Question

Using the Commutative Property, fill in the blanks so that the two algebraic expressions are equivalent.(1 point)
(14)
(
)(x)=(7)
(
)(14)

(its not 1 and 2)

Answer 1

To use the Commutative Property of multiplication, we can rearrange the terms in the expressions. The Commutative Property states that \(a \times b = b \times a\).

Given the expressions \( (14)( \_\_ )(x) = (7)( \_\_ )(14) \), we want to find the blanks so that both sides are equivalent.

1. Starting with the left side \( (14)(\_\_)(x) \):
- We can use \( (1)(x) \), because \(1\) does not change the value.

2. For the right side \( (7)(\_\_)(14) \):
- To make it equivalent, we can use \( (2) \).

So, filling in the blanks:
\[
(14)(1)(x) = (7)(2)(14)
\]

Checking:
- Left: \( (14)(1)(x) = 14x \)
- Right: \( (7)(2)(14) = 14 \times 2 \times 7 = 14 \times 14 = 14^2 \).

Thus, the blanks can be filled with:
\[
(1) \quad \text{and} \quad (2)
\]

However, since the conditions specifically state that "it’s not 1 and 2," let’s consider other possible values that could satisfy the equation using the commutative property.

Since \(14\) is \(2 \times 7\), we could equilibrium by recognizing that on both sides \(14\) must relate with \(7\).
- Instead, fill the first blank with \(7\) to balance it out effectively as:
\[
(14)(7)(x) = (7)(2)(14)
\]
Both sides equate as:
\[
98x = 98
\]

To conclude effectively, plausible fill-ins could be:
\[
(2) \quad \text{and} \quad (7)
\] where \(x\) and \(14\) simply differ on re-ordering equations. Both maintain the properties and validate conditions equitably.