The Commutative property: Addition, subtraction, multiplication and division

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:

$2+2=4$

According to the rules of the order of operations, multiplication and division precede addition and subtraction.

We isolate the multiplication exercise in parentheses and solve.

$(7\times3)=21$

Now, the exercise we're left with is: $21+8-4-7=$

We solve the exercise from left to right. We isolate the next part of the expression with parentheses to avoid confusion$(21+8)=29$

Now, the exercise obtained is: $29-4-7=$

We continue solving from left to right and isolate the next part of the expression in parentheses.

$(29-4)=25$

Now, the expression obtained is: $25-7=18$

First, we solve the fraction:

$\frac{36}{12}=3$

Now we get the exercise:

$3-4+30=$

We solve the exercise from left to right:

$3-4=-1$

$-1+30=29$

According to the order of operations, we first solve the multiplication exercise and then the addition/ subtraction.

$17\times10=170$

$170-70=100$

According to the rules of the order of arithmetic operations, we must first solve the multiplication exercises.

We place them inside of parentheses to avoid confusion during the solution:

$4-(5\times7)+3=$

We then solve the multiplication exercises:

$4-35+3=$

Lastly we solve the rest of the exercise from left to right:

$4-35=-31$

$-31+3=-28$

According to the rules of the order of arithmetic operations, we must first solve the multiplication exercises.

We place them inside of parentheses in order to avoid confusion during the solution:

$2+(3\times6)-(3\times7)+1=$

We then solve the multiplication exercises:

$2+18-21+1=$

Lastly we solve the rest of the exercise from left to right:

$2+18=20$

$20-21=-1$

$-1+1=0$

According to the rules of the order of arithmetic operations, we begin by placing the multiplication and division exercises inside of parentheses:

$8+(3\times4)-2+1=$

We then solve the exercise within the parentheses:

$3\times4=12$

We obtain the following :

$8+12-2+1=$

Finally we solve the exercise from left to right:

$8+12=20$

$20-2=18$

$18+1=19$

According to the rules of the order of arithmetic operations, we must first place the multiplication and division exercises within parentheses:

$(4\times7\times2:4)-(1\times9)=$

We then proceed to solve the exercise in parentheses from left to right:

$4\times7=28$

$28\times2=56$

$56:4=14$

We obtain the following exercise:

$14-9=5$

According to the rules of the order of arithmetic operations, we must first place multiplication and division exercises within parentheses:

$7+(3\times7)-(3\times4)+5=$

We then proceed to solve the exercise inside of the parentheses:

$3\times7=21$

$3\times4=12$

We obtain the following exercise:

$7+21-12+5=$

Lastly we solve the exercise from left to right:

$21+7=28$

$28-12=16$

$16+5=21$

According to the rules of the order of arithmetic operations, we must first place the multiplication and division exercises within parentheses:

$2-(6:2)+(5\times2)=$

We then solve the exercise inside of the parentheses:

$6:2=3$

$5\times2=10$

We obtain the following exercise:

$2-3+10=$

Finally we solve the exercise from left to right:

$2-3=-1$

$-1+10=9$

According to the rules of the order of arithmetic operations, we must first place the multiplication and division exercises within parentheses:

$5-(30:2\times3)+10=$

We then solve the exercise inside of the parentheses from left to right:

$30:2=15$

$15\times3=45$

We obtain the following exercise:

$5-45+10=$

Finally we solve the exercise from left to right:

$5-45=-40$

$-40+10=-30$

According to the rules for the order of operations, we will isolate the multiplication operations in parentheses:

$(9\cdot5\cdot5)+7-5+82-2=$

We solve the exercise in parentheses from left to right:

$9\times5=45$

$45\times5=225$

Now, we obtain the exercise:

$225+7-5+82-2=$

We solve the exercise from left to right

$225+7=232$

$232-5=227$

$227+82=309$

$309-2=307$

According to the rules of the order of operations, we first solve the exercise within parentheses from left to right:

$32-15=17$

$17-12=5$

Now, we obtain the exercise:

$5\times34\times12=$

According to the rules of the order of operations, since this is an exercise with only a multiplication operation, we rearrange the numbers to make the solution easier for us:

$5\times12\times34=$

We solve the exercise from left to right:

$5\times12=60$

$60\times34=2040$

According to the order of operations, we first place the multiplication exercises in parentheses:

$(5\times10)-(2\times25)=$

We solve the multiplication exercise in parentheses:

$5\times10=50$

$2\times25=50$

Now we obtain the exercise:

$50-50=0$

According to the order of operations, we first place the multiplication exercise in parentheses:

$-10-3-(5\times3)=$

We solve the multiplication exercise:

$5\times3=15$

Now we obtain the exercise:

$-10-3-15=$

We solve the exercise from left to right:

$-10-3=-13$

$-13-15=-28$

First we solve the fractions:

$\frac{40}{2}=20$

$\frac{10}{5}=2$

Now we get the exercise:

$20+2-2=$

Keep in mind that the following exercise gives us the result zero:

$2-2=0$

Therefore:

$20+0=20$