The Commutative property - Examples, Exercises and Solutions

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

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

$-5+2=$

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.

$-3$

$10-5-2-3=$

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

$10-5=5$

$5-2=3$

$3-3=0$

$0$

$4-2+2-4=$

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

$4-2=2$

$2+2=4$

$4-4=0$

$0$

$3-2+10-x=$

We solve the exercise from left to right:

$3-2=1$

$1+10=11$

Now we obtain:

$11-x$

$11-x$

$11\times3+7=$

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:

$11\times3=33$

$33+7=40$

$40$

$12\times13+14=$

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:

$156+14=170$

$170$

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

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:

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

$3$

$-2-4+6-1=$

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

$-2-4=-6$

$-6+6=0$

$0-1=-1$

$-1$

$4:2+2=$

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

$4:2=2$

Now we obtain the exercise:

$2+2=4$

$4$

Solve:

$2-3+1$

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

Solve:

$3-4+2+1$

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

Solve:

$-5+4+1-3$

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

$-3$

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

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

Let's keep in mind that:

$7+3=10$

$6+4=10$

Hence we obtain a more manageable exercise to solve:

$10+10=20$

20

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

In order to simplify our calculations, 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 manageable exercise to solve:

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

90