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# 1.12: PEMDAS with Negative Real Numbers

Difficulty Level: At Grade Created by: CK-12

Ginny walked into Math class and saw the following question on the board.

Find the value of the following expression when a=2,b=3,c=1,d=1\begin{align*}a = -2, b = 3, c = -1, d = 1\end{align*}

(4a+c)÷b(bd)÷(ac)\begin{align*}\boxed{(4a+c)\div b-(bd)\div (ac)}\end{align*}

### Guidance

The standard order of operations involve specific steps for performing the mathematical calculations presented in a mathematical statement. These steps are represented by the letters PEMDAS.

P – Parenthesis – Do all the calculations within parenthesis.

E – Exponents – Do all calculations that involve exponents.

M/D – Multiplication/Division – Do all multiplication and division, in the order it occurs, working from left to right.

A/S – Addition/Subtraction – Do all addition and subtraction, in the order it occurs, working from left to right.

These steps do not change whether they are being applied to positive real numbers or to negative real numbers. The rules for adding, subtracting, multiplying and dividing real negative numbers must be applied when evaluating expressions that require PEMDAS to be used.

#### Example A

Perform the following calculations, using PEMDAS.

32÷42×221\begin{align*}32 \div 4^2 \times 2-21\end{align*}

There are no calculations inside parenthesis. The first step is to evaluate the number with the exponent.

32÷16×221(42=4×4=16)\begin{align*}32\div {\color{blue}16}\times 2-21 \quad {\color{blue}(4^2=4 \times 4 =16)}\end{align*}

The next step is to perform any division or multiplication, in the order they occur, working from left to right.

2×221(32÷16=2) Division\begin{align*}{\color{blue}2} \times 2-21 \quad {\color{blue}(32 \div 16 =2)}\ \text{Division}\end{align*}

421(2×2=4) Multiplication\begin{align*}{\color{blue}4}-21 \quad {\color{blue}(2 \times 2 = 4)} \ \text{Multiplication}\end{align*}

The final step is to rewrite the expression as an addition problem and to change the sign of the original number being subtracted.

4+21(421=4+21)\begin{align*}4{\color{blue}+-}21 \quad {\color{blue}(4-21=4+-21)}\end{align*}

When adding two numbers with unlike signs, subtract the numbers and the sign of the number with the greater magnitude will be the sign of the answer.

=17\begin{align*}= {\color{blue}-17}\end{align*}

#### Example B

Find the value of the following expression when a=2,b=3,c=1,d=1\begin{align*}a = -2, b = 3, c = -1, d = 1\end{align*}. Use PEMDAS.

(4a2c2)(3ac3)\begin{align*}(4a^2c^2)-(3ac^3)\end{align*}

Begin by substituting the variables with the given values. Use brackets to group the operations with parenthesis.

[4(2)2(1)2][3(2)(1)3]\begin{align*}[4(-2)^2(-1)^2] - [3(-2)(-1)^3]\end{align*}

In the first set of brackets, do the calculations with exponents.

[4(4)(1)][3(2)(1)3](2)2=(2×2)=4(1)2=(1×1)=1\begin{align*}[4({\color{blue}4})({\color{blue}1})] - [3(-2)(-1)^3] \quad {\color{blue} (-2)^2=(-2 \times -2) = 4} \quad {\color{blue}(-1)^2=(-1 \times -1)=1}\end{align*}

In the second set of brackets, do the calculations with exponents.

[4(4)(1)][3(2)(1)](1)3=(1×1×1)=1\begin{align*}[4(4)(1)]-[3(-2)({\color{blue}-1})] \quad {\color{blue}(-1)^3=(-1 \times -1 \times -1)=-1}\end{align*}

In the first set of brackets, do the multiplication. The brackets can now be removed.

16[3(2)(1)](4×4×1)=16\begin{align*}{\color{blue}16}-[3(-2)({\color{blue}-1})] \quad {\color{blue}(4 \times 4 \times 1)=16}\end{align*}

In the second set of brackets, do the multiplication. The brackets can now be removed.

166(3×2×1)=6\begin{align*}16-{\color{blue}6} \quad {\color{blue}(3 \times -2 \times -1) =6}\end{align*}

Subtract the numbers.

=10\begin{align*}={\color{blue}10}\end{align*}

#### Example C

What is the value of \begin{align*}3-2 \left[\frac{8(-1)-7}{-3(2)-4}\right]\end{align*}?

