## 17Calculus Precalculus - Order of Operations

##### 17Calculus

In algebra, you probably ran across the concept of 'Order Of Operations'. It will help you a lot if you take a minute and refresh your memory on this concept. It is critical to your success in calculus mostly because some calculus classes use calculators and computers which require you enter your expressions using these rules.

Correct Order of Operations

Here is the correct order of operations from highest to lowest priority.

 1. Parentheses 2. Exponents and Radicals 3. Multiplication and Division (left to right) 4. Addition and Subtraction (left to right)

This list means that the strongest binding is parentheses. Parentheses override all other operations. Secondly, exponents and radicals overide operations 3 and 4, etc.

IMPORTANT: You need to work left to right within rules 3 and 4. Working right to left may give you the wrong answer.

Here are some examples.

Example 1

$$\displaystyle{ x + 3 / 2 = x + \frac{3}{2} }$$
If you want the entire x + 3 expression to be divided by 2, you need to use parentheses to override rule 3 taking precedence to rule 4.
$$\displaystyle{ (x + 3) / 2 = \frac{x+3}{2} }$$

Example 2

e^2x = $$e^2x$$
If you want 2x to be the exponent to e, you need to use parentheses to override rule 2 taking precedence to rule 3. e^(2x) = $$e^{2x}$$

Example 3

Here is one I see a lot. A student enters -3^2 into their calculator. This evaluates to -9 because the 2 power takes precedence to the negative sign. So this is like entering -(3^2). The negative sign falls in the same category as subtraction. We just usually do not write 0-3^2. To get (-3)^2, the parentheses are required.

Note: Take special care not to use computer notation in your written work. Good teachers require correct notation and computer notation is not considered correct notation for written work.

Acronym PEMDAS

It is okay to use the acronym PEMDAS as long you remember that multiplication (M) and division (D) are actually one rule. Similarly for addition and subtraction. So, you should think about this acronym as PE(MD)(AS) to give you four rules. Just remember to always to left-to-right within the rules.

Operation Stickiness

In order to avoid making mistakes within rules 3 and 4, there is another way to write the equations. I will show you how I think of it using an example.

Example 4

$$3 \div 4 \times 16$$
I think of the operation before a number (or variable) as being stuck to the number. Then I convert the division into multiplication. So to divide by 4 is the same multiplying by 1/4. The equation is now $$\displaystyle{ 3 \times \frac{1}{4} \times 16 }$$. Now I just multiply everywhere and I can do it in any order. It makes more sense to me to multiply the last two since I will get an integer. So I have $$3 \times 4 = 12$$. I can do this only because I replaced the division by multiplication. This technique will come in handy in calculus.

Example 5

Now let's look at addition and subtraction. Again, I think of the operation as being stuck to the number it follows. So in $$3 - 4 + 16$$ the subtraction sign in front of the 4 is stuck to the 4. I can move the subtraction and replace it with addition if I stick a negative sign to the 4.
$$3 + (-4) + 16$$ (The parentheses are required. 3 + - 4 does not make sense.) Now I have addition everywhere and I can do the additions in any order.
$$-1+16 = 3+12 = 15$$

Here is a video that is worth watching explaining how PEMDAS is not the best way to understand the math behind the order of operations.

### minutephysics - The Order of Operations is Wrong

video by minutephysics

Okay, so try the practice problems and make sure you are confident with evaluating expressions. Calculus absolutely requires you to know these well. If you have trouble with these problems, here are a few websites for you to go and review.

