Asked by nobody
Select the equation that has infinitely many solutions.
A. 3x + 5 = 3x - 5
B. 3 - 3x = -3(x - 1)
C. 4x + 1 = 4(x + 1)
D. 3x = 2x + 1
A. 3x + 5 = 3x - 5
B. 3 - 3x = -3(x - 1)
C. 4x + 1 = 4(x + 1)
D. 3x = 2x + 1
Answers
Answered by
nobody
are you sure this is correct
Answered by
GPT-4o mini
To determine which equation has infinitely many solutions, we need to simplify each equation and see if they lead to a true statement regardless of the value of \( x \) (indicating infinitely many solutions).
**Option A:** \( 3x + 5 = 3x - 5 \)
Subtract \( 3x \) from both sides:
\( 5 = -5 \)
This is a false statement, so there are no solutions.
**Option B:** \( 3 - 3x = -3(x - 1) \)
Distribute on the right side:
\( 3 - 3x = -3x + 3 \)
Add \( 3x \) to both sides:
\( 3 = 3 \)
This is a true statement, so there are infinitely many solutions.
**Option C:** \( 4x + 1 = 4(x + 1) \)
Distribute on the right side:
\( 4x + 1 = 4x + 4 \)
Subtract \( 4x \) from both sides:
\( 1 = 4 \)
This is a false statement, so there are no solutions.
**Option D:** \( 3x = 2x + 1 \)
Subtract \( 2x \) from both sides:
\( x = 1 \)
This has a single solution.
Therefore, the equation that has infinitely many solutions is **Option B: \( 3 - 3x = -3(x - 1) \)**.
**Option A:** \( 3x + 5 = 3x - 5 \)
Subtract \( 3x \) from both sides:
\( 5 = -5 \)
This is a false statement, so there are no solutions.
**Option B:** \( 3 - 3x = -3(x - 1) \)
Distribute on the right side:
\( 3 - 3x = -3x + 3 \)
Add \( 3x \) to both sides:
\( 3 = 3 \)
This is a true statement, so there are infinitely many solutions.
**Option C:** \( 4x + 1 = 4(x + 1) \)
Distribute on the right side:
\( 4x + 1 = 4x + 4 \)
Subtract \( 4x \) from both sides:
\( 1 = 4 \)
This is a false statement, so there are no solutions.
**Option D:** \( 3x = 2x + 1 \)
Subtract \( 2x \) from both sides:
\( x = 1 \)
This has a single solution.
Therefore, the equation that has infinitely many solutions is **Option B: \( 3 - 3x = -3(x - 1) \)**.
Answered by
GPT-4o mini
Let's double-check the analysis of each equation to ensure accuracy:
### Option A: \( 3x + 5 = 3x - 5 \)
Subtract \( 3x \) from both sides:
\[
5 = -5
\]
This statement is false, so there are no solutions.
### Option B: \( 3 - 3x = -3(x - 1) \)
Start by distributing on the right side:
\[
3 - 3x = -3x + 3
\]
Now add \( 3x \) to both sides:
\[
3 = 3
\]
This is a true statement and implies that the equation holds for any \( x \), which means there are infinitely many solutions.
### Option C: \( 4x + 1 = 4(x + 1) \)
Distribute on the right-hand side:
\[
4x + 1 = 4x + 4
\]
Subtract \( 4x \) from both sides:
\[
1 = 4
\]
This statement is false, so there are no solutions.
### Option D: \( 3x = 2x + 1 \)
Subtract \( 2x \) from both sides:
\[
x = 1
\]
This equation has a single solution.
### Conclusion
After reviewing all the options again, it is confirmed that:
- **Option B** is indeed the correct answer as it represents an identity (true for all \( x \)), and thus it has infinitely many solutions.
Thank you for the opportunity to clarify! If you have any further questions, feel free to ask.
### Option A: \( 3x + 5 = 3x - 5 \)
Subtract \( 3x \) from both sides:
\[
5 = -5
\]
This statement is false, so there are no solutions.
### Option B: \( 3 - 3x = -3(x - 1) \)
Start by distributing on the right side:
\[
3 - 3x = -3x + 3
\]
Now add \( 3x \) to both sides:
\[
3 = 3
\]
This is a true statement and implies that the equation holds for any \( x \), which means there are infinitely many solutions.
### Option C: \( 4x + 1 = 4(x + 1) \)
Distribute on the right-hand side:
\[
4x + 1 = 4x + 4
\]
Subtract \( 4x \) from both sides:
\[
1 = 4
\]
This statement is false, so there are no solutions.
### Option D: \( 3x = 2x + 1 \)
Subtract \( 2x \) from both sides:
\[
x = 1
\]
This equation has a single solution.
### Conclusion
After reviewing all the options again, it is confirmed that:
- **Option B** is indeed the correct answer as it represents an identity (true for all \( x \)), and thus it has infinitely many solutions.
Thank you for the opportunity to clarify! If you have any further questions, feel free to ask.
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