#set document(title: "1.9 Properties of Real Numbers", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 1.9#h(0.6em)Properties of Real Numbers #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A more thorough introduction to the topics covered in this section can be found in the #emph[Prealgebra] chapter, #strong[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? #math.equation(block: true, alt: "5 plus 3, 3 plus 5; 8, 8")[$5 + 3 & & & 3 + 5 \ 8 & & & 8$]#math.equation(block: true, alt: "5 plus 3 equals 3 plus 5")[$5 + 3 = 3 + 5$]The results are the same. As we can see, the order in which we add does not matter! What about multiplying #math.equation(block: false, alt: "5 and 3 ?")[$5 #h(0.2em) "and" #h(0.2em) 3 ?$] #math.equation(block: true, alt: "5 times 3, 3 times 5; 15, 15")[$5 · 3 & & & 3 · 5 \ 15 & & & 15$]#math.equation(block: true, alt: "5 times 3 equals 3 times 5")[$5 · 3 = 3 · 5$]Again, the results are the same! The order in which we multiply does not matter! These examples illustrate the #strong[commutative property]. When adding or multiplying, changing the #emph[order] gives the same result. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Commutative Property] #math.equation(block: true, alt: "of Addition, If a , b are real numbers, then, a plus b, equals, b plus a; of Multiplication, If a , b are real numbers, then, a times b, equals, b times a")[$"of Addition" & & & "If" #h(0.2em) a , b #h(0.2em) "are real numbers, then" & & & a + b & = & b + a \ "of Multiplication" & & & "If" #h(0.2em) a , b #h(0.2em) "are real numbers, then" & & & a · b & = & b · a$]When adding or multiplying, changing the #emph[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 #math.equation(block: false, alt: "7 minus 3")[$7 − 3$] give the same result as #math.equation(block: false, alt: "3 minus 7 ?")[$3 − 7 ?$] #math.equation(block: true, alt: "7 minus 3, 3 minus 7; 4, −4; 4 not equal to − 4; 7 minus 3 not equal to 3 minus 7")[$7 − 3 & #h(1em) 3 − 7 \ 4 & #h(1em) −4 \ \ \ #h(0.5em) 4 ≠ "−" 4 \ 7 − 3 ≠ 3 − 7$]The results are not the same. Since changing the order of the subtraction did not give the same result, we know that #emph[subtraction is not commutative]. Let’s see what happens when we divide two numbers. Is #strong[division] commutative? #math.equation(block: true, alt: "12 divided by 4, 4 divided by 12; the fraction 12 over 4, the fraction 4 over 12; 3, the fraction 1 over 3; 3 not equal to the fraction 1 over 3; 12 divided by 4 not equal to 4 divided by 12")[$12 ÷ 4 & #h(1em) 4 ÷ 12 \ frac(12, 4) & #h(1em) frac(4, 12) \ 3 & #h(1em) frac(1, 3) \ 3 ≠ frac(1, 3) \ 12 ÷ 4 ≠ 4 ÷ 12$]The results are not the same. Since changing the order of the division did not give the same result, #emph[division is not commutative]. The commutative properties only apply to addition and multiplication! - Addition and multiplication #emph[are] commutative. - Subtraction and Division #emph[are not] commutative. #linebreak() If you were asked to simplify this expression, how would you do it and what would your answer be? #math.equation(block: true, alt: "7 plus 8 plus 2")[$7 + 8 + 2$]Some people would think #math.equation(block: false, alt: "7 plus 8 is 15")[$7 + 8 #h(0.2em) "is" #h(0.2em) 15$] and then #math.equation(block: false, alt: "15 plus 2 is 17 .")[$15 + 2 #h(0.2em) "is" #h(0.2em) 17 .$] Others might start with #math.equation(block: false, alt: "8 plus 2 makes 10")[$8 + 2 #h(0.2em) "makes" #h(0.2em) 10$] and then #math.equation(block: false, alt: "7 plus 10 makes 17 .")[$7 + 10 #h(0.2em) "makes" #h(0.2em) 17 .$] Either way gives the same result. Remember, we use parentheses as grouping symbols to indicate which operation should be done first. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#linebreak() Add #math.equation(block: false, alt: "7 plus 8")[$7 + 8$]. #linebreak() Add.], [#math.equation(block: false, alt: "open parenthesis 7 plus 8 close parenthesis plus 2; 15 plus 2; 17")[$( 7 + 8 ) + 2 \ 15 + 2 \ 17$]]), [#linebreak() Add #math.equation(block: false, alt: "8 plus 2")[$8 + 2$]. #linebreak() Add.], [#math.equation(block: false, alt: "7 plus open parenthesis 8 plus 2 close parenthesis; 7 plus 10; 17")[$7 + ( 8 + 2 ) \ 7 + 10 \ 17$]], [], [#math.equation(block: false, alt: "open parenthesis 7 plus 8 close parenthesis plus 2 equals 7 plus open parenthesis 8 plus 2 close parenthesis")[$( 7 + 8 ) + 2 = 7 + ( 8 + 2 )$]], )) When adding three numbers, changing the grouping of the numbers gives the same result. This is true for multiplication, too. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#linebreak() Multiply. #math.equation(block: false, alt: "5 times the fraction 1 over 3")[$5 · frac(1, 3)$] #linebreak() Multiply.], [#math.equation(block: false, alt: "open parenthesis 5 times the fraction 1 over 3 close parenthesis times 3; the fraction 5 over 3 times 3; 5")[$#h(2em) ( 5 · frac(1, 3) ) · 3 \ frac(5, 3) · 3 \ 5$]]), [#linebreak() Multiply. #math.equation(block: false, alt: "the fraction 1 over 3 times 3")[$frac(1, 3) · 3$]. #linebreak() Multiply.], [#math.equation(block: false, alt: "5 times open parenthesis the fraction 1 over 3 times 3 close parenthesis; 5 times 1; 5")[$5 · ( frac(1, 3) · 3 ) \ 5 · 1 \ 5$]], [], [#math.equation(block: false, alt: "open parenthesis 5 times the fraction 1 over 3 close parenthesis times 3 equals 5 times open parenthesis the fraction 1 over 3 times 3 close parenthesis")[$( 5 · frac(1, 3) ) · 3 = 5 · ( frac(1, 3) · 3 )$]], )) 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 #strong[associative property]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Associative Property] #math.equation(block: true, alt: "of Addition, If a , b , c are real numbers, then open parenthesis a plus b close parenthesis plus c equals a plus open parenthesis b plus c close parenthesis; of Multiplication, If a , b , c are real numbers, then open parenthesis a times b close parenthesis times c equals a times open parenthesis b times c close parenthesis")[$"of Addition" & & & "If" #h(0.2em) a , b , c #h(0.2em) "are real numbers, then" #h(0.2em) ( a + b ) + c = a + ( b + c ) \ "of Multiplication" & & & "If" #h(0.2em) a , b , c #h(0.2em) "are real numbers, then" #h(0.2em) ( a · b ) · c = a · ( b · c )$]When adding or multiplying, changing the #emph[grouping] gives the same result. ] Let’s think again about multiplying #math.equation(block: false, alt: "5 times the fraction 1 over 3 times 3 .")[$5 · frac(1, 3) · 3 .$] We got the same result both ways, but which way was easier? Multiplying #math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$] and #math.equation(block: false, alt: "3")[$3$] 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 #strong[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. #examplebox("Example 1")[][ Simplify: #math.equation(block: false, alt: "18 p plus 6 q plus 15 p plus 5 q .")[$18 p + 6 q + 15 p + 5 q .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "18 p plus 6 q plus 15 p plus 5 q")[$18 p + 6 q + 15 p + 5 q$]]), [Use the commutative property of addition to re-order so that like terms are together.], [#math.equation(block: false, alt: "18 p plus 15 p plus 6 q plus 5 q")[$18 p + 15 p + 6 q + 5 q$]], [Add like terms.], [#math.equation(block: false, alt: "33 p plus 11 q")[$33 p + 11 q$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "23 r plus 14 s plus 9 r plus 15 s .")[$23 r + 14 s + 9 r + 15 s .$] #solutionbox[ #math.equation(block: true, alt: "32 r plus 29 s")[$32 r + 29 s$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "37 m plus 21 n plus 4 m minus 15 n .")[$37 m + 21 n + 4 m − 15 n .$] #solutionbox[ #math.equation(block: true, alt: "41 m plus 6 n")[$41 m + 6 n$] ] ] When we have to simplify algebraic #strong[expression]s, 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. #examplebox("Example 2")[][ Simplify: #math.equation(block: false, alt: "open parenthesis the fraction 5 over 13 plus the fraction 3 over 4 close parenthesis plus the fraction 1 over 4 .")[$( frac(5, 13) + frac(3, 4) ) + frac(1, 4) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis the fraction 5 over 13 plus the fraction 3 over 4 close parenthesis plus the fraction 1 over 4")[$( frac(5, 13) + frac(3, 4) ) + frac(1, 4)$]]), [Notice that the last 2 terms have a common denominator, so change the grouping.], [#math.equation(block: false, alt: "the fraction 5 over 13 plus open parenthesis the fraction 3 over 4 plus the fraction 1 over 4 close parenthesis")[$frac(5, 13) + ( frac(3, 4) + frac(1, 4) )$]], [Add in parentheses first.], [#math.equation(block: false, alt: "the fraction 5 over 13 plus open parenthesis the fraction 4 over 4 close parenthesis")[$frac(5, 13) + ( frac(4, 4) )$]], [Simplify the fraction.], [#math.equation(block: false, alt: "the fraction 5 over 13 plus 1")[$frac(5, 13) + 1$]], [Add.], [#math.equation(block: false, alt: "1 the fraction 5 over 13")[$1 frac(5, 13)$]], [Convert to an improper fraction.], [#math.equation(block: false, alt: "the fraction 18 over 13")[$frac(18, 13)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "open parenthesis the fraction 7 over 15 plus the fraction 5 over 8 close parenthesis plus the fraction 3 over 8 .")[$( frac(7, 15) + frac(5, 8) ) + frac(3, 8) .