#set document(title: "2.7 Solve Absolute Value Inequalities", 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")) == 2.7#h(0.6em)Solve Absolute Value Inequalities #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Evaluate: #math.equation(block: false, alt: "− | 7 | .")[$"−" | 7 | .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "minus 7")[$− 7$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Fill in #math.equation(block: false, alt: "< , > ,")[$"<" , ">" ,$] or #math.equation(block: false, alt: "equals")[$=$] for each of the following pairs of numbers. #linebreak() ⓐ #math.equation(block: false, alt: "| −8 | ___ minus | −8 |")[$| −8 | "___" − | −8 |$] ⓑ #math.equation(block: false, alt: "12 ___ minus | −12 |")[$12 "___" − | −12 |$] ⓒ #math.equation(block: false, alt: "| −6 | ___ minus 6")[$| −6 | "___" − 6$] ⓓ #math.equation(block: false, alt: "− open parenthesis −15 close parenthesis ___ minus | −15 |")[$"−" ( −15 ) "___" − | −15 |$] #linebreak() If you missed this problem, review . #solutionbox[ a. \>; b. \>; c. \>; d. \> ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "14 minus 2 | 8 minus 3 open parenthesis 4 minus 1 close parenthesis | .")[$14 − 2 | 8 − 3 ( 4 − 1 ) | .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "12")[$12$] ] ] === Solve Absolute Value Equations As we prepare to solve absolute value equations, we review our definition of #strong[absolute value]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Absolute Value] The absolute value of a number is its distance from zero on the number line. The absolute value of a number #emph[n] is written as #math.equation(block: false, alt: "| n |")[$| n |$] and #math.equation(block: false, alt: "| n | greater than or equal to 0")[$| n | ≥ 0$] for all numbers. Absolute values are always greater than or equal to zero. ] We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example: #math.equation(block: false, alt: "−5")[$#h(3em) −5$] is 5 units away from 0, so #math.equation(block: false, alt: "| −5 | equals 5 .")[$| −5 | = 5 .$] #math.equation(block: false, alt: "5")[$#h(3.65em) 5$] is 5 units away from 0, so #math.equation(block: false, alt: "| 5 | equals 5 .")[$| 5 | = 5 .$] illustrates this idea. #figure(figph[The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.], alt: "The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.", caption: [The numbers 5 and #math.equation(block: false, alt: "−5")[$−5$] are both five units away from zero.]) For the equation #math.equation(block: false, alt: "| x | equals 5 ,")[$| x | = 5 ,$] we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and #math.equation(block: false, alt: "−5")[$−5$] are five units from zero on the number line. They are the solutions to the equation. #math.equation(block: true, alt: "If, | x | equals 5; then, x equals −5 or x equals 5")[$"If" & & & & & | x | = 5 \ "then" & & & & & #h(0.3em) x = −5 #h(0.5em) "or" #h(0.5em) x = 5$]The solution can be simplified to a single statement by writing #math.equation(block: false, alt: "x equals ± 5 .")[$x = "±" 5 .$] This is read, “#emph[x] is equal to positive or negative 5”. We can generalize this to the following property for absolute value equations. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Absolute Value Equations] For any algebraic expression, #emph[u], and any positive real number, #emph[a], #math.equation(block: true, alt: "if, | u | equals a; then, u equals − a or u equals a")[$"if" & & & & & | u | = a \ "then" & & & & & #h(0.3em) u = "−" a #h(0.5em) "or" #h(0.5em) u = a$]Remember that an absolute value cannot be a negative number. ] #examplebox("Example 1")[][ Solve: ⓐ #math.equation(block: false, alt: "| x | equals 8")[$| x | = 8$] ⓑ #math.equation(block: false, alt: "| y | equals −6")[$| y | = −6$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] #solutionbox[ ⓐ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "| x | equals 8")[$| x | = 8$]]), [Write the equivalent equations.], [#math.equation(block: false, alt: "x equals −8 or x equals 8")[$x = −8 #h(0.2em) "or" #h(0.4em) x = 8$]], [], [#math.equation(block: false, alt: "x equals ± 8")[$x = "±" 8$]], )) ⓑ #figure(table( columns: 2, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "| y | equals −6")[$| y | = −6$]], [], [No solution], [Since an absolute value is always positive, there are no solutions to this equation.], [], )) ⓒ #figure(table( columns: 2, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$]], [Write the equivalent equations.], [#math.equation(block: false, alt: "z equals −0 or z equals 0")[$z = −0 #h(0.2em) "or" #h(0.2em) z = 0$]], [Since −0 = 0,], [#math.equation(block: false, alt: "z equals 0")[$z = 0$]], [Both equations tell us that #emph[z] = 0 and so there is only one solution.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: ⓐ #math.equation(block: false, alt: "| x | equals 2")[$| x | = 2$] ⓑ #math.equation(block: false, alt: "| y | equals −4")[$| y | = −4$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] #solutionbox[ ⓐ #math.equation(block: false, alt: "± 2")[$"±" 2$] ⓑ no solution ⓒ 0 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: ⓐ #math.equation(block: false, alt: "| x | equals 11")[$| x | = 11$] ⓑ #math.equation(block: false, alt: "| y | equals −5")[$| y | = −5$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] #solutionbox[ ⓐ #math.equation(block: false, alt: "± 11")[$"±" 11$] ⓑ no solution ⓒ 0 ] ] To solve an #strong[absolute value equation], we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations. #examplebox("Example 2")[How to Solve Absolute Value Equations][ Solve #math.equation(block: false, alt: "| 5 x minus 4 | minus 3 equals 8 .")[$| 5 x − 4 | − 3 = 8 .$] #solutionbox[ #figure(figph[Step 1 is to isolate the absolute value expression. The difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Add 3 to both sides. The result is the absolute value of the quantity 5 x minus 4 is equal to 11.], alt: "Step 1 is to isolate the absolute value expression. The difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Add 3 to both sides. The result is the absolute value of the quantity 5 x minus 4 is equal to 11.", caption: none) #figure(figph[Step 2 is to write the equivalent equations, 5 x minus 4 is equal to negative 11 and 5 x minus 4 is equal to 11.], alt: "Step 2 is to write the equivalent equations, 5 x minus 4 is equal to negative 11 and 5 x minus 4 is equal to 11.", caption: none) #figure(figph[Step 3 is to solve each equation. Add 4 to each side. 5 x is equal to negative 7 or 5 x is equal to 15. Divide each side by 5. The result is x is equal to negative seven-fifths or x is equal to 3.], alt: "Step 3 is to solve each equation. Add 4 to each side. 5 x is equal to negative 7 or 5 x is equal to 15. Divide each side by 5. The result is x is equal to negative seven-fifths or x is equal to 3.", caption: none) #figure(figph[Step 4 is to check each solution. Substitute 3 and negative seven-fifths into the original equation, the difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Substitute 3 for x. Is the difference between the absolute value of the quantity 5 times 3 minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity 15 minus 4 and 3 equal to 8? Is the difference between the absolute value of the 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to 3 checks. Substitute negative seven-fifths for x. Is the difference between the absolute value of the quantity 5 times negative seven-fifths minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity negative 7 minus 4 and 3 equal to 8? Is the difference between the absolute value of the negative 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to negative seven-fifths checks.], alt: "Step 4 is to check each solution. Substitute 3 and negative seven-fifths into the original equation, the difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Substitute 3 for x. Is the difference between the absolute value of the quantity 5 times 3 minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity 15 minus 4 and 3 equal to 8? Is the difference between the absolute value of the 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to 3 checks. Substitute negative seven-fifths for x. Is the difference between the absolute value of the quantity 5 times negative seven-fifths minus 4 and 3 equal to 8? Is the difference between the absolute value of the quantity negative 7 minus 4 and 3 equal to 8? Is the difference between the absolute value of the negative 11 and 3 equal to 8? Is 11 minus 3 equal to 8? 