Problem 59
Question
In Exercises 57-60, evaluate the expression. $$ \frac{2}{7}-\frac{1}{3} $$
Step-by-Step Solution
Verified Answer
The result of the subtraction \(\frac{2}{7} - \frac{1}{3}\) is \(-\frac{1}{21}\).
1Step 1: Identify the Least Common Denominator (LCD)
The first step is to identify the least common denominator (LCD) between the two fractions. The denominators are 7 and 3. To find a common denominator, simply multiply the two denominators. So, in this case, the LCD is \(7 \times 3 = 21\).
2Step 2: Convert the fractions
Next, convert each fraction to have the obtained least common denominator. Hence, \[\frac{2}{7} becomes \frac{2 \times 3}{21} = \frac{6}{21}\]and \[\frac{1}{3} becomes \frac{1 \times 7}{21} = \frac{7}{21}\]in order to make both fractions have the same denominator.
3Step 3: Perform the subtraction
Now, subtract the two fractions, keeping the common denominator constant. Hence,\[\frac{6}{21} - \frac{7}{21} = \frac{-1}{21}\]
Key Concepts
Least Common DenominatorConverting FractionsSubtraction of Fractions
Least Common Denominator
When subtracting fractions, it's important they share the same denominator. This shared denominator is known as the Least Common Denominator (LCD). The LCD is the smallest number that both denominators can divide into evenly. In our example, we had the fractions \( \frac{2}{7} \) and \( \frac{1}{3} \).
To find the LCD, you can choose a simple trick if you have small denominators: multiply them together. In this case, \( 7 \times 3 = 21 \). Thus, 21 becomes the Least Common Denominator.
Finding the LCD ensures that both fractions are converted to the same base so that they can be easily added or subtracted. Without the same denominator, the fractions are talking about different "sized" parts, which makes it impossible to directly subtract them.
To find the LCD, you can choose a simple trick if you have small denominators: multiply them together. In this case, \( 7 \times 3 = 21 \). Thus, 21 becomes the Least Common Denominator.
Finding the LCD ensures that both fractions are converted to the same base so that they can be easily added or subtracted. Without the same denominator, the fractions are talking about different "sized" parts, which makes it impossible to directly subtract them.
Converting Fractions
After identifying the Least Common Denominator, the next step is converting each fraction to have that denominator. This involves adjusting the numerators to maintain the same value as the original fractions.
Let's see how it works:
Let's see how it works:
- Start with \( \frac{2}{7} \). To convert it, multiply both the numerator and denominator by 3 to match the denominator of 21. This gives \( \frac{2 \times 3}{21} = \frac{6}{21} \).
- Now, take \( \frac{1}{3} \). Multiply both its numerator and denominator by 7. You'll get \( \frac{1 \times 7}{21} = \frac{7}{21} \).
Subtraction of Fractions
After ensuring fractions have a common denominator, subtraction becomes straightforward. Keep the denominator the same and perform the subtraction on the numerators.
In our exercise, the fractions \( \frac{6}{21} \) and \( \frac{7}{21} \) are ready for subtraction. Since they already share the denominator 21, you subtract the numerators: \( 6 - 7 = -1 \).
So, \( \frac{6}{21} - \frac{7}{21} = \frac{-1}{21} \).
The result shows that the second fraction was larger than the first, hence the subtraction gives a negative fraction. Remember, the denominator remains unchanged in the subtraction process, allowing you to focus only on the numerators.
In our exercise, the fractions \( \frac{6}{21} \) and \( \frac{7}{21} \) are ready for subtraction. Since they already share the denominator 21, you subtract the numerators: \( 6 - 7 = -1 \).
So, \( \frac{6}{21} - \frac{7}{21} = \frac{-1}{21} \).
The result shows that the second fraction was larger than the first, hence the subtraction gives a negative fraction. Remember, the denominator remains unchanged in the subtraction process, allowing you to focus only on the numerators.
Other exercises in this chapter
Problem 59
In Exercises 59-62, solve the system by the method of substitution. $$ \left\\{\begin{aligned} y &=x \\ x+3 y &=20 \end{aligned}\right. $$
View solution Problem 59
In Exercises 59?62, factor the expression. $$ x(3-x)-2(3-x) $$
View solution Problem 60
In Exercises \(59-62\), rewrite the expression as a product. $$ 7^{5} $$
View solution Problem 60
In Exercises 59-62, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} x+y=9 \\ 2 x+2 y=18 \end{array}\right. $$
View solution