All Operations in the Order of Operations: Addition, subtraction, multiplication and division

Complete the exercise:

$4-5\times7+3=$

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:

$4-35=-31$

$-31+3=-28$

-28

$8:2(2+2)=$

Let's start with the part inside the parentheses. 

$2+2=4$
Then we will solve the exercise from left to right 

$8:2=4$
$4 × (4)=16$

The answer: $16$

16

$12:3(1+1)=$

First, we perform the operation inside the parentheses:

$12:3(2)$

When there is no mathematical operation between parentheses and a number, we assume it is a multiplication.

Therefore, we can also write the exercise like this:

$12:3\times2$

Here we solve from left to right:

$12:3\times2=4\times2=8$

8

Complete the exercise:

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

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:

$2+18=20$

$20-21=-1$

$-1+1=0$

0

$20-(1+9:9)=$

First, we solve the exercise in the parentheses

$(1+9:9)=$

According to the order of operations, we first divide and then add:

$1+1=2$

Now we obtain the exercise:

$20-2=18$

$18$

$1+2\times3-7:4=$

According to the rules of the order of arithmetic operations, we must first enclose both the multiplication and division exercises inside of parentheses:

$1+(2\times3)-(7:4)=$

We then solve the exercises within the parentheses:

$2\times3=6$

$7:4=\frac{7}{4}$

We obtain the following:

$1+6-\frac{7}{4}=$

We continue by solving the exercise from left to right:

$1+6=7$

$7-\frac{7}{4}=$

Lastly we break down the numerator of the fraction with a sum exercise as seen below:

$7-(\frac{4+3}{4})$

$7-(\frac{4}{4}+\frac{3}{4})$

$7-(1+\frac{3}{4})$

$7-1\frac{3}{4}=5\frac{1}{4}$

$5\frac{1}{4}$

$19\times(20-4\times5)=$

First, we solve the exercise in the parentheses

$(20-4\times5)=$

According to the order of operations, we first multiply and then subtract:

$20-20=0$

Now we obtain the exercise:

$19\times0=0$

0

$5-2\times\frac{1}{2}+1=$

בשלב הראשון של התרגיל יש לחשב את הכפל.

$2\times\frac{1}{2}=\frac{2}{1}\times\frac{1}{2}=\frac{2}{2}=1$

מכאן ניתן להמשיך לשאר פעולות החיבור והחיסור, מימין לשמאל.

$5-1+1=5$

5

Solve the exercise:

$3:4\cdot(7-1)+3=$

First, we solve the exercise within the parentheses:

$3:4\cdot6+3=$

$\frac{3}{4}\times6+3=$

We multiply:

$\frac{18}{4}+3=$

$4\frac{1}{2}+3=7\frac{1}{2}$

$7\frac{1}{2}$

$12:(4\times2-\frac{9}{3})=$

Given that, according to the rules of the order of operations, parentheses come first, we will first solve the exercise that appears within the parentheses.

$4\times2-\frac{9}{3}=$

We solve the multiplication exercise:

$4\times2=8$

We divide the fraction (numerator by denominator)$\frac{9}{3}=3$

And now the exercise obtained within the parentheses is$8-3=5$

Finally, we divide:$12:5=\frac{12}{5}$

$\frac{12}{5}$

$\frac{21:\sqrt{49}+2}{8-(2+2\times3)}=$

In the numerator we solve the square root exercise:

$\sqrt{49}=7$

In the denominator we solve the exercise within parentheses:

$(2+2\times3)=$

$2+6=8$

The exercise we now have is:

$\frac{21:7+2}{8-8}=$

We solve the exercise in the numerator of fractions from left to right:

$21:7=3$

$3+2=5$

We obtain the exercise:

$\frac{5}{8-8}=\frac{5}{0}$

Since it is impossible for the denominator of the fraction to be 0, it is impossible to solve the exercise.

Cannot be solved

$225:[(26-6:3)\times5]=$

First, we solve the exercise within the innermost parentheses:

$(26-6:3)=$

According to the order of operations, we first divide and then subtract:

$26-2=24$

Now we obtain the exercise:

$225:(24\times5)=$

We solve the multiplication exercise and then divide:

$225:120=1.875$

1.875

$\frac{12-15:3\cdot2}{10:(2+3)}=$

We start by solving the exercise in the numerator and then solve the exercise in the denominator.

We know that multiplication and division operations come before addition and subtraction operations, so first we will divide 15:3 and then multiply the result by 2:

$15:3=5$

$12-5\times2=12-10=2$

The result of the numerator is 2 and now we will solve the exercise that appears in the denominator.

It is known that according to the rules of the order of operations, the exercise that appears between parentheses goes first, so we first solve the exercise$2+3=5$

$$Now, we solve the division exercise:$10:5=2$

The result we get in the denominator is 2.

Finally, divide the numerator by the denominator:

$\frac{2}{2}=1$

1

What is the result of the following equation?

$36-4\div2$

34

$0.6\times(1+2)=$

1.8