1.10: Properties of Real Numbers
- Page ID
- 49857
By the end of this section, you will be able to:
- Use the commutative and associative properties
- Use the identity and inverse properties of addition and multiplication
- Use the properties of zero
- Simplify expressions using the distributive property
A more thorough introduction to the topics covered in this section can be found in the Prealgebra chapter, The Properties of Real Numbers.
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.
\[\begin{array} { l l } { \textbf { of Addition } } & { \text { If } a , b \text { are real numbers, then } \quad a + b = b + a } \\ { \textbf { of Multiplication } } & { \text { If } a , b \text { are real numbers, then } \quad a \cdot b = b \cdot a } \end{array}\]
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.
\[\begin{array} { l l } { \textbf { of Addition } } & { \text { If } a , b , c \text { are real numbers, then } ( a + b ) + c = a + ( b + c ) } \\ { \textbf { of Multiplication } } & { \text { If } a , b , c \text { are real numbers, then } ( a \cdot b ) \cdot c = a \cdot ( b \cdot c ) } \end{array}\]
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.
Simplify: \(18p+6q+15p+5q\).
Solution
\[\begin{array} { l l} {} &{18p+6q+15p+5q}\\ \\{ \text { Use the commutative property of addition } } &{} \\ { \text {to re-order so that like terms are together.} } &{18p+15p+ 6q+5q} \\ \\ {\text{Add like terms.}} &{33p + 11q} \end{array}\]
Simplify: \(23r+14s+9r+15s\).
- Answer
-
\(32r+29s\)
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.
Simplify: \((\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}\)
Solution
\[\begin{array} { l l } {} &{(\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}} \\{ \text { Notice that the last } 2 \text { terms have a } } \\ { \text { common denominator, so change the } } &{\frac { 5 } { 13 } + \left( \frac { 3 } { 4 } + \frac { 1 } { 4 } \right)}\\ { \text { grouping. } } &{}\\ \\ {\text{Add in parentheses first.}} &{\frac{5}{13} + (\frac{4}{4})} \\ \\ {\text{Simplify the fraction.}} &{\frac{5}{13} + 1} \\ \\ {\text{Add.}} &{1\frac{5}{13}} \\ \\ {\text{Convert to an improper fraction.}} &{\frac{18}{13}} \end{array}\]
Simplify: \((\frac{7}{15} + \frac{5}{8}) + \frac{3}{8}\)
- Answer
-
\(1\frac{7}{15}\)
Simplify: \((\frac{2}{9} + \frac{7}{12}) + \frac{5}{12}\)
- Answer
-
\(1\frac{2}{9}\)
Use the associative property to simplify \(6(3x)\).
Solution
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 \(x\).
Use the associative property to simplify \(8(4x)\).
- Answer
-
\(32x\)
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.
\[\begin{array} { l l} { \textbf {of addition}\text{ For any real number } a : } &{ a + 0 = a \quad 0 + a = a } \\ { \textbf{0} \text { is the}\textbf{ additive identity } } \\ {\textbf {of multiplication}\text{ For any real number } a : } &{ a \cdot 1 = a \quad 1 \cdot a = a } \\ { \textbf{1}\text{ is the}\textbf{ multiplicative identity } } \end{array}\]
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?
What number multiplied by 2 gives the multiplicative identity, 1? In other words 2 times what results in 1?
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:
\[\begin{array} { l l l } { \textbf { of addition } } &{ \text { For any real number } a,} &{a + (-a) = 0}\\{} &{-a \text{. is the}\textbf{ additive inverse} \text{ of }a} &{}\\ {} &{ \text { A number and its opposite add to zero. } }&{}\\ \\{ \textbf { of multiplication } } &{ \text { For any real number } a, a\neq 0} &{a\cdot \frac{1}{a} = 1}\\{} &{\frac{1}{a} \text{. is the}\textbf{ multiplicative inverse} \text{ of }a} &{}\\ {} &{ \text { A number and its reciprocal multiply to one. } }&{} \end{array}\]
Find the additive inverse of
- \(\frac{5}{8}\)
- \(0.6\)
- \(-8\)
- \(-\frac{4}{3}\)
Solution
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 \(\frac{4}{3}\).