In the set of brackets, do the multiplication in the numerator first.

\begin{align*}3-2\left[\frac{{\color{blue}-8}-7}{-3(2)-4}\right] \quad {\color{blue}(8 \times -1 = -8)}\end{align*}

In the set of brackets, do the multiplication in the denominator.

\begin{align*}3-2\left[\frac{-8-7}{{\color{blue}-6}-4}\right] \quad ({\color{blue}-3 \times 2 =-6})\end{align*}

Rewrite the expressions in both the numerator and the denominator as addition problems and change the sign of the original numbers being subtracted.

\begin{align*}3-2 \left[\frac{-8{\color{blue}+-}7}{-6{\color{blue}+-}4}\right] \quad {\color{blue}(-8-7=-8+-7)} \quad {\color{blue}(-6-4=-6+-4)}\end{align*}

\begin{align*}3-2\left[{\color{blue}\frac{-15}{-10}}\right] \quad {\color{blue}(-8+-7=-15)} \quad {\color{blue}(-6+-4=-10)}\end{align*}

Multiplication – Multiply the fraction by 2. Remember to multiply numerators and to multiply denominators.

\begin{align*}3- {\color{blue}\frac{-30}{-10}} \quad {\color{blue}\left(\frac{2}{1} \times \frac{-15}{-10} = \frac{-30}{-10}\right)}\end{align*}

Division – The quotient of two integers with like signs is always positive.

\begin{align*}3-{\color{blue}3} \quad {\color{blue} \left(\frac{-30}{-10}=3\right)}\end{align*}

Subtract the numbers.

\begin{align*}={\color{blue}0}\end{align*}

#### Example D

Although the numbers in the following statement are decimals, the standard order of operations does not change. PEMDAS is applied to calculate the answer.

\begin{align*}6.12+8.6 \times (-0.9) -(10.26 \div 3.8)\end{align*}

\begin{align*}6.12+8.6 \times 0.9-{\color{blue}(-2.7)} \quad {\color{blue}(10.26 \div 3.8 = -2.7)} \ \text{Parenthesis}\end{align*}

\begin{align*}6.12+{\color{blue}(-7.74)}-(-2.7) \quad {\color{blue}(8.6 \times -0.9 = -7.74)} \ \text{Multiplication}\end{align*}

Rewrite the expression as an addition problem and change the sign of the original number being subtracted.

\begin{align*}6.12+(-7.74){\color{blue}+}({\color{blue}+}2.7) \quad {\color{blue}(--2.7=++2.7)}\end{align*}

\begin{align*}{\color{blue}8.82}+-7.74 \quad {\color{blue}(6.12+2.7=8.82)}\end{align*} Addition – Add the two positive numbers first.

\begin{align*}={\color{blue}1.08} \quad {\color{blue}(8.82+-7.74=1.08)}\end{align*} Addition – Subtract the numbers and the answer will have the same sign as the number of greater magnitude.

#### Concept Problem Revisited

Find the value of the following expression when \begin{align*}a = -2, b = 3, c = -1, d = 1\end{align*}

\begin{align*}\boxed{(4a+c)\div b-(bd)\div (ac)}\end{align*}

Ginny felt good about answering the problem because she remembered the steps involved in the standard order of operations. When she noticed that the values for two of the variables were negative numbers, she realized that she would have to be careful doing the calculations because she would also have to apply the rules that she had learned for adding, subtracting, multiplying and dividing negative real numbers.

\begin{align*}(4a+c)\div b-(bd) \div(ac)\end{align*}

Ginny began by substituting the variables with the given values.

\begin{align*}(4(-2)+(-1))\div 3-((3)(1))\div((-2)(-1))\end{align*}

To reduce errors in her calculations, Ginny wrote all of the values in parenthesis. The statement now has parenthesis within parenthesis. This may seem confusing and the order in which to perform the operations may become skewed. Ginny asked her teacher about writing the expression another way. Her teacher advised her to replace the outer parenthesis with brackets [ ]. Brackets are another type of grouping symbol. When evaluating an expression that has grouping symbols (parentheses) within grouping symbols (brackets), perform the operations within the innermost set of symbols first. This is not necessary, but it is a good rule to follow.

Ginny rewrote the expression using both brackets and parenthesis.

\begin{align*}[4(-2)+(-1)]\div 3-[(3)(1)] \div [(-2)(-1)]\end{align*} Do calculations in the parenthesis first.

\begin{align*}[{\color{blue}-8}+(-1)]\div 3 - [(3)(1)] \div [(-2)(-1)]\end{align*}

\begin{align*}[-8+(-1)] \div 3-[3]\div[2]\end{align*} Do calculations in the brackets next.

\begin{align*}[{\color{blue}-9}] \div 3-[3] \div [2]\end{align*}

\begin{align*}-9\div 3-3\div 2\end{align*} The brackets can be removed if there are no other calculations inside the brackets.

\begin{align*}{\color{blue}-3}-3 \div 2\end{align*}

\begin{align*}-3- 1.5\end{align*}

\begin{align*}-3- 1.5 \rightarrow -3+ -1.5\end{align*}

\begin{align*}-3+ -1.5\end{align*}

\begin{align*}={\color{blue}-4.5}\end{align*}

Ginny performed all the calculations in the correct order and remembered the rules for operations with negative real numbers.

### Vocabulary

Bracket
A bracket [ ], is a symbol that is used to group numbers in mathematics. A bracket is most often used when the expression has calculations within parenthesis.
Parenthesis
A parenthesis ( ), is a symbol that is used to group numbers in mathematics. The parenthesis shows where an expression starts and stops.
PEMDAS
The letters PEMDAS represent the standard order of operations for calculating mathematical statements.