$] #solutionbox[ #math.equation(block: true, alt: "1 the fraction 7 over 15")[$1 frac(7, 15)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "open parenthesis the fraction 2 over 9 plus the fraction 7 over 12 close parenthesis plus the fraction 5 over 12 .")[$( frac(2, 9) + frac(7, 12) ) + frac(5, 12) .$] #solutionbox[ #math.equation(block: true, alt: "1 the fraction 2 over 9")[$1 frac(2, 9)$] ] ] #examplebox("Example 3")[][ Use the associative property to simplify #math.equation(block: false, alt: "6 open parenthesis 3 x close parenthesis .")[$6 ( 3 x ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "6 open parenthesis 3 x close parenthesis")[$6 ( 3 x )$]]), [Change the grouping.], [#math.equation(block: false, alt: "open parenthesis 6 times 3 close parenthesis x")[$( 6 · 3 ) x$]], [Multiply in the parentheses.], [#math.equation(block: false, alt: "18 x")[$18 x$]], )) Notice that we can multiply #math.equation(block: false, alt: "6 times 3")[$6 · 3$] but we could not multiply 3#emph[x] without having a value for #emph[x]. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the associative property to simplify 8(4#emph[x]). #solutionbox[ 32#emph[x] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the associative property to simplify #math.equation(block: false, alt: "−9 open parenthesis 7 y close parenthesis .")[$−9 ( 7 y ) .$] #solutionbox[ #math.equation(block: true, alt: "−63 y")[$−63 y$] ] ] === 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 #strong[additive identity]. For example, #math.equation(block: true, alt: "13 plus 0, −14 plus 0, 0 plus open parenthesis −8 close parenthesis; 13, minus 14, minus 8")[$13 + 0 & & & −14 + 0 & & & 0 + ( −8 ) \ 13 & & & − 14 & & & − 8$]These examples illustrate the #strong[Identity Property of Addition] that states that for any real number #math.equation(block: false, alt: "a ,")[$a ,$] #math.equation(block: false, alt: "a plus 0 equals a")[$a + 0 = a$] and #math.equation(block: false, alt: "0 plus a equals a .")[$0 + a = a .$] What happens when we multiply any number by one? Multiplying by 1 doesn’t change the value. So we call 1 the #strong[multiplicative identity]. For example, #math.equation(block: true, alt: "43 times 1, minus 27 times 1, 1 times the fraction 3 over 5; 43, minus 27, the fraction 3 over 5")[$43 · 1 & & & − 27 · 1 & & & 1 · frac(3, 5) \ 43 & & & − 27 & & & frac(3, 5)$]These examples illustrate the #strong[Identity Property of Multiplication] that states that for any real number #math.equation(block: false, alt: "a ,")[$a ,$] #math.equation(block: false, alt: "a times 1 equals a")[$a · 1 = a$] and #math.equation(block: false, alt: "1 times a equals a .")[$1 · a = a .$] We summarize the Identity Properties below. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Identity Property] #math.equation(block: true, alt: "of addition For any real number a :, a plus 0 equals a, 0 plus a equals a; 0 is the additive identity; of multiplication For any real number a :, a times 1 equals a, 1 times a equals a; 1 is the multiplicative identity")[$"of addition" #h(0.2em) "For any real number" #h(0.2em) a ":" & & a + 0 = a & & 0 + a = a \ 0 #h(0.2em) "is the" #h(0.2em) "additive identity" & & & & \ \ \ "of multiplication" #h(0.2em) "For any real number" #h(0.2em) a ":" & & a · 1 = a & & 1 · a = a \ 1 #h(0.2em) "is the" #h(0.2em) "multiplicative identity"$] ] #figure(figph[In the top line of this figure, we have the question “What number added to 5 gives the additive identity, 0?” On the following line, we have 5 plus a blank space equals 0. Then it is stated that “We know 5 plus negative 5 equals 0.” On the following line, we have the question “What number added to negative 6 gives the additive identity, 0?” On the following line, we have negative 6 plus a blank space equals 0. Then it is stated that “We know negative 6 plus 6 equals 0.”], alt: "In the top line of this figure, we have the question “What number added to 5 gives the additive identity, 0?” On the following line, we have 5 plus a blank space equals 0. Then it is stated that “We know 5 plus negative 5 equals 0.” On the following line, we have the question “What number added to negative 6 gives the additive identity, 0?” On the following line, we have negative 6 plus a blank space equals 0. Then it is stated that “We know negative 6 plus 6 equals 0.”", caption: none) Notice that in each case, the missing number was the opposite of the number! We call #math.equation(block: false, alt: "− a .")[$"−" a .$] the #strong[additive inverse] of #emph[a]. #emph[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 #strong[Inverse Property of Addition] that states for any real number #math.equation(block: false, alt: "a , a plus open parenthesis − a close parenthesis equals 0 .")[$a , a + ( "−" a ) = 0 .$] Remember, a number and its opposite add to zero. What number multiplied by #math.equation(block: false, alt: "the fraction 2 over 3")[$frac(2, 3)$] gives the multiplicative identity, 1? In other words, #math.equation(block: false, alt: "the fraction 2 over 3")[$frac(2, 3)$] times what results in 1? #figure(figph[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.”], alt: "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.”", caption: none) What number multiplied by 2 gives the multiplicative identity, 1? In other words 2 times what results in 1? #figure(figph[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.”], alt: "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.”", caption: none) Notice that in each case, the missing number was the #strong[reciprocal] of the number! We call #math.equation(block: false, alt: "the fraction 1 over a")[$frac(1, a)$] the #strong[multiplicative inverse] of #emph[a]. #emph[The reciprocal of #math.equation(block: false, alt: "a")[$a$] number is its multiplicative inverse.] A number and its reciprocal multiply to one, which is the multiplicative identity. This leads to the #strong[Inverse Property of Multiplication] that states that for any real number #math.equation(block: false, alt: "a , a not equal to 0 , a times the fraction 1 over a equals 1 .")[$a , a ≠ 0 , a · frac(1, a) = 1 .$] We’ll formally state the inverse properties here: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Property] #figure(table( columns: 3, align: left, inset: 6pt, table.header([#emph[of addition]], [For any real number #math.equation(block: false, alt: "a")[$a$], #linebreak() #math.equation(block: false, alt: "− a")[$"−" a$] is the #emph[additive inverse] of #math.equation(block: false, alt: "a")[$a$]. #linebreak() A number and its opposite add to zero.], [#math.equation(block: false, alt: "a plus open parenthesis − a close parenthesis equals 0")[$a + ( "−" a ) = 0$]]), [#emph[of multiplication]], [For any real number #math.equation(block: false, alt: "a")[$a$], #linebreak() #math.equation(block: false, alt: "the fraction 1 over a")[$frac(1, a)$] is the #emph[multiplicative inverse] of #math.equation(block: false, alt: "a")[$a$]. #linebreak() A number and its reciprocal multiply to one.], [#math.equation(block: false, alt: "a times the fraction 1 over a equals 1")[$a · frac(1, a) = 1$]], )) ] #examplebox("Example 4")[][ Find the additive inverse of ⓐ #math.equation(block: false, alt: "the fraction 5 over 8")[$frac(5, 8)$] ⓑ #math.equation(block: false, alt: "0.6")[$0.6$] ⓒ #math.equation(block: false, alt: "−8")[$−8$] ⓓ #math.equation(block: false, alt: "minus the fraction 4 over 3 .")[$− #h(0.2em) frac(4, 3) .$] #solutionbox[ To find the additive inverse, we find the opposite. + ⓐ The additive inverse of #math.equation(block: false, alt: "the fraction 5 over 8")[$frac(5, 8)$] is the opposite of #math.equation(block: false, alt: "the fraction 5 over 8 .")[$frac(5, 8) .$] The additive inverse of #math.equation(block: false, alt: "the fraction 5 over 8")[$frac(5, 8)$] is #math.equation(block: false, alt: "minus the fraction 5 over 8 .")[$− #h(0.2em) frac(5, 8) .$] #linebreak() #linebreak() + ⓑ The additive inverse of 0.6 is the opposite of 0.6. The additive inverse of 0.6 is #math.equation(block: false, alt: "−0.6 .")[$−0.6 .$] #linebreak() #linebreak() + ⓒ The additive inverse of #math.equation(block: false, alt: "−8")[$−8$] is the opposite of #math.equation(block: false, alt: "−8 .")[$−8 .$] We write the opposite of #math.equation(block: false, alt: "−8")[$−8$] as #math.equation(block: false, alt: "− open parenthesis −8 close parenthesis ,")[$"−" ( −8 ) ,$] and then simplify it to 8. Therefore, the additive inverse of #math.equation(block: false, alt: "−8")[$−8$] is 8. #linebreak() #linebreak() + ⓓ The additive inverse of #math.equation(block: false, alt: "minus the fraction 4 over 3")[$− #h(0.2em) frac(4, 3)$] is the opposite of #math.equation(block: false, alt: "minus the fraction 4 over 3 .")[$− #h(0.2em) frac(4, 3) .$] We write this as #math.equation(block: false, alt: "− open parenthesis minus the fraction 4 over 3 close parenthesis ,")[$"−" ( − #h(0.2em) frac(4, 3) ) ,$] and then simplify to #math.equation(block: false, alt: "the fraction 4 over 3 .")[$frac(4, 3) .$] Thus, the additive inverse of #math.equation(block: false, alt: "minus the fraction 4 over 3")[$− #h(0.2em) frac(4, 3)$] is #math.equation(block: false, alt: "the fraction 4 over 3 .")[$frac(4, 3) .