Problem 29 - Prealgebra

Question

Add and subtract the following mixed numbers as indicated. $$\begin{array}{r}5 \frac{4}{10} \\\\-3 \frac{1}{3} \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

The result is \(2 \frac{1}{15}\).

Step 1: Convert Mixed Numbers to Improper Fractions

First, convert the mixed numbers into improper fractions for easier calculation. For \(5 \frac{4}{10}\), multiply 5 (the whole number) by 10 (the denominator) and add 4, giving a fraction of \(\frac{54}{10}\). Similarly, for \(-3 \frac{1}{3}\), multiply 3 by 3 and add 1, resulting in \(-\frac{10}{3}\). The problem becomes \(\frac{54}{10} - \frac{10}{3}\).

Step 2: Find a Common Denominator

To subtract the fractions \(\frac{54}{10}\) and \(\frac{10}{3}\), find a common denominator. The least common multiple of 10 and 3 is 30. Convert \(\frac{54}{10}\) to \(\frac{162}{30}\) by multiplying the numerator and denominator by 3. Convert \(\frac{10}{3}\) to \(\frac{100}{30}\) by multiplying the numerator and denominator by 10.

Step 3: Subtract the Fractions

Substitute the fractions with the common denominator into the original expression: \(\frac{162}{30} - \frac{100}{30}\). Subtract the numerators \(162 - 100 = 62\), keeping the denominator the same. This gives \(\frac{62}{30}\).

Step 4: Simplify the Result

Simplify \(\frac{62}{30}\) by finding the greatest common divisor, which is 2. Divide both the numerator and denominator by 2 to get \(\frac{31}{15}\).

Step 5: Convert Back to a Mixed Number

Convert \(\frac{31}{15}\) back to a mixed number. Divide 31 by 15, which is 2 with a remainder of 1. Thus, \(\frac{31}{15} = 2 \frac{1}{15}\).

Key Concepts

Improper Fractions

When working with mixed numbers, it's often easier to convert them into improper fractions first. This allows you to handle the whole number and fractional part as a single entity.
For instance:

• Take the mixed number \(5 \frac{4}{10}\). Multiply 5 (whole number) by 10 (denominator) and add 4 (numerator): \(5 \times 10 + 4 = 54\). This gives us the improper fraction \(\frac{54}{10}\).
• Similarly, convert \(-3 \frac{1}{3}\) by multiplying 3 by 3 and adding 1: \(3 \times 3 + 1 = 10\). This results in \(-\frac{10}{3}\).

Improper fractions have numerators larger than their denominators, which can be helpful for performing arithmetic operations like addition and subtraction.

Common Denominator

To subtract fractions, they need to share a common denominator, which allows you to directly subtract the numerators.
The process involves finding a common multiple for the two denominators. In our exercise:

• The denominators are 10 and 3.
• The least common multiple (LCM) is 30.

Now, convert each fraction:

• Multiply the numerator and the denominator of \(\frac{54}{10}\) by 3, giving \(\frac{162}{30}\).
• For \(\frac{10}{3}\), multiply the numerator and denominator by 10, resulting in \(\frac{100}{30}\).

With a common denominator, you're ready to subtract the fractions easily.

Subtracting Fractions

Once you have fractions with a common denominator, the subtraction process is straightforward. You simply subtract the numerators and keep the denominator the same.
Take the fractions from our example:

• \(\frac{162}{30} - \frac{100}{30}\) results in subtracting only the numerators: \(162 - 100 = 62\).

Thus, the result is \(\frac{62}{30}\). After subtracting, it's a good idea to check if the result can be simplified, which brings us to the next step.

Simplifying Fractions

To simplify a fraction, look for the greatest common divisor (GCD) of the numerator and denominator. Simplifying fractions makes them easier to read and interpret.
Here's how to simplify \(\frac{62}{30}\):

• The largest number that divides both 62 and 30 is 2.
• Divide both the numerator and the denominator by 2 to get \(\frac{31}{15}\).

Since \(\frac{31}{15}\) is simplified as much as possible, we can convert it to a mixed number for a clearer result.
The division of 31 by 15 gives 2 with a remainder of 1, leading to the final answer: \(2 \frac{1}{15}\). This shows the importance of simplifying as the last step for clarity in mathematical expressions.