Tri-Fractional Operations: How to Add and Subtract Three Fractions with Different Denominators
When faced with the task of adding or subtracting three fractions with different denominators, it might seem like a complex puzzle. However, with a systematic approach, this puzzle can be solved with ease.
In this guide, we’ll walk you through the steps to add or subtract three fractions, even when they have different denominators.
Step-by-step Guide to Add and Subtract Three Fractions with Different Denominators:
1. Basics of Fractions:
Recall that a fraction consists of a numerator (top number) and a denominator (bottom number). The denominator indicates the total number of equal parts, while the numerator tells us how many of those parts we’re considering.
2. Identifying Different Denominators:
If the fractions you’re working with don’t have the same denominator, they have different denominators. For instance, in the fractions \(\frac{1}{2}\), \(\frac{3}{4}\), and \(\frac{5}{6}\), the denominators 2, 4, and 6 are all different.
3. Finding the Least Common Denominator (LCD):
The LCD is the smallest number into which all the denominators can divide. This ensures that the fractions are of comparable sizes.
4. Adjusting Each Fraction to the LCD:
Multiply the numerator and denominator of each fraction by the necessary factor to achieve the LCD.
5. Performing the Operation:
With the same denominator in place, either add or subtract the numerators of the fractions to get the final result.
Example 1 (Addition):
Add \(\frac{1}{3}\), \(\frac{1}{4}\), and \(\frac{1}{5}\).
Solution:
The LCD for 3, 4, and 5 is 60. Adjusting the fractions:
– \(\frac{1}{3}\) becomes \(\frac{20}{60}\).
– \(\frac{1}{4}\) becomes \(\frac{15}{60}\).
– \(\frac{1}{5}\) becomes \(\frac{12}{60}\).
Adding them up, the result is \(\frac{47}{60}\).
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Example 2 (Subtraction):
Subtract \(\frac{1}{6}\) and \(\frac{1}{8}\) from \(\frac{1}{4}\).
Solution:
The LCD for 6, 8, and 4 is 24. Adjusting the fractions:
– \(\frac{1}{6}\) becomes \(\frac{4}{24}\).
– \(\frac{1}{8}\) becomes \(\frac{3}{24}\).
– \(\frac{1}{4}\) becomes \(\frac{6}{24}\).
Subtracting, the result is \(\frac{6 – 4 – 3}{24} = \(\frac{-1}{24}\).
Practice Questions:
1. Add \(\frac{1}{7}\), \(\frac{2}{9}\), and \(\frac{3}{11}\).
2. Subtract \(\frac{2}{8}\) and \(\frac{3}{12}\) from \(\frac{1}{6}\).
3. Add \(\frac{1}{10}\), \(\frac{2}{15}\), and \(\frac{3}{20}\).
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Answers:
1. \(\frac{293}{693}\)
2. \(\frac{1}{24}\)
3. \(\frac{11}{30}\)
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