$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the additive inverse of: ⓐ #math.equation(block: false, alt: "the fraction 7 over 9")[$frac(7, 9)$] ⓑ #math.equation(block: false, alt: "1.2")[$1.2$] ⓒ #math.equation(block: false, alt: "−14")[$−14$] ⓓ #math.equation(block: false, alt: "minus the fraction 9 over 4 .")[$− #h(0.2em) frac(9, 4) .$] #solutionbox[ ⓐ #math.equation(block: false, alt: "minus the fraction 7 over 9")[$− #h(0.2em) frac(7, 9)$] ⓑ #math.equation(block: false, alt: "−1.2")[$−1.2$] ⓒ #math.equation(block: false, alt: "14")[$14$] ⓓ #math.equation(block: false, alt: "the fraction 9 over 4")[$frac(9, 4)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the additive inverse of: ⓐ #math.equation(block: false, alt: "the fraction 7 over 13")[$frac(7, 13)$] ⓑ #math.equation(block: false, alt: "8.4")[$8.4$] ⓒ #math.equation(block: false, alt: "−46")[$−46$] ⓓ #math.equation(block: false, alt: "minus the fraction 5 over 2 .")[$− #h(0.2em) frac(5, 2) .$] #solutionbox[ ⓐ #math.equation(block: false, alt: "minus the fraction 7 over 13")[$− #h(0.2em) frac(7, 13)$] ⓑ #math.equation(block: false, alt: "−8.4")[$−8.4$] ⓒ #math.equation(block: false, alt: "46")[$46$] ⓓ #math.equation(block: false, alt: "the fraction 5 over 2")[$frac(5, 2)$] ] ] #examplebox("Example 5")[][ Find the multiplicative inverse of ⓐ #math.equation(block: false, alt: "9")[$9$] ⓑ #math.equation(block: false, alt: "minus the fraction 1 over 9")[$− #h(0.2em) frac(1, 9)$] ⓒ #math.equation(block: false, alt: "0.9 .")[$0.9 .$] #solutionbox[ To find the multiplicative inverse, we find the reciprocal. + ⓐ The multiplicative inverse of 9 is the reciprocal of 9, which is #math.equation(block: false, alt: "the fraction 1 over 9 .")[$frac(1, 9) .$] Therefore, the multiplicative inverse of 9 is #math.equation(block: false, alt: "the fraction 1 over 9 .")[$frac(1, 9) .$] + ⓑ The multiplicative inverse of #math.equation(block: false, alt: "minus the fraction 1 over 9")[$− #h(0.2em) frac(1, 9)$] is the reciprocal of #math.equation(block: false, alt: "minus the fraction 1 over 9 ,")[$− #h(0.2em) frac(1, 9) ,$] which is #math.equation(block: false, alt: "−9 .")[$−9 .$] Thus, the multiplicative inverse of #math.equation(block: false, alt: "minus the fraction 1 over 9")[$− #h(0.2em) frac(1, 9)$] is #math.equation(block: false, alt: "−9 .")[$−9 .$] + ⓒ To find the multiplicative inverse of 0.9, we first convert 0.9 to a fraction, #math.equation(block: false, alt: "the fraction 9 over 10 .")[$frac(9, 10) .$] Then we find the reciprocal of the fraction. The reciprocal of #math.equation(block: false, alt: "the fraction 9 over 10")[$frac(9, 10)$] is #math.equation(block: false, alt: "the fraction 10 over 9 .")[$frac(10, 9) .$] So the multiplicative inverse of 0.9 is #math.equation(block: false, alt: "the fraction 10 over 9 .")[$frac(10, 9) .$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the multiplicative inverse of ⓐ #math.equation(block: false, alt: "4")[$4$] ⓑ #math.equation(block: false, alt: "minus the fraction 1 over 7")[$− #h(0.2em) frac(1, 7)$] ⓒ #math.equation(block: false, alt: "0.3")[$0.3$] #solutionbox[ ⓐ #math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$] ⓑ #math.equation(block: false, alt: "−7")[$−7$] ⓒ #math.equation(block: false, alt: "the fraction 10 over 3")[$frac(10, 3)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the multiplicative inverse of ⓐ #math.equation(block: false, alt: "18")[$18$] ⓑ #math.equation(block: false, alt: "minus the fraction 4 over 5")[$− #h(0.2em) frac(4, 5)$] ⓒ #math.equation(block: false, alt: "0.6 .")[$0.6 .$] #solutionbox[ ⓐ #math.equation(block: false, alt: "the fraction 1 over 18")[$frac(1, 18)$] ⓑ #math.equation(block: false, alt: "minus the fraction 5 over 4")[$− #h(0.2em) frac(5, 4)$] ⓒ #math.equation(block: false, alt: "the fraction 5 over 3")[$frac(5, 3)$] ] ] === 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 #strong[product] equal zero. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Multiplication by Zero] For any real number #emph[a]. #math.equation(block: true, alt: "a times 0 equals 0 0 times a equals 0")[$a · 0 = 0 #h(6em) 0 · a = 0$]The product of any real number and 0 is 0. ] What about #strong[division] involving zero? What is #math.equation(block: false, alt: "0 divided by 3 ?")[$0 ÷ 3 ?$] 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, #math.equation(block: true, alt: "0 divided by 3 equals 0")[$0 ÷ 3 = 0$]We can check division with the related multiplication fact. #math.equation(block: true, alt: "12 divided by 6 equals 2 because 2 times 6 equals 12 .")[$12 ÷ 6 = 2 #h(0.2em) "because" #h(0.2em) 2 · 6 = 12 .$]So we know #math.equation(block: false, alt: "0 divided by 3 equals 0")[$0 ÷ 3 = 0$] because #math.equation(block: false, alt: "0 times 3 equals 0 .")[$0 · 3 = 0 .$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Division of Zero] For any real number #emph[a], except #math.equation(block: false, alt: "0 ,")[$0 ,$] #math.equation(block: false, alt: "the fraction 0 over a equals 0")[$frac(0, a) = 0$] and #math.equation(block: false, alt: "0 divided by a equals 0 .")[$0 ÷ a = 0 .$] Zero divided by any real number except zero is zero. ] Now think about dividing #emph[by] zero. What is the result of dividing 4 by 0? Think about the related multiplication fact: #math.equation(block: false, alt: "4 divided by 0 equals ?")[$4 ÷ 0 = ?$] means #math.equation(block: false, alt: "? times 0 equals 4 .")[$? · 0 = 4 .$] 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 #math.equation(block: false, alt: "4 divided by 0")[$4 ÷ 0$] and so we say that division by 0 is undefined. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Division by Zero] For any real number #emph[a], except 0, #math.equation(block: false, alt: "the fraction a over 0")[$frac(a, 0)$] and #math.equation(block: false, alt: "a divided by 0")[$a ÷ 0$] are undefined. Division by zero is undefined. ] We summarize the properties of zero below. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of Zero] #strong[Multiplication by Zero:] For any real number #emph[a], #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "a times 0 equals 0 0 times a equals 0")[$a · 0 = 0 #h(1em) 0 · a = 0$]], [The product of any number and 0 is 0.], )) #strong[Division of Zero, Division by Zero:] For any real number #math.equation(block: false, alt: "a , a not equal to 0")[$a , a ≠ 0$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "the fraction 0 over a equals 0")[$frac(0, a) = 0$]], [Zero divided by any real number except itself is zero.]), [#math.equation(block: false, alt: "the fraction a over 0 is undefined")[$frac(a, 0) #h(0.2em) "is undefined"$]], [Division by zero is undefined.], )) ] #examplebox("Example 6")[][ Simplify: ⓐ #math.equation(block: false, alt: "−8 times 0")[$−8 · 0$] ⓑ #math.equation(block: false, alt: "the fraction 0 over −2")[$frac(0, −2)$] ⓒ #math.equation(block: false, alt: "the fraction −32 over 0 .")[$frac(−32, 0) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([ⓐ #linebreak() The product of any real number and 0 is 0.], [#math.equation(block: false, alt: "−8 times 0; 0")[$−8 · 0 \ 0$]]), [ⓑ #linebreak() The product of any real number and 0 is 0.], [#math.equation(block: false, alt: "the fraction 0 over −2; 0")[$frac(0, −2) \ 0$]], [ⓒ #linebreak() Division by 0 is undefined.], [#math.equation(block: false, alt: "the fraction −32 over 0; Undefined")[$frac(−32, 0) \ "Undefined"$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: ⓐ #math.equation(block: false, alt: "−14 times 0")[$−14 · 0$] ⓑ #math.equation(block: false, alt: "the fraction 0 over −6")[$frac(0, −6)$] ⓒ #math.equation(block: false, alt: "the fraction −2 over 0 .")[$frac(−2, 0) .$] #solutionbox[ ⓐ 0 ⓑ 0 ⓒ undefined ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: ⓐ #math.equation(block: false, alt: "0 open parenthesis −17 close parenthesis")[$0 ( −17 )$] ⓑ #math.equation(block: false, alt: "the fraction 0 over −10")[$frac(0, −10)$] ⓒ #math.equation(block: false, alt: "the fraction −5 over 0 .")[$frac(−5, 0) .$] #solutionbox[ ⓐ 0 ⓑ 0 ⓒ undefined ] ] We will now practice using the properties of identities, inverses, and zero to simplify expressions. #examplebox("Example 7")[][ Simplify: ⓐ #math.equation(block: false, alt: "the fraction 0 over n plus 5 ,")[$frac(0, n + 5) ,$] where #math.equation(block: false, alt: "n not equal to − 5")[$n ≠ "−" 5$] ⓑ #math.equation(block: false, alt: "the fraction 10 minus 3 p over 0 ,")[$frac(10 − 3 p, 0) ,$] where #math.equation(block: false, alt: "10 minus 3 p not equal to 0 .")[$10 − 3 p ≠ 0 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([ⓐ #linebreak() Zero divided by any real number except itself is 0.], [#math.equation(block: false, alt: "the fraction 0 over n plus 5; 0")[$frac(0, n + 5) \ 0$]]), [ⓑ #linebreak() Division by 0 is undefined.], [#math.equation(block: false, alt: "the fraction 10 minus 3 p over 0; Undefined")[$frac(10 − 3 p, 0) \ "Undefined"$]], )) ] ] #examplebox("Example 8")[][ Simplify: #math.equation(block: false, alt: "−84 n plus open parenthesis −73 n close parenthesis plus 84 n .")[$−84 n + ( −73 n ) + 84 n .