Find the additive inverse of
- \(\frac{7}{9}\)
- \(1.2\)
- \(-14\)
- \(-\frac{9}{4}\)
- Answer
-
- \(-\frac{7}{9}\)
- \(-1.2\)
- \(14\)
- \(\frac{9}{4}\)
Find the additive inverse of
- \(\frac{7}{13}\)
- \(8.4\)
- \(-46\)
- \(-\frac{5}{2}\)
- Answer
-
- \(-\frac{7}{13}\)
- \(-8.4\)
- \(46\)
- \(\frac{5}{2}\)
Find the multiplicative inverse of
- \(9\)
- \(-\frac{1}{9}\)
- \(0.9\)
Solution
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 \(\frac{10}{9}\).
Find the multiplicative inverse of
- \(4\)
- \(-\frac{1}{7}\)
- \(0.3\)
- Answer
-
- \(\frac{1}{4}\)
- \(-7\)
- \(\frac{10}{3}\)
Find the multiplicative inverse of
- \(18\)
- \(-\frac{4}{5}\)
- \(0.6\)
- Answer
-
- \(\frac{1}{18}\)
- \(-\frac{5}{4}\)
- \(\frac{5}{3}\)
Use the Properties of Zero
The identity property of addition says that when we add 0 to any number, the result is that same number. What happens when we multiply a number by 0? Multiplying by 0 makes the product equal 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 \(0\cdot 3=0\).
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.
For any real number a, \( \frac{a}{0}\) and \(a\div 0\) are undefined.
Division by zero is undefined.
We summarize the properties of zero below.
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 \(a, a\neq 0\)
\[\begin{array} { l l } { \frac { 0 } { a } = 0 } & { \text { Zero divided by any real number, except itself is zero. } } \\ { \frac { a } { 0 } \text { is undefined } } & { \text { Division by zero is undefined. } } \end{array}\]
Simplify:
- \(-8\cdot 0\)
- \(\frac{0}{-2}\)
- \(\frac{-32}{0}\)
Solution
- \[\begin{array} { cc } { } &{-8\cdot 0}\\{\text{The product of any real number and 0 is 0}} &{0}\end{array}\]
- \[\begin{array} { ll } { } &{\frac{0}{-2}}\\{\text{Zero divided by any real number, except}} &{} \\ {\text{itself, is 0}} &{0}\end{array}\]
- \[\begin{array} { ll } { } &{\frac{-32}{0}}\\ {\text{Division by 0 is undefined.}} &{\text{undefined}} \end{array}\]
Simplify:
- \(-14\cdot 0\)
- \(\frac{0}{-6}\)
- \(\frac{-2}{0}\)
- Answer
-
- \(0\)
- \(0\)
- undefined
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.
Simplify:
- \(\frac{0}{n + 5}\), where \(n\neq −5\)
- \(\frac{10 - 3p}{0}\) where \(10 - 3p \neq 0\)
Solution
- \[\begin{array} { ll } { } &{\frac{0}{n + 5}}\\ {\text { Zero divided by any real number except }} &{0} \\ { \text { itself is } 0.} &{} \end{array}\]
- \[\begin{array} { ll } { } &{\frac{10 - 3p}{0}}\\ {\text { Division by 0 is undefined }} &{\text{undefined}} \end{array}\]
Simplify:
- \(\frac{0}{m + 7}\), where \(m \neq -7\)
- \(\frac{18 - 6c}{0}\), where \(18 - 6c \neq 0\)
- Answer
-
- 0
- undefined
Simplify:
- \(\frac{0}{d - 4}\), where \(d \neq 4\)
- \(\frac{15 - 4q}{0}\), where \(15 - 4q \neq 0\)
- Answer
-
- 0
- undefined
Simplify: \(−84n+(−73n)+84n\).
Solution
\[\begin{array} { l l } { } &{−84n+(−73n)+84n} \\ { \text { Notice that the first and third terms are } } &{}\\ { \text { opposites; use the commutative property of } } &{- 84 n + 84 n + ( - 73 n ) } \\ { \text { addition to re-order the terms. } } &{} \\ \\ { \text { Add left to right. } } &{0 + (-73)}\\ \\{ \text { Add. } } &{-73n} \end{array}\]
Simplify: \(−27a+(−48a)+27a\).
