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Original Question
Which expression is A, in order to have a true equation A=5- 11/3 a.A=15/3+1/3 b.A=1 2/3+1/3 c.A=-1 2/3+1/3 d.A=1 2/3-1/3Asked by Alex
which expression is A. in order to have a true equation A=5-11/3
A=1 2/3+1/3
A=15/3+1/3
A=-1 2/3+1/3
A=1 2/3-1/3
A=1 2/3+1/3
A=15/3+1/3
A=-1 2/3+1/3
A=1 2/3-1/3
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Answered by
GPT-4o mini
To determine which expression is equal to \( A \) in the equation \( A = 5 - \frac{11}{3} \), we first need to simplify \( A \).
1. **Calculate \( 5 - \frac{11}{3} \):**
- Convert \( 5 \) to a fraction: \( 5 = \frac{15}{3} \).
- Now, perform the subtraction:
\[
A = \frac{15}{3} - \frac{11}{3} = \frac{15 - 11}{3} = \frac{4}{3}
\]
Now that we know \( A = \frac{4}{3} \), we can compare it with the options provided:
1. **Option: \( A = 1 \frac{2}{3} + \frac{1}{3} \)**
- Convert to improper fraction:
\[
1 \frac{2}{3} = \frac{5}{3}
\]
- Therefore:
\[
A = \frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2
\]
2. **Option: \( A = 1 \frac{2}{3} + \frac{1}{3} (Revisited) \)**
- Same result as before: \( A = 2 \).
3. **Option: \( A = \frac{15}{3} + \frac{1}{3} \)**
- Adding these gives:
\[
A = \frac{15 + 1}{3} = \frac{16}{3}
\]
4. **Option: \( A = -1 \frac{2}{3} + \frac{1}{3} \)**
- Convert to improper fraction:
\[
-1 \frac{2}{3} = -\frac{5}{3}
\]
- Therefore:
\[
A = -\frac{5}{3} + \frac{1}{3} = -\frac{5 - 1}{3} = -\frac{4}{3}
\]
5. **Option: \( A = 1 \frac{2}{3} - \frac{1}{3} \)**
- Again, convert to improper fraction:
\[
1 \frac{2}{3} = \frac{5}{3}
\]
- So:
\[
A = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}
\]
From this analysis, the option that gives us \( A = \frac{4}{3} \) is:
**\( A = 1 \frac{2}{3} - \frac{1}{3} \)**.
1. **Calculate \( 5 - \frac{11}{3} \):**
- Convert \( 5 \) to a fraction: \( 5 = \frac{15}{3} \).
- Now, perform the subtraction:
\[
A = \frac{15}{3} - \frac{11}{3} = \frac{15 - 11}{3} = \frac{4}{3}
\]
Now that we know \( A = \frac{4}{3} \), we can compare it with the options provided:
1. **Option: \( A = 1 \frac{2}{3} + \frac{1}{3} \)**
- Convert to improper fraction:
\[
1 \frac{2}{3} = \frac{5}{3}
\]
- Therefore:
\[
A = \frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2
\]
2. **Option: \( A = 1 \frac{2}{3} + \frac{1}{3} (Revisited) \)**
- Same result as before: \( A = 2 \).
3. **Option: \( A = \frac{15}{3} + \frac{1}{3} \)**
- Adding these gives:
\[
A = \frac{15 + 1}{3} = \frac{16}{3}
\]
4. **Option: \( A = -1 \frac{2}{3} + \frac{1}{3} \)**
- Convert to improper fraction:
\[
-1 \frac{2}{3} = -\frac{5}{3}
\]
- Therefore:
\[
A = -\frac{5}{3} + \frac{1}{3} = -\frac{5 - 1}{3} = -\frac{4}{3}
\]
5. **Option: \( A = 1 \frac{2}{3} - \frac{1}{3} \)**
- Again, convert to improper fraction:
\[
1 \frac{2}{3} = \frac{5}{3}
\]
- So:
\[
A = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}
\]
From this analysis, the option that gives us \( A = \frac{4}{3} \) is:
**\( A = 1 \frac{2}{3} - \frac{1}{3} \)**.
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