8 is equal to 8, so the solution x is equal to negative seven-fifths checks.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "| 3 x minus 5 | minus 1 equals 6 .")[$| 3 x − 5 | − 1 = 6 .$] #solutionbox[ #math.equation(block: true, alt: "x equals 4 , x equals minus the fraction 2 over 3")[$x = 4 , x = − frac(2, 3)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "| 4 x minus 3 | minus 5 equals 2 .")[$| 4 x − 3 | − 5 = 2 .$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals the fraction 5 over 2")[$x = −1 , x = frac(5, 2)$] ] ] The steps for solving an absolute value equation are summarized here. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve absolute value equations.] + Isolate the absolute value expression. + Write the equivalent equations. + Solve each equation. + Check each solution. ] #examplebox("Example 3")[][ Solve #math.equation(block: false, alt: "2 | x minus 7 | plus 5 equals 9 .")[$2 | x − 7 | + 5 = 9 .$] #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "2 | x minus 7 | plus 5 equals 9")[$2 | x − 7 | + 5 = 9$]], [], [Isolate the absolute value expression.], [#math.equation(block: false, alt: "2 | x minus 7 | equals 4")[$2 | x − 7 | = 4$]], [], [], [#math.equation(block: false, alt: "| x minus 7 | equals 2")[$| x − 7 | = 2$]], [], [Write the equivalent equations.], [#math.equation(block: false, alt: "x minus 7 equals − 2")[$#h(0.6em) x − 7 = "−" 2$] or #math.equation(block: false, alt: "x minus 7 equals 2")[$x − 7 = 2$]], [], [Solve each equation.], [#math.equation(block: false, alt: "x equals 5")[$#h(2.25em) x = 5 #h(0.7em)$] or #math.equation(block: false, alt: "x equals 9")[$#h(1.6em) x = 9$]], [], [Check: #linebreak() #figure(figph[Two solutions for an absolute value equation: checking x=5 and x=9 in 2|x-7|+5=9, both result in a true statement, 9=9, confirming they are valid solutions.], alt: "Two solutions for an absolute value equation: checking x=5 and x=9 in 2|x-7|+5=9, both result in a true statement, 9=9, confirming they are valid solutions.", caption: none)], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "3 | x minus 4 | minus 4 equals 8 .")[$3 | x − 4 | − 4 = 8 .$] #solutionbox[ #math.equation(block: true, alt: "x equals 8 , x equals 0")[$x = 8 , x = 0$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "2 | x minus 5 | plus 3 equals 9 .")[$2 | x − 5 | + 3 = 9 .$] #solutionbox[ #math.equation(block: true, alt: "x equals 8 , x equals 2")[$x = 8 , x = 2$] ] ] Remember, an absolute value is always positive! #examplebox("Example 4")[][ Solve: #math.equation(block: false, alt: "| the fraction 2 over 3 x minus 4 | plus 11 equals 3 .")[$| frac(2, 3) x − 4 | + 11 = 3 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "| the fraction 2 over 3 x minus 4 | plus 11 equals 3")[$| frac(2, 3) x − 4 | + 11 = 3$]]), [Isolate the absolute value term.], [#math.equation(block: false, alt: "| the fraction 2 over 3 x minus 4 | equals −8")[$| frac(2, 3) x − 4 | = −8$]], [An absolute value cannot be negative.], [No solution], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "| the fraction 3 over 4 x minus 5 | plus 9 equals 4 .")[$| frac(3, 4) x − 5 | + 9 = 4 .$] #solutionbox[ No solution ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "| the fraction 5 over 6 x plus 3 | plus 8 equals 6 .")[$| frac(5, 6) x + 3 | + 8 = 6 .$] #solutionbox[ No solution ] ] Some of our absolute value equations could be of the form #math.equation(block: false, alt: "| u | equals | v |")[$| u | = | v |$] where #emph[u] and #emph[v] are algebraic expressions. For example, #math.equation(block: false, alt: "| x minus 3 | equals | 2 x plus 1 | .")