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "−84 n plus open parenthesis −73 n close parenthesis plus 84 n")[$−84 n + ( −73 n ) + 84 n$]]), [Notice that the first and third terms are opposites; use the #linebreak() commutative property of addition to re-order the terms.], [#math.equation(block: false, alt: "−84 n plus 84 n plus open parenthesis −73 n close parenthesis")[$−84 n + 84 n + ( −73 n )$]], [Add left to right.], [#math.equation(block: false, alt: "0 plus open parenthesis −73 n close parenthesis")[$0 + ( −73 n )$]], [Add.], [#math.equation(block: false, alt: "−73 n")[$−73 n$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "−27 a plus open parenthesis −48 a close parenthesis plus 27 a .")[$−27 a + ( −48 a ) + 27 a .$] #solutionbox[ #math.equation(block: true, alt: "−48 a")[$−48 a$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "39 x plus open parenthesis −92 x close parenthesis plus open parenthesis −39 x close parenthesis .")[$39 x + ( −92 x ) + ( −39 x ) .$] #solutionbox[ #math.equation(block: true, alt: "−92 x")[$−92 x$] ] ] Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1. #examplebox("Example 9")[][ Simplify: #math.equation(block: false, alt: "the fraction 7 over 15 times the fraction 8 over 23 times the fraction 15 over 7 .")[$frac(7, 15) · frac(8, 23) · frac(15, 7) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "the fraction 7 over 15 times the fraction 8 over 23 times the fraction 15 over 7")[$frac(7, 15) · frac(8, 23) · frac(15, 7)$]]), [Notice that the first and third terms are reciprocals, so use the #linebreak() commutative property of multiplication to re-order the factors.], [#math.equation(block: false, alt: "the fraction 7 over 15 times the fraction 15 over 7 times the fraction 8 over 23")[$frac(7, 15) · frac(15, 7) · frac(8, 23)$]], [Multiply left to right.], [#math.equation(block: false, alt: "1 times the fraction 8 over 23")[$1 · frac(8, 23)$]], [Multiply.], [#math.equation(block: false, alt: "the fraction 8 over 23")[$frac(8, 23)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "the fraction 9 over 16 times the fraction 5 over 49 times the fraction 16 over 9 .")[$frac(9, 16) · frac(5, 49) · frac(16, 9) .$] #solutionbox[ #math.equation(block: true, alt: "the fraction 5 over 49")[$frac(5, 49)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "the fraction 6 over 17 times the fraction 11 over 25 times the fraction 17 over 6 .")[$frac(6, 17) · frac(11, 25) · frac(17, 6) .$] #solutionbox[ #math.equation(block: true, alt: "the fraction 11 over 25")[$frac(11, 25)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: ⓐ #math.equation(block: false, alt: "the fraction 0 over m plus 7 ,")[$frac(0, m + 7) ,$] where #math.equation(block: false, alt: "m not equal to − 7")[$m ≠ "−" 7$] ⓑ #math.equation(block: false, alt: "the fraction 18 minus 6 c over 0 ,")[$frac(18 − 6 c, 0) ,$] where #math.equation(block: false, alt: "18 minus 6 c not equal to 0 .")[$18 − 6 c ≠ 0 .$] #solutionbox[ ⓐ 0 ⓑ undefined ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: ⓐ #math.equation(block: false, alt: "the fraction 0 over d minus 4 , where d not equal to 4")[$frac(0, d − 4) , "where" #h(0.2em) d ≠ 4$] ⓑ #math.equation(block: false, alt: "the fraction 15 minus 4 q over 0 , where 15 minus 4 q not equal to 0 .")[$frac(15 − 4 q, 0) , "where" #h(0.2em) 15 − 4 q ≠ 0 .$] #solutionbox[ ⓐ 0 ⓑ undefined ] ] #examplebox("Example 10")[][ Simplify: #math.equation(block: false, alt: "the fraction 3 over 4 times the fraction 4 over 3 open parenthesis 6 x plus 12 close parenthesis .")[$frac(3, 4) · frac(4, 3) ( 6 x + 12 ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "the fraction 3 over 4 times the fraction 4 over 3 open parenthesis 6 x plus 12 close parenthesis")[$frac(3, 4) · frac(4, 3) ( 6 x + 12 )$]]), [There is nothing to do in the parentheses, so multiply the #linebreak() two fractions first—notice, they are reciprocals.], [#math.equation(block: false, alt: "1 open parenthesis 6 x plus 12 close parenthesis")[$1 ( 6 x + 12 )$]], [Simplify by recognizing the multiplicative identity.], [#math.equation(block: false, alt: "6 x plus 12")[$6 x + 12$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "the fraction 2 over 5 times the fraction 5 over 2 open parenthesis 20 y plus 50 close parenthesis .")[$frac(2, 5) · frac(5, 2) ( 20 y + 50 ) .$] #solutionbox[ #math.equation(block: true, alt: "20 y plus 50")[$20 y + 50$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "the fraction 3 over 8 times the fraction 8 over 3 open parenthesis 12 z plus 16 close parenthesis .")[$frac(3, 8) · frac(8, 3) ( 12 z + 16 ) .$] #solutionbox[ #math.equation(block: true, alt: "12 z plus 16")[$12 z + 16$] ] ] === 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 #strong[distributive property]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Distributive Property] #math.equation(block: true, alt: "If a , b , c are real numbers, then, a open parenthesis b plus c close parenthesis equals a b plus a c; Also,, open parenthesis b plus c close parenthesis a equals b a plus c a; a open parenthesis b minus c close parenthesis equals a b minus a c; open parenthesis b minus c close parenthesis a equals b a minus c a")[$"If" #h(0.2em) a , b , c #h(0.2em) "are real numbers, then" & & & a ( b + c ) = a b + a c \ \ \ "Also," & & & ( b + c ) a = b a + c a \ & & & a ( b − c ) = a b − a c \ & & & ( b − c ) a = b a − c a$] ] Back to our friends at the movies, we could find the total amount of money they need like this: #math.equation(block: true, alt: "3 open parenthesis 9.25 close parenthesis; 3 open parenthesis 9, plus, 0.25 close parenthesis; 3 open parenthesis 9 close parenthesis, plus, 3 open parenthesis 0.25 close parenthesis; 27, plus, 0.75; 27.75")[$#h(1.3em) 3 ( 9.25 ) \ 3 ( 9 & + & 0.25 ) \ 3 ( 9 ) & + & 3 ( 0.25 ) \ 27 & + & 0.75 \ \ \ #h(1.6em) 27.75$]In algebra, we use the #strong[distributive property] to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression #math.equation(block: false, alt: "3 open parenthesis x plus 4 close parenthesis ,")[$3 ( x + 4 ) ,$] the order of operations says to work in the parentheses first. But we cannot add #emph[x] and 4, since they are not like terms. So we use the distributive property, as shown. #examplebox("Example 11")[][ Simplify: #math.equation(block: false, alt: "3 open parenthesis x plus 4 close parenthesis .")[$3 ( x + 4 ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "3 open parenthesis x plus 4 close parenthesis")[$3 ( x + 4 )$]]), [Distribute.], [#math.equation(block: false, alt: "3 times x plus 3 times 4")[$3 · x + 3 · 4$]], [Multiply.], [#math.equation(block: false, alt: "3 x plus 12")[$3 x + 12$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "4 open parenthesis x plus 2 close parenthesis .")[$4 ( x + 2 ) .$] #solutionbox[ #math.equation(block: true, alt: "4 x plus 8")[$4 x + 8$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "6 open parenthesis x plus 7 close parenthesis .")[$6 ( x + 7 ) .$] #solutionbox[ #math.equation(block: true, alt: "6 x plus 42")[$6 x + 42$] ] ] Some students find it helpful to draw in arrows to remind them how to use the distributive property. Then the first step in would look like this: #figure(figph[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.], alt: "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.", caption: none) #examplebox("Example 12")[][ Simplify: #math.equation(block: false, alt: "8 open parenthesis the fraction 3 over 8 x plus the fraction 1 over 4 close parenthesis .")[$8 ( frac(3, 8) x + frac(1, 4) ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical expression shows 8 multiplied by the sum of (3/8)x and 1/4, with blue arrows illustrating the distributive property where 8 is multiplied by each term inside the parentheses.], alt: "A mathematical expression shows 8 multiplied by the sum of (3/8)x and 1/4, with blue arrows illustrating the distributive property where 8 is multiplied by each term inside the parentheses.", caption: none)]), [Distribute.], [#figure(figph[A mathematical expression displaying the distributive property: 8 multiplied by (3/8)x plus 8 multiplied by (1/4).], alt: "A mathematical expression displaying the distributive property: 8 multiplied by (3/8)x plus 8 multiplied by (1/4).", caption: none)], [Multiply.], [#figure(figph[The mathematical expression '3x+2' is displayed in a clear, dark grey font against a plain white background.], alt: "The mathematical expression '3x+2' is displayed in a clear, dark grey font against a plain white background.", caption: none)], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "6 open parenthesis the fraction 5 over 6 y plus the fraction 1 over 2 close parenthesis .")[$6 ( frac(5, 6) y + frac(1, 2) ) .$] #solutionbox[ #math.equation(block: true, alt: "5 y plus 3")[$5 y + 3$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "12 open parenthesis the fraction 1 over 3 n plus the fraction 3 over 4 close parenthesis .")[$12 ( frac(1, 3) n + frac(3, 4) ) .$] #solutionbox[ #math.equation(block: true, alt: "4 n plus 9")[$4 n + 9$] ] ] Using the distributive property as shown in will be very useful when we solve money applications in later chapters. #examplebox("Example 13")[][ Simplify: #math.equation(block: false, alt: "100 open parenthesis 0.3 plus 0.25 q close parenthesis .")