1.3.3: Associative, Commutative, and Distributive Properties

Last updated
Save as PDF

Page ID: 90372

Leah Griffith, Veronica Holbrook, Johnny Johnson & Nancy Garcia
Rio Hondo College

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

1.3.3 Learning Objectives

Use the commutative and associative properties of addition and multiplication
Use the identity and inverse properties of addition and multiplication
Use the properties of zero
Simplify expressions using the distributive property

Use the Commutative and Associative Properties

Think about adding two numbers, say 5 and 3. The order we add them doesn’t affect the result, does it?

$\begin{array} { cc } { 5 + 3 } & { 3 + 5 } \\ { 8 } & { 8 } \\ { 5 + 3 = } & { 3 + 5 } \end{array}$

The results are the same.

As we can see, the order in which we add does not matter!

What about multiplying 5 and 3?

$\begin{array} { c c } { 5 \cdot 3 } & { 3 \cdot 5 } \\ { 15 } & { 15 } \\ { 5 \cdot 3=} &{3 \cdot 5 } \end{array}$

Again, the results are the same!

The order in which we multiply does not matter!

These examples illustrate the commutative property. When adding or multiplying, changing the order gives the same result.

Definition: Commutative Property

$\begin{array} { l l } { \textbf { of Addition } } & { \text { If } a , b \text { are real numbers, then } \quad a + b = b + a } \\ { \textbf { of Multiplication } } & { \text { If } a , b \text { are real numbers, then } \quad a \cdot b = b \cdot a } \end{array}$

When adding or multiplying, changing the order gives the same result.

The commutative property has to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.

What about subtraction? Does order matter when we subtract numbers? Does 7−3 give the same result as 3−7?

$\begin{array} { c c } { 7 - 3 } & { 3 - 7 } \\ { 4 } & { - 4 } \end{array}$

$\begin{aligned} 4 & \neq - 4 \\ 7 - 3 & \neq 3 - 7 \end{aligned}$

The results are not the same.

Since changing the order of the subtraction did not give the same result, we know that subtraction is not commutative.

Let’s see what happens when we divide two numbers. Is division commutative?

$\begin{array} { cc} { 12 \div 4 } & { 4 \div 12 } \\ { \frac { 12 } { 4 } } & { \frac { 4 } { 12 } } \\ { 3 } & { \frac { 1 } { 3 } } \end{array}$
$\begin{aligned} 3 \neq & \frac { 1 } { 3 } \\ 12 \div 4 & \neq 4 \div 12 \end{aligned}$

The results are not the same.

Since changing the order of the division did not give the same result, division is not commutative. The commutative properties only apply to addition and multiplication!

Addition and multiplication are commutative.
Subtraction and Division are not commutative.

If you were asked to simplify this expression, how would you do it and what would your answer be?

$7 + 8 + 2$

Some people would think $7+8$ is 15 and then $15+2$ is 17. Others might start with $8+2$ makes 10 and then $7+10$ makes 17.

Either way gives the same result. Remember, we use parentheses as grouping symbols to indicate which operation should be done first.

$\begin{array} { ll } { \text{ Add } 7 + 8 . } & { ( 7 + 8 ) + 2 } \\ { \text { Add. } } & { 15 + 2 } \\ { \text { Add. } } & { 17 } \\ \\ { } & { 7 + ( 8 + 2 ) } \\ { \text { Add } 8 + 2 . } & { 7 + 10 } \\ { \text { Add. } } & { 77 } \\\\ { ( 7 + 8 ) + 2 = 7 + ( 8 + 2 ) } \end{array}$

When adding three numbers, changing the grouping of the numbers gives the same result.

This is true for multiplication, too.

