Question

15 2/3 - 21 3/4

Answer 1

To subtract the mixed numbers \(15 \frac{2}{3}\) and \(21 \frac{3}{4}\), we can follow these steps:

1. First, convert both mixed numbers to improper fractions.

For \(15 \frac{2}{3}\):
\[
15 \frac{2}{3} = \left(15 \times 3 + 2\right) / 3 = \frac{45 + 2}{3} = \frac{47}{3}
\]

For \(21 \frac{3}{4}\):
\[
21 \frac{3}{4} = \left(21 \times 4 + 3\right) / 4 = \frac{84 + 3}{4} = \frac{87}{4}
\]

2. Now we need to subtract the two improper fractions:
\[
15 \frac{2}{3} - 21 \frac{3}{4} = \frac{47}{3} - \frac{87}{4}
\]

3. To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12.

Convert both fractions to have the denominator of 12:

For \(\frac{47}{3}\):
\[
\frac{47}{3} = \frac{47 \times 4}{3 \times 4} = \frac{188}{12}
\]

For \(\frac{87}{4}\):
\[
\frac{87}{4} = \frac{87 \times 3}{4 \times 3} = \frac{261}{12}
\]

4. Now we can subtract:
\[
\frac{188}{12} - \frac{261}{12} = \frac{188 - 261}{12} = \frac{-73}{12}
\]

5. The result is \(-\frac{73}{12}\). We can express this as a mixed number:

\(-\frac{73}{12}\) can be written as \(-6 \frac{1}{12}\) because \(73 \div 12 = 6\) with a remainder of \(1\).

Therefore, the final answer is:
\[
-6 \frac{1}{12}
\]