The Commutative Property of Multiplication - Examples, Exercises and Solutions

Since the exercise involves only multiplication, we will use the commutative property to simplify the calculation:

$2\times15\times8=$

Now let's solve the multiplication on the right:

$15\times8=120$

Now we get the expression:

$2\times120=240$

Since the exercise only involves addition, we will use the commutative property to calculate more conveniently:

$5+5+14=$

We will solve the exercise from left to right:

$5+5=10$

$10+14=24$

Since the exercise only involves addition, we will use the commutative property to solve it more conveniently:

$8+2+9=$

Let's solve the exercise from left to right:

$8+2=10$

$10+9=19$

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:

$10\times10\times10=$

$(10\times10)\times10=$

$100\times10=1000$

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.

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

$10-5=5$

$5-2=3$

$3-3=0$

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

$4-2=2$

$2+2=4$

$4-4=0$

We solve the exercise from left to right:

$3-2=1$

$1+10=11$

Now we obtain:

$11-x$

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:

$11\times3=33$

$33+7=40$

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:

$156+14=170$

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:

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

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

$-2-4=-6$

$-6+6=0$

$0-1=-1$

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

$4:2=2$

Now we obtain the exercise:

$2+2=4$

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:

$2+1=3$

$3-3=0$

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:

$6-4=2$