# The Commutative property - Examples, Exercises and Solutions

## What is the commutative property?

The commutative property is an algebraic principle that allows us to "play" with the position that different elements occupy in multiplication and addition exercises without affecting the final result. Our objective in using the commutative property is to make the resolution of the exercise simpler from the point of view of the calculations.

As we have already said, the commutative property can be applied in the case of addition and multiplication.

In other words:

If we change the place of certain elements in the exercise or equation the result will be the same.

In addition operations we can change the place of the addends and arrive at the same result.
That is:
$a+b=b+a$
Same as in algebraic expressions:
$X+número=número+X$

Regardless of the order in which we add the terms and no matter how many addends there are, the result will always be the same.

Commutative property of multiplication:

In multiplication operations we can change the place of the terms and arrive at the same result.
That is:
$a\times b=b\times a$
Same as in algebraic expressions:
$X\times número=número\times X$

Regardless of the order in which we multiply the factors and no matter how many there are in the exercise, the product will always be the same.

Note - The commutative property does not act in this way in subtraction and division operations.

## examples with solutions for the commutative property

### Exercise #1

$5\cdot5\cdot5\cdot2\cdot2\cdot2=?$

### Step-by-Step Solution

We use the substitution property and organize the exercise in the following order:

$5\times2\times5\times2\times5\times2=$

We place parentheses in the exercise:

$(5\times2)\times(5\times2)\times(5\times2)=$

We solve from left to right:

$10\times10\times10=$

$(10\times10)\times10=$

$100\times10=1000$

1000

### Exercise #2

$-5+2=$

### Step-by-Step Solution

If we draw a line that starts at negative five and ends at 5

We will go from the point negative five two steps forward (+2) we will arrive at the number negative 3.

$-3$

### Exercise #3

$10-5-2-3=$

### Step-by-Step Solution

Given that the entire exercise is with subtraction, we solve the exercise from left to right:

$10-5=5$

$5-2=3$

$3-3=0$

$0$

### Exercise #4

$4-2+2-4=$

### Step-by-Step Solution

Given that we are referring to addition and subtraction exercises, we solve the exercise from left to right:

$4-2=2$

$2+2=4$

$4-4=0$

$0$

### Exercise #5

$3-2+10-x=$

### Step-by-Step Solution

We solve the exercise from left to right:

$3-2=1$

$1+10=11$

Now we obtain:

$11-x$

$11-x$

### Exercise #6

$11\times3+7=$

### Step-by-Step Solution

In this exercise, it is not possible to use the substitution property, therefore we solve it as is from left to right according to the order of arithmetic operations.

That is, we first solve the multiplication exercise and then we add:

$11\times3=33$

$33+7=40$

$40$

### Exercise #7

$12\times13+14=$

### Step-by-Step Solution

According to the order of operations, we start with the multiplication exercise and then with the addition.

$12\times13=156$

Now we get the exercise:

$156+14=170$

$170$

### Exercise #8

$\frac{1}{4}\times4+2=$

### Step-by-Step Solution

According to the order of operations, we first solve the multiplication exercise:

We add the 4 in the numerator of the fraction:

$\frac{1\times4}{4}+2=$

We solve the exercise in the numerator of the fraction and obtain:

$\frac{4}{4}+2=1+2=3$

$3$

### Exercise #9

$-2-4+6-1=$

### Step-by-Step Solution

According to the order of operations, we solve the exercise from left to right:

$-2-4=-6$

$-6+6=0$

$0-1=-1$

$-1$

### Exercise #10

$4:2+2=$

### Step-by-Step Solution

According to the order of operations, we first solve the division exercise:

$4:2=2$

Now we obtain the exercise:

$2+2=4$

$4$

### Exercise #11

Solve:

$2-3+1$

### Step-by-Step Solution

We use the substitution property and add parentheses for the addition operation:

$(2+1)-3=$

Now, we solve the exercise according to the order of operations:

$2+1=3$

$3-3=0$

0

### Exercise #12

Solve:

$3-4+2+1$

### Step-by-Step Solution

We will use the substitution property to arrange the exercise a bit more comfortably, we will add parentheses to the addition operation:
$(3+2+1)-4=$
We first solve the addition, from left to right:
$3+2=5$

$5+1=6$
And finally, we subtract:

$6-4=2$

2

### Exercise #13

Solve:

$-5+4+1-3$

### Step-by-Step Solution

According to the order of operations, addition and subtraction are on the same level and, therefore, must be resolved from left to right.

However, in the exercise we can use the substitution property to make solving simpler.

-5+4+1-3

4+1-5-3

5-5-3

0-3

-3

$-3$

### Exercise #14

$7+4+3+6=\text{?}$

### Step-by-Step Solution

To make solving the exercise easier, we try to add numbers that give us a result of 10.

Let's keep in mind that:

$7+3=10$

$6+4=10$

Now, we obtain a more convenient exercise to solve:

$10+10=20$

20

### Exercise #15

$19+34+21+10+6=\text{?}$

### Step-by-Step Solution

To make solving easier, we try to add numbers that give us a round result.

Keep in mind that:

$19+21=40$

$34+6=40$

Now, we get a more convenient exercise to solve:

$40+40+10=80+10=90$