[$100 ( 0.3 + 0.25 q ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical expression shows 100 multiplied by the sum of 0.3 and 0.25q, represented as 100(0.3 + 0.25q). Curved arrows indicate the distributive property, showing 100 multiplying both terms inside the parentheses.], alt: "A mathematical expression shows 100 multiplied by the sum of 0.3 and 0.25q, represented as 100(0.3 + 0.25q). Curved arrows indicate the distributive property, showing 100 multiplying both terms inside the parentheses.", caption: none)]), [Distribute.], [#figure(figph[A mathematical expression: 100(0.3) + 100(0.25q). It shows the sum of two terms, where 100 is multiplied by 0.3 in the first term, and 100 is multiplied by 0.25q in the second term.], alt: "A mathematical expression: 100(0.3) + 100(0.25q). It shows the sum of two terms, where 100 is multiplied by 0.3 in the first term, and 100 is multiplied by 0.25q in the second term.", caption: none)], [Multiply.], [#figure(figph[A mathematical expression '30 + 25q' is displayed in black font on a white background.], alt: "A mathematical expression '30 + 25q' is displayed in black font on a white background.", caption: none)], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "100 open parenthesis 0.7 plus 0.15 p close parenthesis .")[$100 ( 0.7 + 0.15 p ) .$] #solutionbox[ #math.equation(block: true, alt: "70 plus 15 p")[$70 + 15 p$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "100 open parenthesis 0.04 plus 0.35 d close parenthesis .")[$100 ( 0.04 + 0.35 d ) .$] #solutionbox[ #math.equation(block: true, alt: "4 plus 35 d")[$4 + 35 d$] ] ] When we distribute a negative number, we need to be extra careful to get the signs correct! #examplebox("Example 14")[][ Simplify: #math.equation(block: false, alt: "−2 open parenthesis 4 y plus 1 close parenthesis .")[$−2 ( 4 y + 1 ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical expression showing the distributive property, with -2 multiplying the terms (4y + 1) indicated by blue curved arrows.], alt: "A mathematical expression showing the distributive property, with -2 multiplying the terms (4y + 1) indicated by blue curved arrows.", caption: none)]), [Distribute.], [#figure(figph[The mathematical expression reads as negative two times four y plus negative two times one.], alt: "The mathematical expression reads as negative two times four y plus negative two times one.", caption: none)], [Multiply.], [#figure(figph[The image displays the mathematical expression -8y-2 in black text on a white background.], alt: "The image displays the mathematical expression -8y-2 in black text on a white background.", caption: none)], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "−3 open parenthesis 6 m plus 5 close parenthesis .")[$−3 ( 6 m + 5 ) .$] #solutionbox[ #math.equation(block: true, alt: "−18 m minus 15")[$−18 m − 15$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "−6 open parenthesis 8 n plus 11 close parenthesis .")[$−6 ( 8 n + 11 ) .$] #solutionbox[ #math.equation(block: true, alt: "−48 n minus 66")[$−48 n − 66$] ] ] #examplebox("Example 15")[][ Simplify: #math.equation(block: false, alt: "−11 open parenthesis 4 minus 3 a close parenthesis .")[$−11 ( 4 − 3 a ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Distribute.], [#figure(figph[An algebraic expression -11(4-3a) is displayed, with blue arrows demonstrating the distributive property by showing -11 multiplying both terms inside the parentheses.], alt: "An algebraic expression -11(4-3a) is displayed, with blue arrows demonstrating the distributive property by showing -11 multiplying both terms inside the parentheses.", caption: none)]), [Multiply.], [#figure(figph[Mathematical steps showing the simplification of an algebraic expression, starting with -11 \* 4 - (-11) \* 3a and simplifying to -44 - (-33a).], alt: "Mathematical steps showing the simplification of an algebraic expression, starting with -11 * 4 - (-11) * 3a and simplifying to -44 - (-33a).", caption: none)], [Simplify.], [#figure(figph[The image displays the algebraic expression -44 + 33a in clear, dark gray text against a plain white background.], alt: "The image displays the algebraic expression -44 + 33a in clear, dark gray text against a plain white background.", caption: none)], )) Notice that you could also write the result as #math.equation(block: false, alt: "33 a minus 44 .")[$33 a − 44 .$] Do you know why? ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "−5 open parenthesis 2 minus 3 a close parenthesis .")[$−5 ( 2 − 3 a ) .$] #solutionbox[ #math.equation(block: true, alt: "−10 plus 15 a")[$−10 + 15 a$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "−7 open parenthesis 8 minus 15 y close parenthesis .")[$−7 ( 8 − 15 y ) .$] #solutionbox[ #math.equation(block: true, alt: "−56 plus 105 y")[$−56 + 105 y$] ] ] will show how to use the distributive property to find the opposite of an expression. #examplebox("Example 16")[][ Simplify: #math.equation(block: false, alt: "− open parenthesis y plus 5 close parenthesis .")[$"−" ( y + 5 ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis y plus 5 close parenthesis")[$( y + 5 )$]]), [Multiplying by −1 results in the opposite.], [#math.equation(block: false, alt: "−1 open parenthesis y plus 5 close parenthesis")[$−1 ( y + 5 )$]], [Distribute.], [#math.equation(block: false, alt: "−1 times y plus open parenthesis −1 close parenthesis times 5")[$−1 · y + ( −1 ) · 5$]], [Simplify.], [#math.equation(block: false, alt: "− y plus open parenthesis −5 close parenthesis")[$"−" y + ( −5 )$]], [], [#math.equation(block: false, alt: "− y minus 5")[$"−" y − 5$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "− open parenthesis z minus 11 close parenthesis .")[$"−" ( z − 11 ) .$] #solutionbox[ #math.equation(block: true, alt: "− z plus 11")[$"−" z + 11$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "− open parenthesis x minus 4 close parenthesis .")[$"−" ( x − 4 ) .$] #solutionbox[ #math.equation(block: true, alt: "− x plus 4")[$"−" x + 4$] ] ] There will be times when we’ll need to use the #strong[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. #examplebox("Example 17")[][ Simplify: #math.equation(block: false, alt: "8 minus 2 open parenthesis x plus 3 close parenthesis .")[$8 − 2 ( x + 3 ) .$] Be sure to follow the order of operations. Multiplication comes before subtraction, so we will distribute the 2 first and then subtract. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "8 minus 2 open parenthesis x plus 3 close parenthesis")[$8 − 2 ( x + 3 )$]]), [Distribute.], [#math.equation(block: false, alt: "8 minus 2 times x minus 2 times 3")[$8 − 2 · x − 2 · 3$]], [Multiply.], [#math.equation(block: false, alt: "8 minus 2 x minus 6")[$8 − 2 x − 6$]], [Combine like terms.], [#math.equation(block: false, alt: "−2 x plus 2")[$−2 x + 2$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "9 minus 3 open parenthesis x plus 2 close parenthesis .")[$9 − 3 ( x + 2 ) .$] #solutionbox[ #math.equation(block: true, alt: "3 minus 3 x")[$3 − 3 x$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "7 x minus 5 open parenthesis x plus 4 close parenthesis .")[$7 x − 5 ( x + 4 ) .$] #solutionbox[ #math.equation(block: true, alt: "2 x minus 20")[$2 x − 20$] ] ] #examplebox("Example 18")[][ Simplify: #math.equation(block: false, alt: "4 open parenthesis x minus 8 close parenthesis minus open parenthesis x plus 3 close parenthesis .")[$4 ( x − 8 ) − ( x + 3 ) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "4 open parenthesis x minus 8 close parenthesis minus open parenthesis x plus 3 close parenthesis")[$4 ( x − 8 ) − ( x + 3 )$]]), [Distribute.], [#math.equation(block: false, alt: "4 x minus 32 minus x minus 3")[$4 x − 32 − x − 3$]], [Combine like terms.], [#math.equation(block: false, alt: "3 x minus 35")[$3 x − 35$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "6 open parenthesis x minus 9 close parenthesis minus open parenthesis x plus 12 close parenthesis .")[$6 ( x − 9 ) − ( x + 12 ) .$] #solutionbox[ #math.equation(block: true, alt: "5 x minus 66")[$5 x − 66$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "8 open parenthesis x minus 1 close parenthesis minus open parenthesis x plus 5 close parenthesis .")[$8 ( x − 1 ) − ( x + 5 ) .