[$| x − 3 | = | 2 x + 1 | .$] How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, #emph[u], and a positive real number, #emph[a], if #math.equation(block: false, alt: "| u | equals a ,")[$| u | = a ,$] then #math.equation(block: false, alt: "u equals − a")[$u = "−" a$] or #math.equation(block: false, alt: "u equals a .")[$u = a .$] This tells us that #math.equation(block: true, alt: "if, | u | equals | v |; then, u equals − v, or u equals v")[$"if" & #h(0.3em) | u | = | v | & \ "then" & #h(0.3em) u = "−" v & #h(0.5em) "or" #h(0.5em) u = v$]This leads us to the following property for equations with two absolute values. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Equations with Two Absolute Values] For any algebraic expressions, #emph[u] and #emph[v], #math.equation(block: true, alt: "if, | u | equals | v |; then, u equals − v or u equals v")[$"if" & & & & & | u | = | v | \ "then" & & & & & #h(0.3em) u = "−" v #h(0.5em) "or" #h(0.5em) u = v$] ] When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed. #examplebox("Example 5")[][ Solve: #math.equation(block: false, alt: "| 5 x minus 1 | equals | 2 x plus 3 | .")[$| 5 x − 1 | = | 2 x + 3 | .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "| 5 x minus 1 | equals | 2 x plus 3 |")[$| 5 x − 1 | = | 2 x + 3 |$]]), [Write the equivalent equations. #linebreak() Solve each equation.], [#math.equation(block: false, alt: "5 x minus 1, equals, − open parenthesis 2 x plus 3 close parenthesis, or , 5 x minus 1, equals, 2 x plus 3; 5 x minus 1, equals, −2 x minus 3, or , 3 x minus 1, equals, 3; 7 x minus 1, equals, −3, 3 x, equals, 4; 7 x, equals, −2, x, equals, the fraction 4 over 3; x, equals, minus the fraction 2 over 7, or , x, equals, the fraction 4 over 3")[$& & & #h(3em) 5 x − 1 & = & "−" ( 2 x + 3 ) & #h(1em) "or" #h(1em) & 5 x − 1 & = & 2 x + 3 \ & & & #h(3em) 5 x − 1 & = & −2 x − 3 & #h(1em) "or" #h(1em) & 3 x − 1 & = & 3 \ & & & #h(3em) 7 x − 1 & = & −3 & & 3 x & = & 4 \ & & & #h(3em) 7 x & = & −2 & & x & = & frac(4, 3) \ & & & #h(3em) x & = & − frac(2, 7) & #h(1em) "or" #h(1em) & x & = & frac(4, 3)$]], [Check.], [], [We leave the check to you.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "| 7 x minus 3 | equals | 3 x plus 7 | .")[$| 7 x − 3 | = | 3 x + 7 | .$] #solutionbox[ #math.equation(block: true, alt: "x equals minus the fraction 2 over 5 ,")[$x = − frac(2, 5) ,$] #math.equation(block: true, alt: "x equals the fraction 5 over 2")[$x = frac(5, 2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "| 6 x minus 5 | equals | 3 x plus 4 | .")[$| 6 x − 5 | = | 3 x + 4 | .$] #solutionbox[ #math.equation(block: true, alt: "x equals 3 ,")[$x = 3 ,$] #math.equation(block: true, alt: "x equals the fraction 1 over 9")[$x = frac(1, 9)$] ] ] === Solve Absolute Value Inequalities with “Less Than” Let’s look now at what happens when we have an #strong[absolute value inequality]. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work. Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation #math.equation(block: false, alt: "| x | equals 5 ,")[$| x | = 5 ,$] we saw that both 5 and #math.equation(block: false, alt: "−5")[$−5$] are five units from zero on the number line. They are the solutions to the equation. #math.equation(block: true, alt: "| x | equals 5; x equals −5 or x equals 5")[$| x | = 5 \ x = −5 #h(4em) "or" #h(4em) x = 5$]What about the inequality #math.equation(block: false, alt: "| x | less than or equal to 5 ?")[$| x | ≤ 5 ?$] Where are the numbers whose distance is less than or equal to 5? We know #math.equation(block: false, alt: "−5")[$−5$] and 5 are both five units from zero. All the numbers between #math.equation(block: false, alt: "−5")[$−5$] and 5 are less than five units from zero. #figure(figph[The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.], alt: "The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.", caption: none) In a more general way, we can see that if #math.equation(block: false, alt: "| u | less than or equal to a ,")[$| u | ≤ a ,$] then #math.equation(block: false, alt: "− a less than or equal to u less than or equal to a .")