$\begin{array} { ll } { } & { (5\cdot \frac{1}{3})\cdot 3 } \\ { \text { Multiply. } 5\cdot \frac{1}{3} } & { \frac{5}{3}\cdot 3 } \\ { \text { Multiply. } } & { 5 } \\ \\ { } & { 5\cdot (\frac{1}{3}\cdot 3) } \\ { \text { Multiply. } \frac{1}{3}\cdot 3 } & { 5\cdot 1 } \\ { \text { Multiply. } } & { 5 } \\ \\ { (5\cdot \frac{1}{3})\cdot 3 = 5\cdot (\frac{1}{3}\cdot 3) } \end{array}$

When multiplying three numbers, changing the grouping of the numbers gives the same result.

You probably know this, but the terminology may be new to you. These examples illustrate the associative property.

Definition: Associative property

$\begin{array} { l l } { \textbf { of Addition } } & { \text { If } a , b , c \text { are real numbers, then } ( a + b ) + c = a + ( b + c ) } \\ { \textbf { of Multiplication } } & { \text { If } a , b , c \text { are real numbers, then } ( a \cdot b ) \cdot c = a \cdot ( b \cdot c ) } \end{array}$

When adding or multiplying, changing the grouping gives the same result.

Let’s think again about multiplying $5\cdot \frac{1}{3}\cdot 3$. We got the same result both ways, but which way was easier? Multiplying $\frac{1}{3}$ and 3 first, as shown above on the right side, eliminates the fraction in the first step. Using the associative property can make the math easier!

The associative property has to do with grouping. If we change how the numbers are grouped, the result will be the same. Notice it is the same three numbers in the same order—the only difference is the grouping.

We saw that subtraction and division were not commutative. They are not associative either.

When simplifying an expression, it is always a good idea to plan what the steps will be. In order to combine like terms in the next example, we will use the commutative property of addition to write the like terms together.

Example 1

Simplify: $18p+6q+15p+5q$.

Answer: $\begin{array} { l l} {} &{18p+6q+15p+5q}\\ \\{ \text { Use the commutative property of addition } } &{} \\ { \text {to re-order so that like terms are together.} } &{18p+15p+ 6q+5q} \\ \\ {\text{Add like terms.}} &{33p + 11q} \end{array}$

Try it Now 1

Simplify: $23r+14s+9r+15s$.

Answer: $32r+29s$

Try it Now 2

Simplify: $37m+21n+4m−15n$.

Answer: $41m+6n$

When we have to simplify algebraic expressions, we can often make the work easier by applying the commutative or associative property first, instead of automatically following the order of operations. When adding or subtracting fractions, combine those with a common denominator first.

Example 2

Simplify: $(\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}$

Answer: $\begin{array} { l l } {} &{(\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}} \\{ \text { Notice that the last } 2 \text { terms have a } } \\ { \text { common denominator, so change the } } &{\frac { 5 } { 13 } + \left( \frac { 3 } { 4 } + \frac { 1 } { 4 } \right)}\\ { \text { grouping. } } &{}\\ \\ {\text{Add in parentheses first.}} &{\frac{5}{13} + (\frac{4}{4})} \\ \\ {\text{Simplify the fraction.}} &{\frac{5}{13} + 1} \\ \\ {\text{Add.}} &{1\frac{5}{13}} \\ \\ {\text{Convert to an improper fraction.}} &{\frac{18}{13}} \end{array}$

Try It Now 3

Simplify: $(\frac{7}{15} + \frac{5}{8}) + \frac{3}{8}$

Answer: $1\frac{7}{15}$

Try it Now 4

Simplify: $(\frac{2}{9} + \frac{7}{12}) + \frac{5}{12}$

Answer: $1\frac{2}{9}$

Example 3

Use the associative property to simplify $6(3x)$.

Answer

Use the associative property of multiplication, $(a\cdot b)\cdot c=a\cdot (b\cdot c)$, to change the grouping.

\[\begin{array} { ll } {} &{ 6 ( 3 x ) } \\ { \text { Change the grouping. } } &{(6\cdot 3)x} \\ { \text { Multiply in the parentheses. } } &{18} \end{array}\]

Notice that we can multiply $6\cdot 3$ but we could not multiply $3x$ without having a value for $x$.

