Lesson Objectives

- Demonstrate an understanding of operations with integers
- Demonstrate an understanding of the order of operations
- Learn how to use the order of operations for problems with integers

## Order of Operations with Integers

In our lesson on the order of operations,
we learned the correct priority or order for each operation, when faced with multiple operations in a problem. Let’s review the order of operations:

Example 1: Evaluate each:

(-9 - (-3) x 7) ÷ |-3

Example 2: Evaluate each:

[(2

- Work inside of any parentheses or grouping symbols
- Perform all exponent operations
- Multiply or divide (working left to right)
- Add or subtract (working left to right)

### PEMDAS

- P - Parentheses or grouping symbols
- E - Exponents
- M/D - Multiply/Divide (working left to right)
- A/S - Addition/Subtraction (working left to right)

Example 1: Evaluate each:

(-9 - (-3) x 7) ÷ |-3

^{2}+ 5|- First, we look for parentheses or grouping symbols, once inside we will reapply the order of operations.
- In this problem, we have parentheses and absolute value bars. These are both grouping symbols. We will start by working inside of each.
- (-9 - (-3) x 7) ÷ |-3
^{2}+ 5| - Inside of our parentheses, we have subtraction and multiplication. Multiplication has a higher priority. We will perform (-3) x 7 first.
- (-3) x 7 = -21, let's replace this in our problem.
- (-9 - (-21)) ÷ |-3
^{2}+ 5| - Continuing inside of the parentheses, we only have subtraction. We will perform -9 - (-21) next.
- -9 - (-21) = 12, let's replace this in our problem.
- 12 ÷ |-3
^{2}+ 5| - Next, we look at our other grouping symbol, the absolute value operation.
- 12 ÷ |-3
^{2}+ 5| - Inside of the absolute value bars, we have an exponent operation and addition. The exponent operation has the highest priority. We will
perform -3
^{2}first. - -3
^{2}= -9, let's replace this in our problem. - 12 ÷ |-9 + 5|
- Continuing inside of the absolute value bars, we only have addition. We will perform -9 + 5 next.
- -9 + 5 = -4, let's replace this in our problem.
- 12 ÷ |-4|
- Let's now take the absolute value of -4, this will give us +4. Let's replace this in our problem.
- 12 ÷ 4
- Last, we have one single division operation left, let's divide 12 by 4 and get our final answer of 3.
- 12 ÷ 4 = 3

^{2}+ 5| = 3Example 2: Evaluate each:

[(2

^{5}- 2^{4}÷ 2^{2}) ÷ |-5 x 3 + 8|] ÷ |-5^{2}+ 7 x 5 - 6|- First, we look for parentheses or grouping symbols, once inside we will reapply the order of operations.
- In this problem, we have brackets "[ ]", parentheses, and absolute value bars. These are all grouping symbols. We will start by working inside of each.
- Let's start by working inside the brackets. Once inside we will reapply our order of operations.
- [(2
^{5}- 2^{4}÷ 2^{2}) ÷ |-5 x 3 + 8|] ÷ |-5^{2}+ 7 x 5 - 6| - Inside of the brackets, we have parentheses and absolute value bars. We will start with the parentheses.
- Inside of parentheses, we have 2
^{5}- 2^{4}÷ 2^{2}. Here exponents are a higher priority than subtraction or division. Let's start by performing all exponent operations. - 2
^{5}= 32, 2^{4}= 16, 2^{2}= 4, let's replace these in our problem. - [(32 - 16 ÷ 4) ÷ |-5 x 3 + 8|] ÷ |-5
^{2}+ 7 x 5 - 6| - Now we have subtraction and division, we will divide before we subtract. We will perform 16 ÷ 4 next.
- 16 ÷ 4 = 4, let's replace this in our problem.
- [(32 - 4) ÷ |-5 x 3 + 8|] ÷ |-5
^{2}+ 7 x 5 - 6| - Now we have only subtraction. We will perform 32 - 4 next.
- 32 - 4 = 28, let's replace this in our problem.
- [28 ÷ |-5 x 3 + 8|] ÷ |-5
^{2}+ 7 x 5 - 6| - Continuing inside of our brackets, we have an absolute value operation. Inside, we have multiplication and addition. We will multiply before we add. We will perform -5 x 3 next.
- -5 x 3 = -15, let's replace this in our problem.
- [28 ÷ |-15 + 8|] ÷ |-5
^{2}+ 7 x 5 - 6| - Now we have only addition inside of the absolute value bars, we will perform -15 + 8 next.
- -15 + 8 = -7, let's replace this in our problem
- [28 ÷ |-7|] ÷ |-5
^{2}+ 7 x 5 - 6| - Now we can take the absolute value of -7, which is 7. Let's replace this in our problem.
- [28 ÷ 7] ÷ |-5
^{2}+ 7 x 5 - 6| - Lastly, we will divide 28 by 7.
- 28 ÷ 7 = 4, let's replace this in our problem.
- 4 ÷ |-5
^{2}+ 7 x 5 - 6| - Now we will simplify inside of the absolute value operation on the right.
- 4 ÷ |-5
^{2}+ 7 x 5 - 6| - Inside of the absolute value operation, we have an exponent operation, multiplication, addition, and subtraction.
We will begin with our exponent operation -5
^{2}. - -5
^{2}= -25, let's replace this in our problem. - 4 ÷ |-25 + 7 x 5 - 6|
- Continuing inside of the absolute value bars, we have addition, multiplication, and subtraction. We will multiply 7 x 5 next.
- 7 x 5 = 35, let's replace this in our problem.
- 4 ÷ |-25 + 35 - 6|
- Continuing inside of the absolute value bars, we have addition and subtraction. We perform addition and subtraction working left to right. Since the addition operation is the leftmost, we will perform -25 + 35 next.
- -25 + 35 = 10, let's replace this in our problem.
- 4 ÷ |10 - 6|
- Now we only have subtraction left inside of the absolute value bars. We will perform 10 - 6 as our next operation.
- 10 - 6 = 4
- 4 ÷ |4|
- Now we will take the absolute value of 4, which is 4. Let's replace this in our problem.
- 4 ÷ 4
- Our last operation will be the division of 4 by 4, which will give us our final answer of 1.
- 4 ÷ 4 = 1

^{5}- 2^{4}÷ 2^{2}) ÷ |-5 x 3 + 8|] ÷ |-5^{2}+ 7 x 5 - 6| = 1
Ready for more?

Watch the Step by Step Video Lesson
Take the Practice Test