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!