Answer :
To solve the problem of finding a common denominator for the fractions [tex]\(\frac{3}{4}\)[/tex] and [tex]\(\frac{7}{8}\)[/tex], follow these steps:
1. Identify the denominators: The fractions have denominators 4 and 8.
2. Find the least common denominator (LCD): The LCD of 4 and 8 is 8, since it is the smallest number that both denominators can divide into without leaving a remainder.
3. Adjust the fractions:
- The fraction [tex]\(\frac{3}{4}\)[/tex] needs to be converted to have a denominator of 8. To do this:
- Multiply both the numerator and the denominator by 2 (because [tex]\( \frac{8}{4} = 2 \)[/tex]):
[tex]\[
\frac{3 \times 2}{4 \times 2} = \frac{6}{8}
\][/tex]
- The fraction [tex]\(\frac{7}{8}\)[/tex] already has a denominator of 8, so it remains [tex]\(\frac{7}{8}\)[/tex].
4. Compare the options: We need to find which option matches [tex]\(\frac{6}{8}\)[/tex] and [tex]\(\frac{7}{8}\)[/tex] converted to an equivalent representation if necessary.
- Option A: [tex]\(\frac{12}{16}\)[/tex] and [tex]\(\frac{15}{16}\)[/tex]
- Option B: [tex]\(\frac{15}{20}\)[/tex] and [tex]\(\frac{18}{20}\)[/tex]
- Option C: [tex]\(\frac{9}{12}\)[/tex] and [tex]\(\frac{21}{24}\)[/tex]
- Option D: [tex]\(\frac{24}{32}\)[/tex] and [tex]\(\frac{28}{32}\)[/tex]
5. Verify with calculations:
- If we simplify [tex]\(\frac{24}{32}\)[/tex], we get [tex]\(\frac{6}{8}\)[/tex].
- [tex]\(\frac{28}{32}\)[/tex] simplifies to [tex]\(\frac{7}{8}\)[/tex].
Therefore, Option D: [tex]\(\frac{24}{32}\)[/tex] and [tex]\(\frac{28}{32}\)[/tex] correctly represents the fractions with a common denominator. So, the answer is Option D.
1. Identify the denominators: The fractions have denominators 4 and 8.
2. Find the least common denominator (LCD): The LCD of 4 and 8 is 8, since it is the smallest number that both denominators can divide into without leaving a remainder.
3. Adjust the fractions:
- The fraction [tex]\(\frac{3}{4}\)[/tex] needs to be converted to have a denominator of 8. To do this:
- Multiply both the numerator and the denominator by 2 (because [tex]\( \frac{8}{4} = 2 \)[/tex]):
[tex]\[
\frac{3 \times 2}{4 \times 2} = \frac{6}{8}
\][/tex]
- The fraction [tex]\(\frac{7}{8}\)[/tex] already has a denominator of 8, so it remains [tex]\(\frac{7}{8}\)[/tex].
4. Compare the options: We need to find which option matches [tex]\(\frac{6}{8}\)[/tex] and [tex]\(\frac{7}{8}\)[/tex] converted to an equivalent representation if necessary.
- Option A: [tex]\(\frac{12}{16}\)[/tex] and [tex]\(\frac{15}{16}\)[/tex]
- Option B: [tex]\(\frac{15}{20}\)[/tex] and [tex]\(\frac{18}{20}\)[/tex]
- Option C: [tex]\(\frac{9}{12}\)[/tex] and [tex]\(\frac{21}{24}\)[/tex]
- Option D: [tex]\(\frac{24}{32}\)[/tex] and [tex]\(\frac{28}{32}\)[/tex]
5. Verify with calculations:
- If we simplify [tex]\(\frac{24}{32}\)[/tex], we get [tex]\(\frac{6}{8}\)[/tex].
- [tex]\(\frac{28}{32}\)[/tex] simplifies to [tex]\(\frac{7}{8}\)[/tex].
Therefore, Option D: [tex]\(\frac{24}{32}\)[/tex] and [tex]\(\frac{28}{32}\)[/tex] correctly represents the fractions with a common denominator. So, the answer is Option D.
