Question

the commutative property practice
complete this assessment to review what youve learned. it will not count toward your grade.
using the commutative property, fill in the blanks to prove the two algebraic expressions are equivalent. (1 point)
$y + \boldsymbol{\times} + x + 7 = 4 + \boldsymbol{\times} + 7 + x$
incorrect

• the commutative property of addition states that changing the order of the addends does not affect the sum.
• the commutative property of addition states that changing the order of the addends does not affect the sum.

Explanation:

Step1: Analyze the left - hand side expression

The left - hand side expression is \(y+\square+x + 7\). We want to match the structure of the right - hand side expression \(4+\square+7 + x\) using the Commutative Property of Addition (which states that \(a + b=b + a\) for addition). Looking at the right - hand side, the first term is 4. So, to make the left - hand side have a term of 4 (by using the Commutative Property to re - order the terms), the blank on the left should be 4. Let's check: If we put 4 in the left blank, the left - hand side becomes \(4+x + 7+y\) (wait, no, the original left - hand side is \(y+\square+x + 7\), if we put 4 in the blank, it's \(y + 4+x + 7\)). Then, using the Commutative Property of Addition, we can re - order the terms \(y+4+x + 7=4 + y+7 + x\)? Wait, no, the right - hand side is \(4+\square+7 + x\). Wait, maybe I misread. Let's look again. The right - hand side is \(4+\square+7 + x\), and the left - hand side is \(y+\square+x + 7\). Wait, no, the correct approach: The Commutative Property of Addition says that the order of addends can be changed. So, for the left - hand side \(y+\square+x + 7\) and the right - hand side \(4+\square+7 + x\), we need to find the missing number. Let's assume the blank is a number. Let's look at the terms. On the right - hand side, the non - variable, non - 7 term is 4. On the left - hand side, the non - variable, non - 7 term (before re - ordering) should be 4. Wait, maybe the left - hand side was supposed to be \(4+\square+x + 7\) and the right - hand side is \(4+\square+7 + x\)? No, the original left - hand side has a \(y\)? Wait, no, maybe there is a typo, and the \(y\) is a mistake, or maybe I missee. Wait, the problem is to make \(y+\square+x + 7=4+\square+7 + x\) using the Commutative Property. Wait, no, that can't be. Wait, maybe the \(y\) is a 4. Maybe it's a typo, and the left - hand side is \(4+\square+x + 7\). Let's proceed with the Commutative Property. The Commutative Property of Addition: \(a + b=b + a\). So, for the expression \(4+x + 7\) (if we correct the left - hand side's \(y\) to 4), we can re - order the terms \(x\) and 7. So \(4+x + 7=4 + 7+x\) (by Commutative Property, \(x+7 = 7 + x\)). Wait, the blanks: Let's look at the original problem again. The left - hand side is \(y+\square+x + 7\), the right - hand side is \(4+\square+7 + x\). Wait, maybe the first blank (on the left) is 4, and the second blank (on both sides) is \(y\)? No, that doesn't make sense. Wait, maybe the \(y\) is a mistake, and it's supposed to be 4. Let's assume that the left - hand side is \(4+\square+x + 7\) and the right - hand side is \(4+\square+7 + x\). Then, using the Commutative Property of Addition, \(4+x + 7=4 + 7+x\) because \(x+7 = 7 + x\) (Commutative Property of Addition: \(a + b=b + a\), here \(a=x\), \(b = 7\)). So the blank on the left (the first blank) should be 4, and the blank on the right (the second blank) should be \(x\)? Wait, no, the blanks are the same? Wait, the problem has two blanks, both with \(x\) crossed out. Wait, maybe the blanks are for numbers. Wait, let's start over. The Commutative Property of Addition: For any real numbers \(a\) and \(b\), \(a + b=b + a\). So, we have an expression on the left: \(y+\underline{\quad}+x + 7\) and on the right: \(4+\underline{\quad}+7 + x\). We need to find the value of the underlined part. Let's equate the two expressions. \(y+\underline{\quad}+x + 7=4+\underline{\quad}+7 + x\). Subtract \(\underline{\quad}+x + 7\) from both sides: \(y = 4\). So the first blank (on the left) should be 4, and the second blank (on both sides) should be the same (maybe \(y\) was a typo, and it's 4). So, if we put 4 in the left blank, then the left - hand side is \(4 + 4+x + 7\)? No, that's not right. Wait, I think the correct approach is that the two expressions are \(4+x + 7\) and \(4+7 + x\), and the blank is 4 on the left and 4 on the right? No, the blanks are the same. Wait, the problem says "fill in the blanks to prove the two algebraic expressions are equivalent". So, the left - hand side is \(y+\square+x + 7\), the right - hand side is \(4+\square+7 + x\). Using the Commutative Property, we can see that the first blank (on the left) should be 4 (to match the 4 on the right), and then by the Commutative Property of Addition, \(4+x + 7=4+7 + x\) (re - ordering \(x\) and 7). So the blank on the left is 4, and the blank on the right is also 4? Wait, no, the blanks are the same. So the answer is that the blank should be 4.

Step2: Confirm with the Commutative Property

The Commutative Property of Addition states that \(a + b=b + a\). So, for the expression \(4+x + 7\), we can re - order the addends \(x\) and 7 to get \(4+7 + x\). So, if we fill the blank with 4, then \(4+x + 7=4+7 + x\) (after re - ordering using the Commutative Property), which shows the two expressions are equivalent.

Answer:

The blank should be filled with 4. So the left - hand side blank is 4 and the right - hand side blank is also 4. (Assuming the \(y\) is a typo and should be 4)