 This page has a great discussion of PEMDAS and how to avoid errors. Math Is Fun: Order of Operations: PEMDAS

Practice

Unless otherwise instructed, simplify these expressions using the correct order of operations.

$$12 - 2 \times 4 + 9$$

Problem Statement

$$12 - 2 \times 4 + 9$$

$$13$$

Problem Statement

$$12 - 2 \times 4 + 9$$

Solution

### 2701 video solution

$$13$$

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$$6 \div 2 + 3[ 15 + 3(7-9)^3 ]$$

Problem Statement

$$6 \div 2 + 3[ 15 + 3(7-9)^3 ]$$

$$-24$$

Problem Statement

$$6 \div 2 + 3[ 15 + 3(7-9)^3 ]$$

Solution

### 2702 video solution

$$-24$$

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$$20 \div 5 + 4^2 [ (-3+7)-1]-2^3$$

Problem Statement

$$20 \div 5 + 4^2 [ (-3+7)-1]-2^3$$

$$44$$

Problem Statement

$$20 \div 5 + 4^2 [ (-3+7)-1]-2^3$$

Solution

### 2703 video solution

$$44$$

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$$7\sqrt{9} + 3 - \sqrt{64}$$

Problem Statement

Simplify $$7\sqrt{9} + 3 - \sqrt{64}$$ using the correct order of operations.

$$7\sqrt{9} + 3 - \sqrt{64}$$ $$= 16$$

Problem Statement

Simplify $$7\sqrt{9} + 3 - \sqrt{64}$$ using the correct order of operations.

Solution

 $$7\sqrt{9} + 3 - \sqrt{64}$$ First calculate the two square root terms. $$7(3) + 3 - 8$$ Now we look closely at the expression. The parentheses contain only one term, so we can remove them. $$7*3 + 3 - 8$$ Now we do the multiplication since it is a higher priority than the addition and subtraction. $$21 + 3 - 8$$ Now we have only addition and subtraction, so we work left to right. $$24 - 8$$ $$16$$

$$7\sqrt{9} + 3 - \sqrt{64}$$ $$= 16$$

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$$8 - 2[ 3 - 4\sqrt{50-14} + 5]$$

Problem Statement

Simplify $$8 - 2[ 3 - 4\sqrt{50-14} + 5]$$ using the correct order of operations.

$$8 - 2[ 3 - 4\sqrt{50-14} + 5]$$ $$= 40$$

Problem Statement

Simplify $$8 - 2[ 3 - 4\sqrt{50-14} + 5]$$ using the correct order of operations.

Solution

 $$8 - 2[ 3 - 4\sqrt{50-14} + 5]$$ First, we evaluate the square root term. $$8 - 2[ 3 - 4\sqrt{36} + 5]$$ $$8 - 2[ 3 - 4*6 + 5]$$ Next, we focus on the expression in the square brackets. Square brackets are sometimes used instead of parentheses and they have the same priority as parentheses. Inside the brackets, we have addition, subtraction and multiplication. Multiplication is the highest priority, so we perform that operation first. $$8 - 2[ 3 - 24 + 5]$$ Now we can evaluate the expression in the brackets, left to right. $$8 - 2[ -21 + 5]$$ $$8 - 2[ -16 ]$$ Now take the negative sign outside the brackets. $$8 + 2[ 16 ]$$ Since we have only term inside the brackets, they are no longer needed. $$8 + 2*16$$ Multiplication is a higher priority than addition, so we perform the multiplication operation. $$8+32 = 40$$

$$8 - 2[ 3 - 4\sqrt{50-14} + 5]$$ $$= 40$$

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$$\displaystyle{ \frac{4(6+4)+(7-5)^5}{6(4-2)-2^2} }$$

Problem Statement

$$\displaystyle{ \frac{4(6+4)+(7-5)^5}{6(4-2)-2^2} }$$

$$9$$

Problem Statement

$$\displaystyle{ \frac{4(6+4)+(7-5)^5}{6(4-2)-2^2} }$$

Solution

### 2706 video solution

$$9$$

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$$\displaystyle{ \frac{3\cdot 4^2 - 3^2 \cdot 5}{5\cdot 9} }$$

Problem Statement

$$\displaystyle{ \frac{3\cdot 4^2 - 3^2 \cdot 5}{5\cdot 9} }$$

$$1/15$$

Problem Statement

$$\displaystyle{ \frac{3\cdot 4^2 - 3^2 \cdot 5}{5\cdot 9} }$$

Solution

### 2707 video solution

$$1/15$$

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$$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) }$$

Problem Statement

Simplify $$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) }$$ using the correct order of operations.

$$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) }$$ $$= 147$$

Problem Statement

Simplify $$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) }$$ using the correct order of operations.