Try it Now 5

Use the associative property to simplify $8(4x)$.

Answer: $32x$

Try it Now 6

Use the associative property to simplify $-9(7y)$.

Answer: $-63y$

Use the Identity and Inverse Properties of Addition and Multiplication

What happens when we add 0 to any number? Adding 0 doesn’t change the value. For this reason, we call 0 the additive identity.

For example,

$\begin{array} { c c c } { 13 + 0 } & { - 14 + 0 } & { 0 + ( - 8 ) } \\ { 13 } & { - 14 } & { - 8 } \end{array}$

These examples illustrate the Identity Property of Addition that states that for any real number $a$, $a+0=a$ and $0+a=a$.

What happens when we multiply any number by one? Multiplying by 1 doesn’t change the value. So we call 1 the multiplicative identity.

For example,

$\begin{array} { r r r } { 43 \cdot 1 } & { - 27 \cdot 1 } & { 1 \cdot \frac { 3 } { 5 } } \\ { 43 } & { - 27 } & { \frac { 3 } { 5 } } \end{array}$

These examples illustrate the Identity Property of Multiplication that states that for any real number $a$, $a\cdot 1=a$ and $1\cdot a=a$.

We summarize the Identity Properties below.

Definition: Identity property

$\begin{array} { l l} { \textbf {of addition}\text{ For any real number } a : } &{ a + 0 = a \quad 0 + a = a } \\ { \textbf{0} \text { is the}\textbf{ additive identity } } \\ {\textbf {of multiplication}\text{ For any real number } a : } &{ a \cdot 1 = a \quad 1 \cdot a = a } \\ { \textbf{1}\text{ is the}\textbf{ multiplicative identity } } \end{array}$

Figure $\PageIndex{1}$

Notice that in each case, the missing number was the opposite of the number!

We call $−a$. The additive inverse of a. The opposite of a number is its additive inverse. A number and its opposite add to zero, which is the additive identity. This leads to the Inverse Property of Addition that states for any real number $a, a+(−a)=0$. Remember, a number and its opposite add to zero.

What number multiplied by $\frac{2}{3}$ gives the multiplicative identity, 1? In other words, $\frac{2}{3}$ times what results in 1?

We have the statement that 2/3 times a blank space equals 1. Then it is stated that “We know 2/3 times 3/2 equals 1.”

Figure $\PageIndex{2}$

What number multiplied by 2 gives the multiplicative identity, 1? In other words 2 times what results in 1?

We have the statement that 2 times a blank space equals 1. Then it is stated that “We know 2 times 1/2 equals 1.”

Figure $\PageIndex{3}$

Notice that in each case, the missing number was the reciprocal of the number!

We call $\frac{1}{a}$ the multiplicative inverse of a. The reciprocal of aa number is its multiplicative inverse. A number and its reciprocal multiply to one, which is the multiplicative identity. This leads to the Inverse Property of Multiplication that states that for any real number $a, a\neq 0, a\cdot \frac{1}{a}=1$.

We’ll formally state the inverse properties here:

Definition: Inverse property

$\begin{array} { l l l } { \textbf { of addition } } &{ \text { For any real number } a,} &{a + (-a) = 0}\\{} &{-a \text{. is the}\textbf{ additive inverse} \text{ of }a} &{}\\ {} &{ \text { A number and its opposite add to zero. } }&{}\\ \\{ \textbf { of multiplication } } &{ \text { For any real number } a, a\neq 0} &{a\cdot \frac{1}{a} = 1}\\{} &{\frac{1}{a} \text{. is the}\textbf{ multiplicative inverse} \text{ of }a} &{}\\ {} &{ \text { A number and its reciprocal multiply to zero. } }&{} \end{array}$

Example 4

Find the additive inverse of

$\frac{5}{8}$
$0.6$
$-8$
$-\frac{4}{3}$

Answer

To find the additive inverse, we find the opposite.