$] #solutionbox[ #math.equation(block: true, alt: "7 x minus 13")[$7 x − 13$] ] ] All the properties of real numbers we have used in this chapter are summarized in . #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Commutative Property]], []), [#strong[of addition] If #math.equation(block: false, alt: "a , b")[$a , b$] are real numbers, then #linebreak() #linebreak()   #strong[of multiplication] If #math.equation(block: false, alt: "a , b")[$a , b$] are real numbers, then], [#math.equation(block: false, alt: "a plus b equals b plus a")[$a + b = b + a$] #linebreak() #linebreak() #math.equation(block: false, alt: "a times b equals b times a")[$a · b = b · a$]], [#strong[Associative Property]], [], [#strong[of addition] If #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are real numbers, then #linebreak() #linebreak()   #strong[of multiplication] If #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are real numbers, then], [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis plus c equals a plus open parenthesis b plus c close parenthesis")[$( a + b ) + c = a + ( b + c )$] #linebreak() #linebreak() #math.equation(block: false, alt: "open parenthesis a times b close parenthesis times c equals a times open parenthesis b times c close parenthesis")[$( a · b ) · c = a · ( b · c )$]], [#strong[Distributive Property]], [], [If #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are real numbers, then], [#math.equation(block: false, alt: "a open parenthesis b plus c close parenthesis equals a b plus a c")[$a ( b + c ) = a b + a c$]], [#strong[Identity Property]], [], [#strong[of addition] For any real number #math.equation(block: false, alt: "a :")[$a ":"$] #linebreak()    0 is the #strong[additive identity] #linebreak() #linebreak()   #strong[of multiplication] For any real number #math.equation(block: false, alt: "a :")[$a ":"$] #linebreak()    #math.equation(block: false, alt: "1")[$1$] is the #strong[multiplicative identity]], [#math.equation(block: false, alt: "a plus 0 equals a; 0 plus a equals a")[$a + 0 = a \ 0 + a = a$] #linebreak() #linebreak() #math.equation(block: false, alt: "a times 1 equals a; 1 times a equals a")[$a · 1 = a \ 1 · a = a$]], [#strong[Inverse Property]], [], [#strong[of addition] For any real number #math.equation(block: false, alt: "a ,")[$a ,$] #linebreak()    #math.equation(block: false, alt: "− a")[$"−" a$] is the #strong[additive inverse] of #math.equation(block: false, alt: "a")[$a$] #linebreak() #linebreak()   #strong[of multiplication] For any real number #math.equation(block: false, alt: "a , a not equal to 0")[$a , a ≠ 0$] #linebreak()    #math.equation(block: false, alt: "the fraction 1 over a")[$frac(1, a)$] is the #strong[multiplicative inverse] of #math.equation(block: false, alt: "a .")[$a .$]], [#math.equation(block: false, alt: "a plus open parenthesis − a close parenthesis equals 0")[$a + ( "−" a ) = 0$] #linebreak() #linebreak() #linebreak() #math.equation(block: false, alt: "a times the fraction 1 over a equals 1")[$a · frac(1, a) = 1$]], [#strong[Properties of Zero]], [], [For any real number #emph[a], #linebreak() #linebreak() #linebreak() #linebreak()   For any real number #math.equation(block: false, alt: "a , a not equal to 0")[$a , a ≠ 0$] #linebreak() #linebreak()   For any real number #math.equation(block: false, alt: "a , a not equal to 0")[$a , a ≠ 0$]], [#math.equation(block: false, alt: "a times 0 equals 0; 0 times a equals 0")[$a · 0 = 0 \ 0 · a = 0$] #linebreak() #linebreak() #math.equation(block: false, alt: "the fraction 0 over a equals 0")[$frac(0, a) = 0$] #linebreak() #linebreak() #math.equation(block: false, alt: "the fraction a over 0")[$frac(a, 0)$] is undefined], )) === Key Concepts - #strong[Commutative Property of] - #strong[Addition:] If #math.equation(block: false, alt: "a , b")[$a , b$] are real numbers, then #math.equation(block: false, alt: "a plus b equals b plus a .")[$a + b = b + a .$] - #strong[Multiplication:] If #math.equation(block: false, alt: "a , b")[$a , b$] are real numbers, then #math.equation(block: false, alt: "a times b equals b times a .")[$a · b = b · a .$] When adding or multiplying, changing the #emph[order] gives the same result. - #strong[Associative Property of] - #strong[Addition:] If #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are real numbers, then #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis plus c equals a plus open parenthesis b plus c close parenthesis .")[$( a + b ) + c = a + ( b + c ) .$] - #strong[Multiplication:] If #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are real numbers, then #math.equation(block: false, alt: "open parenthesis a times b close parenthesis times c equals a times open parenthesis b times c close parenthesis .")[$( a · b ) · c = a · ( b · c ) .$] #linebreak() When adding or multiplying, changing the #emph[grouping] gives the same result. - #strong[Distributive Property:] If #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are real numbers, then - #math.equation(block: false, alt: "a open parenthesis b plus c close parenthesis equals a b plus a c")[$a ( b + c ) = a b + a c$] - #math.equation(block: false, alt: "open parenthesis b plus c close parenthesis a equals b a plus c a")[$( b + c ) a = b a + c a$] - #math.equation(block: false, alt: "a open parenthesis b minus c close parenthesis equals a b minus a c")[$a ( b − c ) = a b − a c$] - #math.equation(block: false, alt: "open parenthesis b minus c close parenthesis a equals b a minus c a")[$( b − c ) a = b a − c a$] - #strong[Identity Property] - #strong[of Addition:] For any real number #math.equation(block: false, alt: "a : a plus 0 equals a 0 plus a equals a")[$a ":" #h(0.2em) a + 0 = a #h(0.6em) 0 + a = a$] #linebreak() #strong[0] is the #strong[additive identity] - #strong[of Multiplication:] For any real number #math.equation(block: false, alt: "a : a times 1 equals a 1 times a equals a")[$a ":" #h(0.2em) a · 1 = a #h(0.6em) 1 · a = a$] #linebreak() #math.equation(block: false, alt: "1")[$1$] is the #strong[multiplicative identity] - #strong[Inverse Property] - #strong[of Addition:] For any real number #math.equation(block: false, alt: "a , a plus open parenthesis − a close parenthesis equals 0 .")[$a , a + ( "−" a ) = 0 .$] A number and its #emph[opposite] add to zero. #math.equation(block: false, alt: "− a")[$"−" a$] is the #strong[additive inverse] of #math.equation(block: false, alt: "a .")[$a .$] - #strong[of Multiplication:] For any real number #math.equation(block: false, alt: "a , open parenthesis a not equal to 0 close parenthesis a times the fraction 1 over a equals 1 .")[$a , ( a ≠ 0 ) #h(0.2em) a · frac(1, a) = 1 .$] A number and its #emph[reciprocal] multiply to one. #math.equation(block: false, alt: "the fraction 1 over a")[$frac(1, a)$] is the #strong[multiplicative inverse] of #math.equation(block: false, alt: "a .")[$a .$] - #strong[Properties of Zero] - For any real number #math.equation(block: false, alt: "a ,")[$a ,$] #linebreak() #math.equation(block: false, alt: "a times 0 equals 0 0 times a equals 0")[$a · 0 = 0 #h(0.5em) 0 · a = 0$] – The product of any real number and 0 is 0. - #math.equation(block: false, alt: "the fraction 0 over a equals 0")[$frac(0, a) = 0$] for #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$] – Zero divided by any real number except zero is zero. - #math.equation(block: false, alt: "the fraction a over 0")[$frac(a, 0)$] is undefined – Division by zero is undefined. ==== Practice Makes Perfect #strong[Use the Commutative and Associative Properties] In the following exercises, use the associative property to simplify. 3(4#emph[x]) #solutionbox[ 12#emph[x] ] 4(7#emph[m]) #math.equation(block: true, alt: "open parenthesis y plus 12 close parenthesis plus 28")[$( y + 12 ) + 28$] #solutionbox[ #math.equation(block: true, alt: "y plus 40")[$y + 40$] ] #math.equation(block: true, alt: "open parenthesis n plus 17 close parenthesis plus 33")[$( n + 17 ) + 33$] In the following exercises, simplify. #math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 7 over 8 plus open parenthesis minus the fraction 1 over 2 close parenthesis")[$frac(1, 2) + frac(7, 8) + ( − #h(0.2em) frac(1, 2) )$] #solutionbox[ #math.equation(block: true, alt: "the fraction 7 over 8")[$frac(7, 8)$] ] #math.equation(block: true, alt: "the fraction 2 over 5 plus the fraction 5 over 12 plus open parenthesis minus the fraction 2 over 5 close parenthesis")[$frac(2, 5) + frac(5, 12) + ( − #h(0.2em) frac(2, 5) )$] #math.equation(block: true, alt: "the fraction 3 over 20 times the fraction 49 over 11 times the fraction 20 over 3")[$frac(3, 20) · frac(49, 11) · frac(20, 3)$] #solutionbox[ #math.equation(block: true, alt: "the fraction 49 over 11")[$frac(49, 11)$] ] #math.equation(block: true, alt: "the fraction 13 over 18 times the fraction 25 over 7 times the fraction 18 over 13")[$frac(13, 18) · frac(25, 7) · frac(18, 13)$] #math.equation(block: true, alt: "−24 times 7 times the fraction 3 over 8")[$−24 · 7 ⋅ frac(3, 8)$] #solutionbox[ #math.equation(block: true, alt: "−63")[$−63$] ] #math.equation(block: true, alt: "−36 times 11 times the fraction 4 over 9")[$−36 · 11 · frac(4, 9)$] #math.equation(block: true, alt: "open parenthesis the fraction 5 over 6 plus the fraction 8 over 15 close parenthesis plus the fraction 7 over 15")[$( frac(5, 6) + frac(8, 15) ) + frac(7, 15)$] #solutionbox[ #math.equation(block: true, alt: "1 the fraction 5 over 6")[$1 frac(5, 6)$] ] #math.equation(block: true, alt: "open parenthesis the fraction 11 over 12 plus the fraction 4 over 9 close parenthesis plus the fraction 5 over 9")[$( frac(11, 12) + frac(4, 9) ) + frac(5, 9)$] 17(0.25)(4) #solutionbox[ 17 ] 36(0.2)(5) \[2.48(12)\](0.5) #solutionbox[ 14.88 ] \[9.731(4)\](0.75) 7(4#emph[a]) #solutionbox[ 28#emph[a] ] 9(8#emph[w]) #math.equation(block: true, alt: "−15 open parenthesis 5 m close parenthesis")[$−15 ( 5 m )$] #solutionbox[ #math.equation(block: true, alt: "−75 m")[$−75 m$] ] #math.equation(block: true, alt: "−23 open parenthesis 2 n close parenthesis")[$−23 ( 2 n )$] #math.equation(block: true, alt: "12 open parenthesis the fraction 5 over 6 p close parenthesis")[$12 ( frac(5, 6) p )$] #solutionbox[ 10#emph[p] ] #math.equation(block: true, alt: "20 open parenthesis the fraction 3 over 5 q close parenthesis")[$20 ( frac(3, 5) q )$] #math.equation(block: true, alt: "43 m plus open parenthesis −12 n close parenthesis plus open parenthesis −16 m close parenthesis plus open