- Answer
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\(−48a\)
Simplify: \(39x+(−92x)+(−39x)\).
- Answer
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\(−92x\)
Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1.
Simplify: \(\frac{7}{15}\cdot\frac{8}{23}\cdot\frac{15}{7}\)
Solution
\[\begin{array} { l l } { } &{\frac{7}{15}\cdot\frac{8}{23}\cdot\frac{15}{7}} \\ { \text { Notice that the first and third terms are } } &{}\\ { \text { reciprocals, so use the commutative } } &{\frac{7}{15}\cdot\frac{15}{7}\cdot\frac{8}{23}} \\ { \text { property of multiplication to re-order the } } &{} \\ { \text { factors. } } &{}\\ \\{ \text { Multiply left to right. } } &{1\cdot\frac{8}{23}} \\\\{\text{Multiply.}} &{\frac{8}{23}}\end{array}\]
Simplify: \(\frac{9}{16}\cdot\frac{5}{49}\cdot\frac{16}{9}\)
- Answer
-
\(\frac{5}{49}\)
Simplify: \(\frac{6}{17}\cdot\frac{11}{25}\cdot\frac{17}{6}\)
- Answer
-
\(\frac{11}{25}\)
Simplify: \(\frac{3}{4}\cdot\frac{4}{3}(6x + 12)\)
Solution
\[\begin{array} { l l } { } &{\frac{3}{4}\cdot\frac{4}{3}(6x + 12)} \\ { \text { There is nothing to do in the parentheses, } } &{}\\ { \text { so multiply the two fractions first—notice, } } &{1(6x + 12)} \\ { \text { they are reciprocals. } } &{} \\ \\{ \text { Simplify by recognizing the multiplicative } } &{} \\{\text{ identity.}} &{6x + 12} \end{array}\]
Simplify: \(\frac{2}{5}\cdot\frac{5}{2}(20y + 50)\)
- Answer
-
\(20y + 50\)
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.
\[\begin{array} { rr } {\text { If } a , b , c \text { are real numbers, then }} &{a ( b + c ) = a b + a c} \\ \\{ \text { Also,} } &{( b + c ) a = b a + c a} \\ {} &{a ( b - c ) = a b - a c } &{} \\{} &{( b - c ) a = b a - c a } \end{array}\]
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 \(\PageIndex{31}\).
Simplify: \(3(x+4)\).
Solution
\[\begin{array} { l l } { } & { 3 ( x + 4 ) } \\ { \text { Distribute. } } & { 3 \cdot x + 3 \cdot 4 } \\ { \text { Multiply. } } & { 3 x + 12 } \end{array}\]
Simplify: \(4(x+2)\).
- Answer
-
\(4x + 8\)
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:
Simplify: \(8(\frac{3}{8}x+\frac{1}{4})\).
Solution
 |
|
| Distribute. |  |
| Multiply. |  |
Simplify: \(6(\frac{5}{6}y+\frac{1}{2})\).
- Answer
-
\(5y + 3\)
Simplify: \(12(\frac{1}{3}n+\frac{3}{4})\).
- Answer
-
\(4n + 9\)
Using the distributive property as shown in Exercise \(\PageIndex{37}\) will be very useful when we solve money applications in later chapters.
Simplify: \(100(0.3+0.25q)\).
Solution
 |
|
| Distribute. |  |
| Multiply. |  |
Simplify: \(100(0.7+0.15p)\).
- Answer
-
\(70 + 15p\)
Simplify: \(100(0.04+0.35d)\).
- Answer
-
\(4 + 35d\)
When we distribute a negative number, we need to be extra careful to get the signs correct!
Simplify: \(−2(4y+1)\).
Solution
 |
|
| Distribute. |  |
| Multiply. |  |
Simplify: \(−3(6m+5)\).
- Answer
-
\(−18m-15)\)
Simplify: \(−6(8n+11)\).
- Answer
-
\(−48n- 66)\)
Simplify: \(−11(4-3a)\).