The additive inverse of $\frac{5}{8}$ is the opposite of $\frac{5}{8}$. The additive inverse of $\frac{5}{8}$ is $-\frac{5}{8}$
The additive inverse of $0.6$ is the opposite of $0.6$. The additive inverse of $0.6$ is $-0.6$.
The additive inverse of $-8$ is the opposite of $-8$. We write the opposite of $-8$ as $-(-8)$, and then simplify it to $8$. Therefore, the additive inverse of $-8$ is $8$.
The additive inverse of $-\frac{4}{3}$ is the opposite of $-\frac{4}{3}$. We write this as $-(-\frac{4}{3})$, and then simplify to $\frac{4}{3}$. Thus, the additive inverse of $-\frac{4}{3}$ is $\frac{4}{3}$.

Try it Now 7

Find the additive inverse of

$\frac{7}{9}$
$1.2$
$-14$
$-\frac{9}{4}$

Answer

$-\frac{7}{9}$
$-1.2$
$14$
$\frac{9}{4}$

Try it Now 8

Find the additive inverse of

$\frac{7}{13}$
$8.4$
$-46$
$-\frac{5}{2}$

Answer

$-\frac{7}{13}$
$-8.4$
$46$
$\frac{5}{2}$

Example 5

Find the multiplicative inverse of

$9$
$-\frac{1}{9}$
$0.9$

Answer

To find the multiplicative inverse, we find the reciprocal.

The multiplicative inverse of $9$ is the reciprocal of $9$, which is $\frac{1}{9}$. Therefore, the multiplicative inverse of $9$ is $\frac{1}{9}$.
The multiplicative inverse of $-\frac{1}{9}$ is the reciprocal of $-\frac{1}{9}$, which is $−9$. Thus, the multiplicative inverse of $-\frac{1}{9}$ is $-9$.
To find the multiplicative inverse of $0.9$, we first convert $0.9$ to a fraction, $\frac{9}{10}$. Then we find the reciprocal of the fraction. The reciprocal of $\frac{9}{10}$ is $\frac{10}{9}$. So the multiplicative inverse of $0.9$ is $\frac{10}{9}$.

Try it Now 9

Find the multiplicative inverse of

$4$
$-\frac{1}{7}$
$0.3$

Answer

$\frac{1}{4}$
$-7$
$\frac{10}{3}$

Try it Now 10

Find the multiplicative inverse of

$18$
$-\frac{4}{5}$
$0.6$

Answer

$\frac{1}{18}$
$-\frac{5}{4}$
$\frac{5}{3}$

Definition: Multiplication by zero

For any real number a.

$a \cdot 0 = 0 \quad 0 \cdot a = 0$

The product of any real number and 0 is 0.

What about division involving zero? What is $0\div 3$? Think about a real example: If there are no cookies in the cookie jar and 3 people are to share them, how many cookies does each person get? There are no cookies to share, so each person gets 0 cookies. So,

$0\div 3 = 0$

We can check division with the related multiplication fact.

$12 \div 6 = 2 \text { because } 2 \cdot 6 = 12$

So we know $0\div 3=0$ because $0\cdot 3=0$.

Definition: Division of Zero

For any real number a, except $0, \frac{0}{a}=0$ and $0\div a=0$.

Zero divided by any real number except zero is zero.

Now think about dividing by zero. What is the result of dividing 4 by 0? Think about the related multiplication fact: $4\div 0=?$ means $?\cdot 0=4$. Is there a number that multiplied by 0 gives 4? Since any real number multiplied by 0 gives 0, there is no real number that can be multiplied by 0 to obtain 4.

We conclude that there is no answer to $4\div 0$ and so we say that division by 0 is undefined.

Definition: Division by Zero

For any real number a, except $0, \frac{a}{0}$ and $a\div 0$ are undefined.

Division by zero is undefined.

We summarize the properties of zero below.