parenthesis −9 n close parenthesis")[$43 m + ( −12 n ) + ( −16 m ) + ( −9 n )$] #solutionbox[ #math.equation(block: true, alt: "27 m plus open parenthesis −21 n close parenthesis")[$27 m + ( −21 n )$] ] #math.equation(block: true, alt: "−22 p plus 17 q plus open parenthesis −35 p close parenthesis plus open parenthesis −27 q close parenthesis")[$−22 p + 17 q + ( −35 p ) + ( −27 q )$] #math.equation(block: true, alt: "the fraction 3 over 8 g plus the fraction 1 over 12 h plus the fraction 7 over 8 g plus the fraction 5 over 12 h")[$frac(3, 8) g + frac(1, 12) h + frac(7, 8) g + frac(5, 12) h$] #solutionbox[ #math.equation(block: true, alt: "the fraction 5 over 4 g plus the fraction 1 over 2 h")[$frac(5, 4) g + frac(1, 2) h$] ] #math.equation(block: true, alt: "the fraction 5 over 6 a plus the fraction 3 over 10 b plus the fraction 1 over 6 a plus the fraction 9 over 10 b")[$frac(5, 6) a + frac(3, 10) b + frac(1, 6) a + frac(9, 10) b$] #math.equation(block: true, alt: "6.8 p plus 9.14 q plus open parenthesis −4.37 p close parenthesis plus open parenthesis −0.88 q close parenthesis")[$6.8 p + 9.14 q + ( −4.37 p ) + ( −0.88 q )$] #solutionbox[ #math.equation(block: true, alt: "2.43 p plus 8.26 q")[$2.43 p + 8.26 q$] ] #math.equation(block: true, alt: "9.6 m plus 7.22 n plus open parenthesis −2.19 m close parenthesis plus open parenthesis −0.65 n close parenthesis")[$9.6 m + 7.22 n + ( −2.19 m ) + ( −0.65 n )$] #strong[Use the Identity and Inverse Properties of Addition and Multiplication] In the following exercises, find the additive inverse of each number. ⓐ #math.equation(block: false, alt: "the fraction 2 over 5")[$frac(2, 5)$] ⓑ 4.3 #linebreak() ⓒ #math.equation(block: false, alt: "−8")[$−8$] #linebreak() ⓓ #math.equation(block: false, alt: "minus the fraction 10 over 3")[$− #h(0.2em) frac(10, 3)$] #solutionbox[ ⓐ #math.equation(block: false, alt: "minus the fraction 2 over 5")[$− #h(0.2em) frac(2, 5)$] ⓑ #math.equation(block: false, alt: "−4.3")[$−4.3$] ⓒ 8 ⓓ #math.equation(block: false, alt: "the fraction 10 over 3")[$frac(10, 3)$] ] ⓐ #math.equation(block: false, alt: "the fraction 5 over 9")[$frac(5, 9)$] #linebreak() ⓑ 2.1 ⓒ #math.equation(block: false, alt: "−3")[$−3$] ⓓ #math.equation(block: false, alt: "minus the fraction 9 over 5")[$− #h(0.2em) frac(9, 5)$] ⓐ #math.equation(block: false, alt: "minus the fraction 7 over 6")[$− #h(0.2em) frac(7, 6)$] ⓑ #math.equation(block: false, alt: "−0.075")[$−0.075$] ⓒ 23 ⓓ #math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$] #solutionbox[ ⓐ #math.equation(block: false, alt: "the fraction 7 over 6")[$frac(7, 6)$] ⓑ 0.075 ⓒ #math.equation(block: false, alt: "−23")[$−23$] ⓓ #math.equation(block: false, alt: "minus the fraction 1 over 4")[$− #h(0.2em) frac(1, 4)$] ] ⓐ #math.equation(block: false, alt: "minus the fraction 8 over 3")[$− #h(0.2em) frac(8, 3)$] ⓑ #math.equation(block: false, alt: "−0.019")[$−0.019$] ⓒ 52 ⓓ #math.equation(block: false, alt: "the fraction 5 over 6")[$frac(5, 6)$] In the following exercises, find the multiplicative inverse of each number. ⓐ 6 ⓑ #math.equation(block: false, alt: "minus the fraction 3 over 4")[$− #h(0.2em) frac(3, 4)$] ⓒ 0.7 #solutionbox[ ⓐ #math.equation(block: false, alt: "the fraction 1 over 6")[$frac(1, 6)$] ⓑ #math.equation(block: false, alt: "minus the fraction 4 over 3")[$− #h(0.2em) frac(4, 3)$] ⓒ #math.equation(block: false, alt: "the fraction 10 over 7")[$frac(10, 7)$] ] ⓐ 12 ⓑ #math.equation(block: false, alt: "minus the fraction 9 over 2")[$− #h(0.2em) frac(9, 2)$] ⓒ 0.13 ⓐ #math.equation(block: false, alt: "the fraction 11 over 12")[$frac(11, 12)$] ⓑ #math.equation(block: false, alt: "−1.1")[$−1.1$] ⓒ #math.equation(block: false, alt: "−4")[$−4$] #solutionbox[ ⓐ #math.equation(block: false, alt: "the fraction 12 over 11")[$frac(12, 11)$] ⓑ #math.equation(block: false, alt: "minus the fraction 10 over 11")[$− #h(0.2em) frac(10, 11)$] ⓒ #math.equation(block: false, alt: "minus the fraction 1 over 4")[$− #h(0.2em) frac(1, 4)$] ] ⓐ #math.equation(block: false, alt: "the fraction 17 over 20")[$frac(17, 20)$] ⓑ #math.equation(block: false, alt: "−1.5")[$−1.5$] ⓒ #math.equation(block: false, alt: "−3")[$−3$] #strong[Use the Properties of Zero] In the following exercises, simplify. #math.equation(block: true, alt: "the fraction 0 over 6")[$frac(0, 6)$] #solutionbox[ 0 ] #math.equation(block: true, alt: "the fraction 3 over 0")[$frac(3, 0)$] #math.equation(block: true, alt: "0 divided by the fraction 11 over 12")[$0 ÷ frac(11, 12)$] #solutionbox[ 0 ] #math.equation(block: true, alt: "the fraction 6 over 0")[$frac(6, 0)$] #math.equation(block: true, alt: "the fraction 0 over 3")[$frac(0, 3)$] #solutionbox[ 0 ] #math.equation(block: true, alt: "0 times the fraction 8 over 15")[$0 · frac(8, 15)$] #math.equation(block: true, alt: "open parenthesis −3.14 close parenthesis open parenthesis 0 close parenthesis")[$( −3.14 ) ( 0 )$] #solutionbox[ 0 ] #math.equation(block: true, alt: "the fraction the fraction 1 over 10 over 0")[$frac(frac(1, 10), 0)$] #strong[Mixed Practice] In the following exercises, simplify. #math.equation(block: true, alt: "19 a plus 44 minus 19 a")[$19 a + 44 − 19 a$] #solutionbox[ 44 ] #math.equation(block: true, alt: "27 c plus 16 minus 27 c")[$27 c + 16 − 27 c$] #math.equation(block: true, alt: "10 open parenthesis 0.1 d close parenthesis")[$10 ( 0.1 d )$] #solutionbox[ d ] #math.equation(block: true, alt: "100 open parenthesis 0.01 p close parenthesis")[$100 ( 0.01 p )$] #math.equation(block: false, alt: "the fraction 0 over u minus 4.99 ,")[$frac(0, u − 4.99) ,$] where #math.equation(block: false, alt: "u not equal to 4.99")[$u ≠ 4.99$] #solutionbox[ 0 ] #math.equation(block: false, alt: "the fraction 0 over v minus 65.1 ,")[$frac(0, v − 65.1) ,$] where #math.equation(block: false, alt: "v not equal to 65.1")[$v ≠ 65.1$] #math.equation(block: false, alt: "0 divided by open parenthesis x minus the fraction 1 over 2 close parenthesis ,")[$0 ÷ ( x − #h(0.2em) frac(1, 2) ) ,$] where #math.equation(block: false, alt: "x not equal to the fraction 1 over 2")[$x ≠ frac(1, 2)$] #solutionbox[ 0 ] #math.equation(block: false, alt: "0 divided by open parenthesis y minus the fraction 1 over 6 close parenthesis ,")[$0 ÷ ( y − #h(0.2em) frac(1, 6) ) ,$] where #math.equation(block: false, alt: "y not equal to the fraction 1 over 6")[$y ≠ frac(1, 6)$] #math.equation(block: false, alt: "the fraction 32 minus 5 a over 0 ,")[$frac(32 − 5 a, 0) ,$] where #linebreak() #math.equation(block: false, alt: "32 minus 5 a not equal to 0")[$32 − 5 a ≠ 0$] #solutionbox[ undefined ] #math.equation(block: false, alt: "the fraction 28 minus 9 b over 0 ,")[$frac(28 − 9 b, 0) ,$] where #linebreak() #math.equation(block: false, alt: "28 minus 9 b not equal to 0")[$28 − 9 b ≠ 0$] #math.equation(block: false, alt: "open parenthesis the fraction 3 over 4 plus the fraction 9 over 10 m close parenthesis divided by 0")[$( frac(3, 4) + frac(9, 10) m ) ÷ 0$] where #linebreak() #math.equation(block: false, alt: "the fraction 3 over 4 plus the fraction 9 over 10 m not equal to 0")[$frac(3, 4) + frac(9, 10) m ≠ 0$] #solutionbox[ undefined ] #math.equation(block: false, alt: "open parenthesis the fraction 5 over 16 n minus the fraction 3 over 7 close parenthesis divided by 0")[$( frac(5, 16) n − #h(0.2em) frac(3, 7) ) ÷ 0$] where #linebreak() #math.equation(block: false, alt: "the fraction 5 over 16 n minus the fraction 3 over 7 not equal to 0")[$frac(5, 16) n − #h(0.2em) frac(3, 7) ≠ 0$] #math.equation(block: true, alt: "15 times the fraction 3 over 5 open parenthesis 4 d plus 10 close parenthesis")[$15 · frac(3, 5) ( 4 d + 10 )$] #solutionbox[ #math.equation(block: true, alt: "36 d plus 90")[$36 d + 90$] ] #math.equation(block: true, alt: "18 times the fraction 5 over 6 open parenthesis 15 h plus 24 close parenthesis")[$18 · frac(5, 6) ( 15 h + 24 )$] #strong[Simplify Expressions Using the Distributive Property] In the following exercises, simplify using the distributive property. #math.equation(block: true, alt: "8 open parenthesis 4 y plus 9 close parenthesis")[$8 ( 4 y + 9 )$] #solutionbox[ #math.equation(block: true, alt: "32 y plus 72")[$32 y + 72$] ] #math.equation(block: true, alt: "9 open parenthesis 3 w plus 7 close parenthesis")[$9 ( 3 w + 7 )$] #math.equation(block: true, alt: "6 open parenthesis c minus 13 close parenthesis")[$6 ( c − 13 )$] #solutionbox[ #math.equation(block: true, alt: "6 c minus 78")[$6 c − 78$] ] #math.equation(block: true, alt: "7 open parenthesis y minus 13 close parenthesis")[$7 ( y − 13 )$] #math.equation(block: true, alt: "the fraction 1 over 4 open parenthesis 3 q plus 12 close parenthesis")[$frac(1, 4) ( 3 q + 12 )$] #solutionbox[ #math.equation(block: true, alt: "the fraction 3 over 4 q plus 3")[$frac(3, 4) q + 3$] ] #math.equation(block: true, alt: "the fraction 1 over 5 open parenthesis 4 m plus 20 close parenthesis")[$frac(1, 5) ( 4 m + 20 )$] #math.equation(block: true, alt: "9 open parenthesis the fraction 5 over 9 y minus the fraction 1 over 3 close parenthesis")[$9 ( frac(5, 9) y − #h(0.2em) frac(1, 3) )$] #solutionbox[ #math.equation(block: true, alt: "5 y minus 3")[$5 y − 3$] ] #math.equation(block: true, alt: "10 open parenthesis the fraction 3 over 10 x minus the fraction 2 over 5 close parenthesis")[$10 ( frac(3, 10) x − #h(0.2em) frac(2, 5) )$] #math.equation(block: true, alt: "12 open parenthesis the fraction 1 over 4 plus the fraction 2 over 3 r close parenthesis")[$12 ( frac(1, 4) + frac(2, 3) r )$] #solutionbox[ #math.equation(block: true, alt: "3 plus 8 r")[$3 + 8 r$] ] #math.equation(block: true, alt: "12 open parenthesis the fraction 1 over 6 plus the fraction 3 over 4 s close parenthesis")[$12 ( frac(1, 6) + frac(3, 4) s )$] #math.equation(block: true, alt: "r open parenthesis s minus 18 close parenthesis")[$r ( s − 18 )$] #solutionbox[ #math.equation(block: true, alt: "r s minus 18 r")[$r s − 18 r$] ] #math.equation(block: true, alt: "u open parenthesis v minus 10 close parenthesis")[$u ( v − 10 )$] #math.equation(block: true, alt: "open parenthesis y plus 4 close parenthesis p")[$( y + 4 ) p$] #solutionbox[ #math.equation(block: true, alt: "y p plus 4 p")[$y p + 4 p$] ] #math.equation(block: true, alt: "open parenthesis a plus 7 close parenthesis x")[$( a + 7 ) x$] #math.equation(block: true, alt: "−7 open parenthesis 4 p plus 1 close parenthesis")[$−7 ( 4 p + 1 )$] #solutionbox[ #math.equation(block: true, alt: "−28 p minus 7")[$−28 p − 7$] ] #math.equation(block: true, alt: "−9 open parenthesis 9 a plus 4 close parenthesis")[$−9 ( 9 a + 4 )$] #math.equation(block: true, alt: "−3 open parenthesis x minus 6 close parenthesis")[$−3 ( x − 6 )$] #solutionbox[ #math.equation(block: true, alt: "−3 x plus 18")[$−3 x + 18$] ] #math.equation(block: true, alt: "−4 open parenthesis q minus 7 close parenthesis")[$−4 ( q − 7 )$] #math.equation(block: true, alt: "− open parenthesis 3 x minus 7 close parenthesis")[$"−" ( 3 x − 7 )$] #solutionbox[ #math.equation(block: true, alt: "−3 x plus 7")[$−3 x + 7$] ] #math.equation(block: true, alt: "− open parenthesis 5 p minus 4 close parenthesis")[$"−" ( 5 p − 4 )$] #math.equation(block: true, alt: "16 minus 3 open parenthesis y plus 8 close parenthesis")[$16 − 3 ( y + 8 )$] #solutionbox[ #math.equation(block: true, alt: "−3 y minus 8")[$−3 y − 8$] ] #math.equation(block: true, alt: "18 minus 4 open parenthesis x plus 2 close parenthesis")[$18 − 4 ( x + 2 )$] #math.equation(block: true, alt: "4 minus 11 open parenthesis 3 c minus 2 close parenthesis")[$4 − 11 ( 3 c − 2 )$] #solutionbox[ #math.equation(block: true, alt: "−33 c plus 26")[$−33 c + 26$] ] #math.equation(block: true, alt: "9 minus 6 open parenthesis 7 n minus 5 close parenthesis")[$9 − 6 ( 7 n − 5 )$] #math.equation(block: true, alt: "22 minus open parenthesis a plus 3 close parenthesis")[$22 − ( a + 3 )$] #solutionbox[ #math.equation(block: true, alt: "− a plus 19")[$"−" a + 19$] ] #math.equation(block: true, alt: "8 minus open parenthesis r minus 7 close parenthesis")[$8 − ( r − 7 )$] #math.equation(block: true, alt: "open parenthesis 5 m minus 3 close parenthesis minus open parenthesis m plus 7 close parenthesis")[$( 5 m − 3 ) − ( m + 7 )$] #solutionbox[ #math.equation(block: true, alt: "4 m minus 10")[$4 m − 10$] ] #math.equation(block: true, alt: "open parenthesis 4 y minus 1 close parenthesis minus open parenthesis y minus 2 close parenthesis")[$( 4 y − 1 ) − ( y − 2 )$] #math.equation(block: true, alt: "5 open parenthesis 2 n plus 9 close parenthesis plus 12 open parenthesis n minus 3 close parenthesis")[$5 ( 2 n + 9 ) + 12 ( n − 3 )$] #solutionbox[ #math.equation(block: true, alt: "22 n plus 9")[$22 n + 9$] ] #math.equation(block: true, alt: "9 open parenthesis 5 u plus 8 close parenthesis plus 2 open parenthesis u minus 6 close parenthesis")[$9 ( 5 u + 8 ) + 2 ( u − 6 )$] #math.equation(block: true, alt: "9 open parenthesis 8 x minus 3 close parenthesis minus open parenthesis −2 close parenthesis")[$9 ( 8 x − 3 ) − ( −2 )$] #solutionbox[ #math.equation(block: true, alt: "72 x minus 25")[$72 x − 25$] ] #math.equation(block: true, alt: "4 open parenthesis 6 x minus 1 close parenthesis minus open parenthesis −8 close parenthesis")[$4 ( 6 x − 1 ) − ( −8 )$] #math.equation(block: true, alt: "14 open parenthesis c minus 1 close parenthesis minus 8 open parenthesis c minus 6 close parenthesis")[$14 ( c − 1 ) − 8 ( c − 6 )$] #solutionbox[ #math.equation(block: true, alt: "6 c plus 34")[$6 c + 34$] ] #math.equation(block: true, alt: "11 open parenthesis n minus 7 close parenthesis minus 5 open parenthesis n minus 1 close parenthesis")[$11 ( n − 7 ) − 5 ( n − 1 )$] #math.equation(block: true, alt: "6 open parenthesis 7 y plus 8 close parenthesis minus open parenthesis 30 y minus 15 close parenthesis")[$6 ( 7 y + 8 ) − ( 30 y − 15 )$] #solutionbox[ #math.equation(block: true, alt: "12 y plus 63")[$12 y + 63$] ] #math.equation(block: true, alt: "7 open parenthesis 3 n plus 9 close parenthesis minus open parenthesis 4 n minus 13 close parenthesis")[$7 ( 3 n + 9 ) − ( 4 n − 13 )$] ==== Everyday Math #strong[Insurance copayment] Carrie had to have 5 fillings done. Each filling cost \$80. Her dental insurance required her to pay 20% of the cost as a copay. Calculate Carrie’s copay: + ⓐ First, by multiplying 0.20 by 80 to find her copay for each filling and then multiplying your answer by 5 to find her total copay for 5 fillings. + ⓑ Next, by multiplying \[5(0.20)\](80) + ⓒ Which of the properties of real numbers says that your answers to parts (a), where you multiplied 5\[(0.20)(80)\] and (b), where you multiplied \[5(0.20)\](80), should be equal? #solutionbox[ ⓐ \$80 ⓑ \$80 ⓒ answers will vary ] #strong[Cooking time] Helen bought a 24-pound turkey for her family’s Thanksgiving dinner and wants to know what time to put the turkey in to the oven. She wants to allow 20 minutes per pound cooking time. Calculate the length of time needed to roast the turkey: + ⓐ First, by multiplying #math.equation(block: false, alt: "24 times 20")[$24 · 20$] to find the total number of minutes and then multiplying the answer by #math.equation(block: false, alt: "the fraction 1 over 60")[$frac(1, 60)$] to convert minutes into hours. + ⓑ Next, by multiplying #math.equation(block: false, alt: "24 open parenthesis 20 times the fraction 1 over 60 close parenthesis .")[$24 ( 20 · frac(1, 60) ) .$] + ⓒ Which of the properties of real numbers says that your answers to parts (a), where you multiplied #math.equation(block: false, alt: "open parenthesis 24 times 20 close parenthesis the fraction 1 over 60 ,")[$( 24 · 20 ) frac(1, 60) ,$] and (b), where you multiplied #math.equation(block: false, alt: "24 open parenthesis 20 times the fraction 1 over 60 close parenthesis ,")[$24 ( 20 · frac(1, 60) ) ,$] should be equal? #strong[Buying by the case] Trader Joe’s grocery stores sold a bottle of wine they called “Two Buck Chuck” for \$1.99. They sold a case of 12 bottles for \$23.88. To find the cost of 12 bottles at \$1.99, notice that 1.99 is #math.equation(block: false, alt: "2 minus 0.01 .")[$2 − 0.01 .$] + ⓐ Multiply 12(1.99) by using the distributive property to multiply #math.equation(block: false, alt: "12 open parenthesis 2 minus 0.01 close parenthesis .")[$12 ( 2 − 0.01 ) .$] + ⓑ Was it a bargain to buy “Two Buck Chuck” by the case? #solutionbox[ ⓐ \$23.88 ⓑ no, the price is the same ] #strong[Multi-pack purchase] Adele’s shampoo sells for \$3.99 per bottle at the grocery store. At the warehouse store, the same shampoo is sold as a 3 pack for \$10.49. To find the cost of 3 bottles at \$3.99, notice that 3.99 is #math.equation(block: false, alt: "4 minus 0.01 .")[$4 − 0.01 .$] + ⓐ Multiply 3(3.99) by using the distributive property to multiply #math.equation(block: false, alt: "3 open parenthesis 4 minus 0.01 close parenthesis .")[$3 ( 4 − 0.01 ) .$] + ⓑ How much would Adele save by buying 3 bottles at the warehouse store instead of at the grocery store? ==== Writing Exercises In your own words, state the commutative property of addition. #solutionbox[ Answers may vary ] What is the difference between the additive inverse and the multiplicative inverse of a number? Simplify #math.equation(block: false, alt: "8 open parenthesis x minus the fraction 1 over 4 close parenthesis")[$8 ( x − #h(0.2em) frac(1, 4) )$] using the distributive property and explain each step. #solutionbox[ Answers may vary ] Explain how you can multiply 4(\$5.97) without paper or calculator by thinking of \$5.97 as #math.equation(block: false, alt: "6 minus 0.03")[$6 − 0.03$] and then using the distributive property. ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This is a table that has five rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “use the commutative and associative properties,” “use the identity and inverse properties of addition and multiplication,” “use the properties of zero,” and “simplify expressions using the distributive property.” The rest of the cells are blank.], alt: "This is a table that has five rows and four columns. In the first row, which is a header row, the cells read from left to right “I can…,” “Confidently,” “With some help,” and “No-I don’t get it!” The first column below “I can…” reads “use the commutative and associative properties,” “use the identity and inverse properties of addition and multiplication,” “use the properties of zero,” and “simplify expressions using the distributive property.” The rest of the cells are blank.", caption: none) ⓑ After reviewing this checklist, what will you do to become confident for all objectives?