Question

6(y+3) and 6⋅y+3 6 left parenthesis y plus 3 right parenthesis and 6 times y plus 3 −5(a−1) and −5⋅a−5 negative 5 left parenthesis a minus 1 right parenthesis and negative 5 times a minus 5 8⋅c+40 and 8(c−5) 8 times c plus 40 and 8 left parenthesis c minus 5 right parenthesis 12⋅t+6 and 12(t+1/2)

Answer 1

Let's analyze and simplify each of the pairs you provided.

1. **For 6(y + 3) and 6y + 3:**
- **6(y + 3)** can be expanded as:
\[
6y + 18
\]
- **6y + 3** remains unchanged.

They are not equivalent since \( 6(y + 3) = 6y + 18 \) and \( 6y + 3 \) are different.

2. **For -5(a - 1) and -5a - 5:**
- **-5(a - 1)** can be expanded as:
\[
-5a + 5
\]
- **-5a - 5** remains unchanged.

Thus, they are also not equivalent since \( -5(a - 1) = -5a + 5 \) and \( -5a - 5 \) are different.

3. **For 8c + 40 and 8(c - 5):**
- **8(c - 5)** can be expanded as:
\[
8c - 40
\]
- **8c + 40** remains unchanged.

They are not equivalent since \( 8(c - 5) = 8c - 40 \) and \( 8c + 40 \) are different.

4. **For 12t + 6 and 12(t + 1/2):**
- **12(t + 1/2)** can be expanded as:
\[
12t + 6
\]
- **12t + 6** remains unchanged.

They are equivalent since \( 12(t + 1/2) = 12t + 6 \).

### Summary:
1. \( 6(y + 3) \neq 6y + 3 \)
2. \( -5(a - 1) \neq -5a - 5 \)
3. \( 8c + 40 \neq 8(c - 5) \)
4. \( 12t + 6 = 12(t + 1/2) \)

If you have any questions or need further assistance, feel free to ask!