Definition: Properties of Zero

Multiplication by Zero: For any real number a,

\[a \cdot 0 = 0 \quad 0 \cdot a = 0 \quad \text { The product of any number and } 0 \text { is } 0\]

Division of Zero, Division by Zero: For any real number $a, a\neq 0$

$\begin{array} { l l } { \frac { 0 } { a } = 0 } & { \text { Zero divided by any real number, except itself is zero. } } \\ { \frac { a } { 0 } \text { is undefined } } & { \text { Division by zero is undefined. } } \end{array}$

Example 6

Simplify:

$-8\cdot 0$
$\frac{0}{-2}$
$\frac{-32}{0}$

Answer

$\begin{array} { cc } { } &{-8\cdot 0}\\{\text{The product of any real number and 0 is 0}} &{0}\end{array}$
$\begin{array} { ll } { } &{\frac{0}{-2}}\\{\text{Zero divided by any real number, except}} &{} \\ {\text{itself, is 0}} &{0}\end{array}$
$\begin{array} { ll } { } &{\frac{-32}{0}}\\ {\text{Division by 0 is undefined.}} &{\text{undefined}} \end{array}$

Try it Now 11

Simplify:

$-14\cdot 0$
$\frac{0}{-6}$
$\frac{-2}{0}$

Answer

$0$
$0$
undefined

Try it Now 12

Simplify:

$0(-17)$
$\frac{0}{-10}$
$\frac{-5}{0}$

Answer

$0$
$0$
undefined

We will now practice using the properties of identities, inverses, and zero to simplify expressions.

Example 7

Simplify:

$\frac{0}{n + 5}$, where $n\neq −5$
$\frac{10 - 3p}{0}$ where $10 - 3p \neq 0$

Answer

$\begin{array} { ll } { } &{\frac{0}{n + 5}}\\ {\text { Zero divided by any real number except }} &{0} \\ { \text { itself is } 0.} &{} \end{array}$
$\begin{array} { ll } { } &{\frac{10 - 3p}{0}}\\ {\text { Division by 0 is undefined }} &{\text{undefined}} \end{array}$

Example 8

Simplify: $−84n+(−73n)+84n$.

Answer: $\begin{array} { l l } { } &{−84n+(−73n)+84n} \\ { \text { Notice that the first and third terms are } } &{}\\ { \text { opposites; use the commutative property of } } &{- 84 n + 84 n + ( - 73 n ) } \\ { \text { addition to re-order the terms. } } &{} \\ \\ { \text { Add left to right. } } &{0 + (-73)}\\ \\{ \text { Add. } } &{-73n} \end{array}$

Try it Now 13

Simplify: $−27a+(−48a)+27a$.

Answer: $−48a$

Try it Now 14

Simplify: $39x+(−92x)+(−39x)$.

Answer: $−92x$

Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1.

Example 9

Simplify: $\frac{7}{15}\cdot\frac{8}{23}\cdot\frac{15}{7}$

Answer: $\begin{array} { l l } { } &{\frac{7}{15}\cdot\frac{8}{23}\cdot\frac{15}{7}} \\ { \text { Notice that the first and third terms are } } &{}\\ { \text { reciprocals, so use the commutative } } &{\frac{7}{15}\cdot\frac{15}{7}\cdot\frac{8}{23}} \\ { \text { property of multiplication to re-order the } } &{} \\ { \text { factors. } } &{}\\ \\{ \text { Multiply left to right. } } &{1\cdot\frac{8}{23}} \\\\{\text{Multiply.}} &{\frac{8}{23}}\end{array}$

Try it Now 15

Simplify: $\frac{9}{16}\cdot\frac{5}{49}\cdot\frac{16}{9}$

Answer: $\frac{5}{49}$

Try it Now 16

Simplify: $\frac{6}{17}\cdot\frac{11}{25}\cdot\frac{17}{6}$

Answer: $\frac{11}{25}$

Try it Now 17

Simplify:

$\frac{0}{m + 7}$, where $m \neq -7$
$\frac{18 - 6c}{0}$, where $18 - 6c \neq 0$

Answer

0
undefined

Try it Now 18

Simplify:

$\frac{0}{d - 4}$, where $d \neq 4$
$\frac{15 - 4q}{0}$, where $15 - 4q \neq 0$

Answer

0
undefined

Example 10

Simplify: $\frac{3}{4}\cdot\frac{4}{3}(6x + 12)$

Answer: $\begin{array} { l l } { } &{\frac{3}{4}\cdot\frac{4}{3}(6x + 12)} \\ { \text { There is nothing to do in the parentheses, } } &{}\\ { \text { so multiply the two fractions first—notice, } } &{1(6x + 12)} \\ { \text { they are reciprocals. } } &{} \\ \\{ \text { Simplify by recognizing the multiplicative } } &{} \\{\text{ identity.}} &{6x + 12} \end{array}$

Try it Now 19

Simplify: $\frac{2}{5}\cdot\frac{5}{2}(20y + 50)$

Answer: $20y + 50$

Try it Now 20

Simplify: $\frac{3}{8}\cdot\frac{8}{3}(12z + 16)$

Answer: $12z + 16$

Simplify Expressions Using the Distributive Property

Suppose that three friends are going to the movies. They each need $9.25—that’s 9 dollars and 1 quarter—to pay for their tickets. How much money do they need all together?

You can think about the dollars separately from the quarters. They need 3 times $9 so $27, and 3 times 1 quarter, so 75 cents. In total, they need $27.75. If you think about doing the math in this way, you are using the distributive property.

Definition: Distributive property

$\begin{array} { rr } {\text { If } a , b , c \text { are real numbers, then }} &{a ( b + c ) = a b + a c} \\ \\{ \text { Also,} } &{( b + c ) a = b a + c a} \\ {} &{a ( b - c ) = a b - a c } &{} \\{} &{( b - c ) a = b a - c a } \end{array}$

Back to our friends at the movies, we could find the total amount of money they need like this:

$\begin{array} { c } { 3 ( 9.25 ) } \\ { 3 ( 9 + 0.25 ) } \\ { 3 ( 9 ) + 3 ( 0.25 ) } \\ { 27 + 0.75 } \\ \\ { 27.75 } \end{array}$

In algebra, we use the distributive property to remove parentheses as we simplify expressions.

For example, if we are asked to simplify the expression $3(x+4)$, the order of operations says to work in the parentheses first. But we cannot add x and 4, since they are not like terms. So we use the distributive property, as shown in Exercise $\PageIndex{31}$.

Example 11

Simplify: $3(x+4)$.

Answer: $\begin{array} { l l } { } & { 3 ( x + 4 ) } \\ { \text { Distribute. } } & { 3 \cdot x + 3 \cdot 4 } \\ { \text { Multiply. } } & { 3 x + 12 } \end{array}$

Try it Now 21

Simplify: $4(x+2)$.

Answer: $4x + 8$

Try it Now 22

Simplify: $6(x+7)$.

Answer: $6x + 42$

Some students find it helpful to draw in arrows to remind them how to use the distributive property. Then the first step in Exercise $\PageIndex{31}$ would look like this:

We have the expression 3 times (x plus 4) with two arrows coming from the 3. One arrow points to the x, and the other arrow points to the 4.

Example 12

Simplify: $8(\frac{3}{8}x+\frac{1}{4})$.

Answer


Distribute.
Multiply.

Try it Now 23

Simplify: $6(\frac{5}{6}y+\frac{1}{2})$.

Answer: $5y + 3$

Using the distributive property will be very useful when we solve money applications in later chapters.

Example 13

Simplify: $100(0.3+0.25q)$.

Answer


Distribute.
Multiply.

Try it Now 24

Simplify: $100(0.7+0.15p)$.

Answer: $70 + 15p$

When we distribute a negative number, we need to be extra careful to get the signs correct!

Example 14

Simplify: $−2(4y+1)$.

Answer


Distribute.
Multiply.