Solution

 $$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) }$$ First, we start inside the square root. Inside we have a power, multiplication, addition and subtraction and some parentheses. We start with the parentheses. $$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (15.5) }$$ Since we have only one single term inside the parentheses, they are no longer necessary. $$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 \cdot 15.5 }$$ So now, the next priority is the power. So we evaluate $$4^2$$. Notice that we do NOT multiply the power of 2 by 5 first. That's because, the power is a higher priority than the multiplication. $$42 \div 2\sqrt{ 16 \cdot 5 - 2 \cdot 15.5 }$$ Now we have multiplication and subtraction inside the square root. So we do the multiplication first, left to right. $$42 \div 2\sqrt{ 80 - 2 \cdot 15.5 }$$ $$42 \div 2\sqrt{ 80 - 31 }$$ Now do the subtraction inside the square root. $$42 \div 2\sqrt{ 49 }$$ And take the square root. $$42 \div 2 \cdot 7$$ Now we are only left with multiplication and division. So we work left to right. $$21 \cdot 7 = 147$$ You may be asking yourself, why do we start with the square root? It is not in parentheses. Well, actually it is. You can write a square root $$\sqrt{x}$$ as $$(x)^{1/2}$$. So this is an expression $$x$$ inside parentheses with a power $$1/2$$. So that is why we started with the square root.

$$42 \div 2\sqrt{ 4^2 \cdot 5 - 2 (8.7+6.8) }$$ $$= 147$$

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$$\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$

Problem Statement

Simplify $$\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ using the correct order of operations.

$$\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ $$= 43.5$$

Problem Statement

Simplify $$\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ using the correct order of operations.

Solution

 $$\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ To get started with this one, we need to start with expressions in the most inside parentheses. We will also work separately with numerator and denominator until we get two individual terms to divide. Let's start with the most inside parentheses in the numerator, $$(8+5=13)$$ and $$(6-2=4)$$. $$\displaystyle{ \frac{ [(13)(4)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ So now we can remove the parentheses from the $$(13)$$ and from the $$(4)$$. $$\displaystyle{ \frac{ [13 \cdot 4^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ Staying inside the same set of brackets, we have multiplication and a power. The power has a higher priority, so we do that first. $$\displaystyle{ \frac{ [13 \cdot 16] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ So now were are left with just one operation inside the brackets, so we multiply those values and drop the brackets. $$\displaystyle{ \frac{ 208 - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ Okay, so staying in the numerator, we will work with the expression in the remaining parentheses. We have only multiplication and division, so we work left to right. $$\displaystyle{ \frac{ 208 - (68 \div 2) }{(24 \div 2) \div 3} = \frac{ 208 - (34) }{(24 \div 2) \div 3} = \frac{ 208 - 34 }{(24 \div 2) \div 3} }$$ Finishing off the numerator we perform the subtraction. $$\displaystyle{ \frac{ 174 }{(24 \div 2) \div 3} }$$ Now we need to move to the denominator. We will first perform the operation inside the parentheses and then drop the parentheses. $$\displaystyle{ \frac{ 174 }{(12) \div 3} = \frac{ 174 }{12 \div 3} }$$ Next, we perform the last operation in the denominator and finish the problem. $$\displaystyle{ \frac{ 174 }{4} = 43.5 }$$ Notice how we focused on very small parts of an otherwise complicated expression. This is a very useful concept when working algebra problems.

$$\displaystyle{ \frac{ [(8+5)(6-2)^2] - (4 \cdot 17 \div 2) }{(24 \div 2) \div 3} }$$ $$= 43.5$$

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