Parentheses

First, let's solve only what's inside the parentheses, regardless of whether it involves addition and subtraction, multiplication, or division.
Note that the expression inside the parentheses must be solved according to the correct order of operations - (additional parentheses if any, multiplication and division, then addition and subtraction).
If there are no parentheses at all in the expression, we'll move on to the next step.

Multiplication and division in order.
After opening the parentheses ,if there were any, we proceed to rewrite the exercise in order to avoid confusion. We will solve the multiplication and division in order from left to right.
Once again rewrite the exercise after solving the multiplication and division.

Addition and subtraction in order.
We add and subtract from left to right.
Pay attention! After solving the expression inside of the parentheses, we replace the entire expression in the parentheses with its result and remove the parentheses. Another option is to draw a small arc above the parentheses and write the result above them.

And now? Let's practice!!

Solve the following exercise:
$2\cdot4+(6+2)+1=$

Solution:
Upon observation, we immediately see that the exercise contains parentheses. Parentheses are the first operation in the order of operations, so we'll start with them first.
The expression in the parentheses is $6+2$
We'll solve it and obtain the following $6+2=8$
The result $8$ that we obtained will be written instead of the entire expression $(6+2)$ and we'll proceed to remove the parentheses.
Rewrite the exercise with the new result as seen below:
$2\cdot4+8+1=$
Great! We removed the parentheses.
Now we'll solve the exercise according to the correct order of operations - first multiplication and division, then addition and subtraction.
As shown below:
$2\cdot4=8$
We'll once again rewrite the exercise:
$8+8+1=$
Now we'll solve the addition exercise as shown below:
$8+8+1=17$

Another exercise:
$(7+3\cdot6):5=$

Solution:
Upon observation we can see that there are parentheses. Therefore, we'll start with the expression inside the parentheses first.
Note that inside the parentheses there is both multiplication and addition, and according to the order of operations, we must start with the multiplication first.
Let's rewrite the expression inside of the parentheses:
$(7+3\cdot6)=$
First, we'll solve
$3\cdot6=18$
Now we'll add the $7$
$7+18=25$
The result is $25$! This is the result of the expression inside the parentheses. Proceed to remove the parentheses and insert this value back into the original expression as follows.
$25:5=5$
The final result of the exercise is $5$!

Another exercise!
$(120:2)+(2\cdot2)+(1+3)\cdot0=$

Solution:
Wow! An exercise with lots of parentheses!
How do we approach it?
Simply solve all of the expressions inside of the parentheses, one at a time.

Solve the exercises inside of the parentheses in order.
Let's begin:
$120:2=60$
Rewrite:
$60+(2\cdot2)+(1+3)\cdot0=$
Continue to solve:
$2\cdot2=4$
Rewrite once more:

$60+4+(1+3)\cdot0=$
Continue to solve:
$1+3=4$
Rewrite:
$60+4+4\cdot0=$
Now we have a multiplication operation, let's handle it
$4\cdot0=0$
We obtain the following :
$60+4=64$
$64$ is the final result of the exercise!

One last tip!
It's not always advisable to rewrite the entire expression at each step, due to the fact that if there are multiple parentheses in the same expression the solution may look overly complicated.
If it's an expression with simple parentheses, you have the option to draw small arcs above them and write the result there instead.
For example in the expression we solved:
$(120:2)+(2\cdot2)+(1+3)\cdot0=$
We can place small arcs above each set of parentheses and write the results above them as follows:

Now all we need to do is to rewrite the exercise one last time:
$60+4+4\cdot0=$
Here too we can draw a small arc above the multiplication exercise:

Lastly solve the exercise and here is the result$64$.