Try it Now 25

Simplify: $−3(6m+5)$.

Answer: $−18m-15)$

Exercise $\PageIndex{46}$ will show how to use the distributive property to find the opposite of an expression.

Example 15

Simplify: $−(y+5)$.

Answer: \[\begin{array} { ll } {} &{-(y + 5)} \\ \\{ \text {Multiplying by -1 results in the opposite.} } &{-1( y + 5 )} \\ \\ {\text{Distribute.}} &{-1\cdot y + (-1)\cdot 5}\\ \\{\text{Simplify.}} &{-y + (-5)} \\ \\ {} &{-y - 5} \end{array}\]

Try it Now 26

Simplify: $−(z-11)$.

Answer: $-z + 11$

There will be times when we’ll need to use the distributive property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiplication using the distributive property, which removes the parentheses. The next two examples will illustrate this.

Example 16

Simplify: $8−2(x + 3)$.

Be sure to follow the order of operations. Multiplication comes before subtraction, so we will distribute the 2 first and then subtract.

Answer: \[\begin{array} { ll } {} &{8−2(x + 3)} \\ \\{ \text {Distribute.} } &{8−2\cdot x -2\cdot 3} \\ \\ {\text{Multiply.}} &{8 - 2x - 6}\\ \\{\text{Combine like terms.}} &{-2x + 2} \end{array}\]

Try it Now 27

Simplify: $9−3(x + 2)$.

Answer: $3 - 3x$

All the properties of real numbers we have used in this chapter are summarized in below.

Commutative Property
of addition If a,b are real numbers, then of multiplication If a,b are real numbers, then	$a+b=b+a$ $a\cdot b=b\cdot a$
Associative Property
of addition If a,b,c are real numbers, then of multiplication If a,b,c are real numbers, then	$(a+b)+c=a+(b+c)$ $(a\cdot b)\cdot c=a\cdot (b\cdot c)$
Distributive Property
If a,b,c are real numbers, then	$a(b+c)=ab+ac$
Identity Property
of addition For any real number a: 0 is the additive identity of multiplication For any real number a: 1 is the multiplicative identity	$a+0=a$ $0+a=a$ $a·1=a$ $1·a=a$
Inverse Property
of addition For any real number a, $−a$ is the additive inverse of a of multiplication For any real number $a,a\neq 0$ $\frac{1}{a}$ is the multiplicative inverse of a	$a+(−a)=0$ $a\cdot\frac{1}{a}=1$
Properties of Zero
For any real number a, For any real number $a,a\neq 0$ For any real number $a,a\neq 0$	$a\cdot 0=0$ $0\cdot a=0$ $\frac{0}{a} = 0$ $\frac{a}{0}$ is undefined

Commutative Property
of addition If a,b are real numbers, then of multiplication If a,b are real numbers, then	\(a+b=b+a\) \(a\cdot b=b\cdot a\)
Associative Property
of addition If a,b,c are real numbers, then of multiplication If a,b,c are real numbers, then	\((a+b)+c=a+(b+c)\) \((a\cdot b)\cdot c=a\cdot (b\cdot c)\)
Distributive Property
If a,b,c are real numbers, then	\(a(b+c)=ab+ac\)
Identity Property
of addition For any real number a: 0 is the additive identity of multiplication For any real number a: 1 is the multiplicative identity	\(a+0=a\) \(0+a=a\) \(a·1=a\) \(1·a=a\)
Inverse Property
of addition For any real number a, \(−a\) is the additive inverse of a of multiplication For any real number \(a,a\neq 0\) \(\frac{1}{a}\) is the multiplicative inverse of a	\(a+(−a)=0\) \(a\cdot\frac{1}{a}=1\)
Properties of Zero
For any real number a, For any real number \(a,a\neq 0\) For any real number \(a,a\neq 0\)	\(a\cdot 0=0\) \(0\cdot a=0\) \(\frac{0}{a} = 0\) \(\frac{a}{0}\) is undefined