commutative Property Addition Worksheets for 1st and 2nd Grade - Free Printable

Problem Description:

The task involves solving addition problems using the Commutative Property of Addition. The commutative property states that the order in which numbers are added does not affect the sum. Mathematically, this is expressed as:

\[
a + b = b + a
\]

For problems with three addends, the property can be extended to show that rearranging the order of the numbers does not change the result. For example:

\[
a + b + c = b + a + c = c + b + a
\]

The worksheet provides several problems with three addends and asks to solve them in two different orders.

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Solution Approach:

1. Understand the Commutative Property: Rearrange the order of the addends in each problem.
2. Solve Each Problem Twice: Solve the problem in the given order and then in a different order to verify that the sum remains the same.
3. Verify the Results: Ensure both solutions yield the same answer.

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Step-by-Step Solutions:

#### Problem 1: \( 4 + 5 + 2 \)
- Original Order: \( 4 + 5 + 2 \)
\[
4 + 5 = 9, \quad 9 + 2 = 11
\]
So, \( 4 + 5 + 2 = 11 \).

- Rearranged Order: \( 5 + 2 + 4 \)
\[
5 + 2 = 7, \quad 7 + 4 = 11
\]
So, \( 5 + 2 + 4 = 11 \).

- Answer: Both orders give the same result: \( \boxed{11} \).

#### Problem 2: \( 3 + 5 + 0 \)
- Original Order: \( 3 + 5 + 0 \)
\[
3 + 5 = 8, \quad 8 + 0 = 8
\]
So, \( 3 + 5 + 0 = 8 \).

- Rearranged Order: \( 5 + 0 + 3 \)
\[
5 + 0 = 5, \quad 5 + 3 = 8
\]
So, \( 5 + 0 + 3 = 8 \).

- Answer: Both orders give the same result: \( \boxed{8} \).

#### Problem 3: \( 3 + 2 + 7 \)
- Original Order: \( 3 + 2 + 7 \)
\[
3 + 2 = 5, \quad 5 + 7 = 12
\]
So, \( 3 + 2 + 7 = 12 \).

- Rearranged Order: \( 2 + 7 + 3 \)
\[
2 + 7 = 9, \quad 9 + 3 = 12
\]
So, \( 2 + 7 + 3 = 12 \).

- Answer: Both orders give the same result: \( \boxed{12} \).

#### Problem 4: \( 7 + 2 + 8 \)
- Original Order: \( 7 + 2 + 8 \)
\[
7 + 2 = 9, \quad 9 + 8 = 17
\]
So, \( 7 + 2 + 8 = 17 \).

- Rearranged Order: \( 2 + 8 + 7 \)
\[
2 + 8 = 10, \quad 10 + 7 = 17
\]
So, \( 2 + 8 + 7 = 17 \).

- Answer: Both orders give the same result: \( \boxed{17} \).

#### Problem 5: \( 8 + 5 + 1 \)
- Original Order: \( 8 + 5 + 1 \)
\[
8 + 5 = 13, \quad 13 + 1 = 14
\]
So, \( 8 + 5 + 1 = 14 \).

- Rearranged Order: \( 5 + 1 + 8 \)
\[
5 + 1 = 6, \quad 6 + 8 = 14
\]
So, \( 5 + 1 + 8 = 14 \).

- Answer: Both orders give the same result: \( \boxed{14} \).

#### Problem 6: \( 7 + 6 + 9 \)
- Original Order: \( 7 + 6 + 9 \)
\[
7 + 6 = 13, \quad 13 + 9 = 22
\]
So, \( 7 + 6 + 9 = 22 \).

- Rearranged Order: \( 6 + 9 + 7 \)
\[
6 + 9 = 15, \quad 15 + 7 = 22
\]
So, \( 6 + 9 + 7 = 22 \).

- Answer: Both orders give the same result: \( \boxed{22} \).

#### Problem 7: \( 3 + 9 + 11 \)
- Original Order: \( 3 + 9 + 11 \)
\[
3 + 9 = 12, \quad 12 + 11 = 23
\]
So, \( 3 + 9 + 11 = 23 \).

- Rearranged Order: \( 9 + 11 + 3 \)
\[
9 + 11 = 20, \quad 20 + 3 = 23
\]
So, \( 9 + 11 + 3 = 23 \).

- Answer: Both orders give the same result: \( \boxed{23} \).

#### Problem 8: \( 4 + 6 + 7 \)
- Original Order: \( 4 + 6 + 7 \)
\[
4 + 6 = 10, \quad 10 + 7 = 17
\]
So, \( 4 + 6 + 7 = 17 \).

- Rearranged Order: \( 6 + 7 + 4 \)
\[
6 + 7 = 13, \quad 13 + 4 = 17
\]
So, \( 6 + 7 + 4 = 17 \).

- Answer: Both orders give the same result: \( \boxed{17} \).

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Final Answers:

\[
\boxed{11, 8, 12, 17, 14, 22, 23, 17}
\]

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Explanation:

The commutative property of addition ensures that the sum remains unchanged regardless of the order in which the numbers are added. By solving each problem in two different orders, we verified this property for all given problems. This exercise reinforces the understanding of the commutative property and its application in arithmetic.