Solution
| Distribute. |  |
| Multiply. |  |
| Simplify. |  |
Notice that you could also write the result as \(33a−44\). Do you know why?
Simplify: \(−5(2-3a)\).
- Answer
-
\(10+ 15a\)
Simplify: \(−7(8-15y)\).
- Answer
-
\(-56 + 105y\)
Exercise \(\PageIndex{46}\) will show how to use the distributive property to find the opposite of an expression.
Simplify: \(−(y+5)\).
Solution
\[\begin{array} { ll } {} &{-(y + 5)} \\ \\{ \text {Multiplying by -1 results in the opposite.} } &{-1( y + 5 )} \\ \\ {\text{Distribute.}} &{-1\cdot y + (-1)\cdot 5}\\ \\{\text{Simplify.}} &{-y + (-5)} \\ \\ {} &{-y - 5} \end{array}\]
Simplify: \(−(z-11)\).
- Answer
-
\(-z + 11\)
Simplify: \(−(x -4)\).
- Answer
-
\(-x + 4\)
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 multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.
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.
Solution
\[\begin{array} { ll } {} &{8−2(x + 3)} \\ \\{ \text {Distribute.} } &{8−2\cdot x -2\cdot 3} \\ \\ {\text{Multiply.}} &{8 - 2x - 6}\\ \\{\text{Combine like terms.}} &{-2x + 2} \end{array}\]
Simplify: \(9−3(x + 2)\).
- Answer
-
\(3 - 3x\)
Simplify: \(7x−5(x + 4)\).
- Answer
-
\(2x - 20\)
Simplify: \(4(x - 8)−(x + 3)\).
Solution
\[\begin{array} { ll } {} &{4(x - 8)−(x + 3)} \\ \\{ \text {Distribute.} } &{4x - 32 - x - 3} \\ \\{\text{Combine like terms.}} &{3x - 35} \end{array}\]
Simplify: \(6(x - 9)−(x + 12)\).
- Answer
-
\(5x - 66\)
Simplify: \(8(x - 1)-(x + 5)\).
- Answer
-
\(7x - 13\)
All the properties of real numbers we have used in this chapter are summarized in Table \(\PageIndex{1}\).
| 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: of multiplication For any real number a: |
\(a+0=a\) \(0+a=a\) \(1·a=a\) |
| 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 \(a,a\neq 0\) |
\(a\cdot 0=0\) \(0\cdot a=0\) \(\frac{0}{a} = 0\) |
Key Concepts
- Commutative Property of
- Addition: If a,b are real numbers, then \(a+b=b+a\).
- Multiplication: If a,b are real numbers, then \(a\cdot b=b\cdot a\). When adding or multiplying, changing the order gives the same result.
- Associative Property of
- Addition: If a,b,c are real numbers, then \((a+b)+c=a+(b+c)\).
- Multiplication: If a,b,c are real numbers, then \((a\cdot b)\cdot c=a\cdot (b\cdot c)\).
When adding or multiplying, changing the grouping gives the same result.
- Distributive Property: If a,b,c are real numbers, then
- \(a(b+c)=ab+ac\)
- \((b+c)a=ba+ca\)
- \(a(b-c)=ab-ac\)
- \((b+c)a=ba+ca\)
- Identity Property
- of Addition: For any real number a: \(a+0=a\)
0 is the additive identity - of Multiplication: For any real number a: \(a\cdot 1=a \quad 1·a=a\)
11 is the multiplicative identity
- of Addition: For any real number a: \(a+0=a\)
- Inverse Property
- of Addition: For any real number \(a, a+(−a)=0\). A number and its opposite add to zero. \(−a\) is the additive inverse of a.
- of Multiplication: For any real number \(a,(a\neq 0)a\cdot\frac{1}{a}=1\). A number and its reciprocal multiply to one. \(\frac{1}{a}\) is the multiplicative inverse of a.
- Properties of Zero
- For any real number a,
\(a\cdot 0=0 \quad 0·a=0\) – The product of any real number and 0 is 0. - \(\frac{0}{a}=0\) for \(a\neq 0\) – Zero divided by any real number except zero is zero.
- \(\frac{a}{0}\) is undefined – Division by zero is undefined.
- For any real number a,