P - Parenthesis E - Exponents M - Multiplication D - Division A - Addition S - Subtraction

### Guided Practice

1. Perform the following operations using PEMDAS \begin{align*}8 \times -9 +19 \div (-30+11)-14 \times (-1)^2\end{align*}

2. Determine the answer to by using the rules for the \begin{align*}\left(\frac{-12-6}{6+3}\right)+\left(\frac{-36}{-4}\right)+(-8 \times 2)\end{align*} standard order of operations.

3. A formula from geometry is \begin{align*}V=\frac{h}{6}(B+4M+b)\end{align*}. Find \begin{align*}V\end{align*} when \begin{align*}h = -15, B = 12,M = 8, b = 4\end{align*}.

1.

\begin{align*} & 8 \times -9+19 \div (-30+11)-14 \times (-1)^2\\ & 8 \times 9 +19 \div {\color{blue}-19}-14 \times (-1)^2 \ \text{Do the calculations in parenthesis.}\\ & 8 \times9+19 \div {\color{blue}-19}-14 \times (-1)^2 \ \text{Do the calculations with exponents.}\\ & {\color{blue}72}+19 \div -19-14\times 1 \ \text{Do the first multiplication.}\\ & 72+{\color{blue}-1}-14 \times 1 \ \text{Do the division.}\\ & 72+-1-{\color{blue}14}\ \text{Do the second multiplication.}\\ & {\color{blue}71}-14 \ \text{Do the addition.}\\ & ={\color{blue}57} \ \text{Do the subtraction.}\end{align*}

2.

\begin{align*}& \left(\frac{-12-6}{6+3}\right)+\left(\frac{-36}{-4}\right)+(-8 \times 2)\\ & \text{Do the calculations in parenthesis.}\\ & \left({\color{blue}\frac{-18}{9}}\right)+\left(\frac{-36}{-4}\right)+(-8 \times 2)\\ & {\color{blue}-2}+\left(\frac{-36}{-4}\right)+(-8 \times 2)\\ & -2+{\color{blue}9}+(-8 \times 2)\\ & -2+9+{\color{blue}-16}\\ & \text{Add the numbers}\\ & ={\color{blue}-9}\end{align*}

3.

\begin{align*}& V=\frac{h}{6}(B+4M+b).\\ & \text{Substitute the values for the variables}\\ & V=\frac{-15}{6}(12+4(8)+4)\\ & \text{Do the calculations in parenthesis}\\ & V=\frac{-15}{6}(12+{\color{blue}32}+4)\\ & V=\frac{-15}{6}(\color{blue}48)\\ & \text{Do the division}\\ & V={\color{blue}-2.5}(48)\\ &\text{Do the multiplication}\\ & V={\color{blue}-120}\end{align*}

### Practice

1. If \begin{align*}a = -3, b = 1\end{align*} and \begin{align*}c = 2\end{align*}, what is the value of \begin{align*}2a^3-3b+2c^2\end{align*}? _______
1. 221
2. 59
3. -49
4. 43
2. Evaluate \begin{align*}a-bc\end{align*} when \begin{align*}a=\frac{1}{2}, b=\frac{1}{3}\end{align*} and \begin{align*}c=\frac{5}{4}\end{align*}: _______
1. 1
2. \begin{align*}\frac{1}{12}\end{align*}
3. \begin{align*}\frac{2}{5}\end{align*}
4. \begin{align*}\frac{5}{2}\end{align*}
3. Evaluate: \begin{align*}a(-b^2-a^2)\end{align*} when \begin{align*}a=-3\end{align*} and \begin{align*}b=4\end{align*} _______
1. 75
2. 2
3. 3
4. 4
4. Simplify the following: \begin{align*}[3-[5-(6-8)]+4]-2\end{align*} _______
1. -6
2. -2
3. 2
4. -19
5. Perform the following operations and evaluate: \begin{align*}(5\times 3-7)^2 \div 4+9\end{align*} _______
1. 109
2. 5
3. 11
4. 25
6. Which expression has the greatest value if \begin{align*}a=-2\end{align*} and \begin{align*}b=3\end{align*}?
1. \begin{align*}3[a^2+b^2-2ab-2(a^2-b^2)]\end{align*}
2. \begin{align*} 3(a-b)-2(b-a)+3(a-2b)\end{align*}
3. \begin{align*} b(a-b)-a(b-a)-ab\end{align*}

Perform the indicated calculations, using PEMDAS to determine the answer.

1. \begin{align*}\frac{6-(24-14)}{-10-[2-(-4)^2]}\end{align*}
2. \begin{align*}\frac{[12-(-15+6)]\times 4 -16}{-4}\end{align*}
3. \begin{align*}3\left(-2 \times \frac{20}{-4 \times -1}-5-7\right)\end{align*}
4. \begin{align*}-3 \times 5 -[3(3+9)-9\times -3]\end{align*}
5. \begin{align*}4^2-(4+8-2-4-2\times 7)\end{align*}

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Dec 19, 2012
Apr 29, 2014

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