \frac{3}{6} \) by multiplying the numerator and the denominator by 3.

• \( \frac{1}{3} = \frac{2}{6} \)
• \( \frac{1}{2} = \frac{3}{6} \)

These conversions allow the sum of \( \frac{2}{6} + \frac{3}{6} \) to be easily calculated as \( \frac{5}{6} \). Understanding how to form equivalent fractions is crucial for successfully performing many fraction operations such as addition, subtraction, and even complex algebraic equations.