Question
Which expressions are equivalent to
\[x+x+x + 2\] ?
Choose 2 answers:
Choose 2 answers:
(Choice A)
\[3x + 2\]
A
\[3x + 2\]
(Choice B)
\[3 + 2x\]
B
\[3 + 2x\]
(Choice C)
\[3(x+2)\]
C
\[3(x+2)\]
(Choice D)
\[2(x+1)+x\]
D
\[2(x+1)+x\]
(Choice E)
\[5x\]
E
\[5x\]
\[x+x+x + 2\] ?
Choose 2 answers:
Choose 2 answers:
(Choice A)
\[3x + 2\]
A
\[3x + 2\]
(Choice B)
\[3 + 2x\]
B
\[3 + 2x\]
(Choice C)
\[3(x+2)\]
C
\[3(x+2)\]
(Choice D)
\[2(x+1)+x\]
D
\[2(x+1)+x\]
(Choice E)
\[5x\]
E
\[5x\]
Answers
GPT-4o mini
To simplify the expression \(x + x + x + 2\), we can combine the \(x\) terms.
There are three \(x\) terms, so:
\[
x + x + x = 3x
\]
Adding the constant \(2\), we have:
\[
3x + 2
\]
Now, we'll check each of the answer choices to see if they are equivalent to \(3x + 2\):
**Choice A: \(3x + 2\)**
This is equal to the original expression.
**Choice B: \(3 + 2x\)**
This is not equivalent to \(3x + 2\).
**Choice C: \(3(x + 2)\)**
Distributing gives \(3x + 6\), which is not equal to \(3x + 2\).
**Choice D: \(2(x + 1) + x\)**
Distributing \(2\) gives \(2x + 2 + x = 3x + 2\), which matches our expression.
**Choice E: \(5x\)**
This is not equivalent to \(3x + 2\).
The equivalent expressions are:
- **Choice A: \(3x + 2\)**
- **Choice D: \(2(x + 1) + x\)**
So the two correct answers are **Choice A** and **Choice D**.
There are three \(x\) terms, so:
\[
x + x + x = 3x
\]
Adding the constant \(2\), we have:
\[
3x + 2
\]
Now, we'll check each of the answer choices to see if they are equivalent to \(3x + 2\):
**Choice A: \(3x + 2\)**
This is equal to the original expression.
**Choice B: \(3 + 2x\)**
This is not equivalent to \(3x + 2\).
**Choice C: \(3(x + 2)\)**
Distributing gives \(3x + 6\), which is not equal to \(3x + 2\).
**Choice D: \(2(x + 1) + x\)**
Distributing \(2\) gives \(2x + 2 + x = 3x + 2\), which matches our expression.
**Choice E: \(5x\)**
This is not equivalent to \(3x + 2\).
The equivalent expressions are:
- **Choice A: \(3x + 2\)**
- **Choice D: \(2(x + 1) + x\)**
So the two correct answers are **Choice A** and **Choice D**.