Question
Find the difference.
4 1/4 - 2 1/8
Simplify your answer completely.
[__] [__]/[__]
4 1/4 - 2 1/8
Simplify your answer completely.
[__] [__]/[__]
Answers
GPT-4o mini
To find the difference between \(4 \frac{1}{4}\) and \(2 \frac{1}{8}\), we can follow these steps:
1. Convert the mixed numbers to improper fractions.
For \(4 \frac{1}{4}\):
\[
4 \frac{1}{4} = \frac{4 \times 4 + 1}{4} = \frac{16 + 1}{4} = \frac{17}{4}
\]
For \(2 \frac{1}{8}\):
\[
2 \frac{1}{8} = \frac{2 \times 8 + 1}{8} = \frac{16 + 1}{8} = \frac{17}{8}
\]
2. Find a common denominator to subtract the fractions. The least common denominator (LCD) of 4 and 8 is 8.
Convert \(\frac{17}{4}\) to a fraction with a denominator of 8:
\[
\frac{17}{4} = \frac{17 \times 2}{4 \times 2} = \frac{34}{8}
\]
3. Now subtract \(\frac{17}{8}\) from \(\frac{34}{8}\):
\[
\frac{34}{8} - \frac{17}{8} = \frac{34 - 17}{8} = \frac{17}{8}
\]
4. Now, convert \(\frac{17}{8}\) back to a mixed number. Divide 17 by 8:
\[
17 \div 8 = 2 \quad \text{(this gives a whole number of 2 and a remainder of 1)}
\]
So,
\[
\frac{17}{8} = 2 \frac{1}{8}
\]
Thus, the simplified difference is:
\[
\boxed{2} \quad \frac{1}{8}
\]
1. Convert the mixed numbers to improper fractions.
For \(4 \frac{1}{4}\):
\[
4 \frac{1}{4} = \frac{4 \times 4 + 1}{4} = \frac{16 + 1}{4} = \frac{17}{4}
\]
For \(2 \frac{1}{8}\):
\[
2 \frac{1}{8} = \frac{2 \times 8 + 1}{8} = \frac{16 + 1}{8} = \frac{17}{8}
\]
2. Find a common denominator to subtract the fractions. The least common denominator (LCD) of 4 and 8 is 8.
Convert \(\frac{17}{4}\) to a fraction with a denominator of 8:
\[
\frac{17}{4} = \frac{17 \times 2}{4 \times 2} = \frac{34}{8}
\]
3. Now subtract \(\frac{17}{8}\) from \(\frac{34}{8}\):
\[
\frac{34}{8} - \frac{17}{8} = \frac{34 - 17}{8} = \frac{17}{8}
\]
4. Now, convert \(\frac{17}{8}\) back to a mixed number. Divide 17 by 8:
\[
17 \div 8 = 2 \quad \text{(this gives a whole number of 2 and a remainder of 1)}
\]
So,
\[
\frac{17}{8} = 2 \frac{1}{8}
\]
Thus, the simplified difference is:
\[
\boxed{2} \quad \frac{1}{8}
\]