[$"−" a ≤ u ≤ a .$] #figure(figph[The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.], alt: "The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.", caption: none) This result is summarized here. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Absolute Value Inequalities with #math.equation(block: false, alt: "less than")[$<$] or #math.equation(block: false, alt: "less than or equal to")[$≤$]] For any algebraic expression, #emph[u], and any positive real number, #emph[a], #math.equation(block: true, alt: "if, | u | less than a ,, then − a less than u less than a; if, | u | less than or equal to a ,, then − a less than or equal to u less than or equal to a")[$"if" & & & & | u | < a , & & & & & "then" #h(0.5em) "−" a < u < a \ "if" & & & & | u | ≤ a , & & & & & "then" #h(0.5em) "−" a ≤ u ≤ a$] ] After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution. #examplebox("Example 6")[][ Solve #math.equation(block: false, alt: "| x | less than 7 .")[$| x | < 7 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical inequality is shown in the center of a white background, which states the absolute value of x is less than 7, written as |x| \< 7.], alt: "A mathematical inequality is shown in the center of a white background, which states the absolute value of x is less than 7, written as |x| < 7.", caption: none)]), [Write the equivalent inequality.], [#figure(figph[A mathematical inequality is displayed with the expression -7 \< |X| \< 7. The characters are rendered in a standard mathematical font on a white background.], alt: "A mathematical inequality is displayed with the expression -7 < |X| < 7. The characters are rendered in a standard mathematical font on a white background.", caption: none)], [Graph the solution.], [#figure(figph[A number line illustrates the open interval (-7, 7), where all real numbers between -7 and 7 are included, but the endpoints -7 and 7 are excluded. The interval is highlighted in teal.], alt: "A number line illustrates the open interval (-7, 7), where all real numbers between -7 and 7 are included, but the endpoints -7 and 7 are excluded. The interval is highlighted in teal.", caption: none)], [Write the solution using interval notation.], [#figure(figph[The image displays the mathematical ordered pair or interval notation '(-7, 7)' centered on a plain white background.], alt: "The image displays the mathematical ordered pair or interval notation '(-7, 7)' centered on a plain white background.", caption: none)], )) Check: To verify, check a value in each section of the number line showing the solution. Choose numbers such as #math.equation(block: false, alt: "−8 ,")[$−8 ,$] 1, and 9. #figure(figph[The figure is a number line with a left parenthesis at negative 7, a right parenthesis at 7 and shading between the parentheses. The values negative 8, 1, and 9 are marked with points. The absolute value of negative 8 is less than 7 is false. It does not satisfy the absolute value of x is less than 7. The absolute value of 1 is less than 7 is true. It does satisfy the absolute value of x is less than 7. The absolute value of 9 is less than 7 is false. It does not satisfy the absolute value of x is less than 7.], alt: "The figure is a number line with a left parenthesis at negative 7, a right parenthesis at 7 and shading between the parentheses. The values negative 8, 1, and 9 are marked with points. The absolute value of negative 8 is less than 7 is false. It does not satisfy the absolute value of x is less than 7. The absolute value of 1 is less than 7 is true. It does satisfy the absolute value of x is less than 7. The absolute value of 9 is less than 7 is false. It does not satisfy the absolute value of x is less than 7.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "| x | less than 9 .")[$| x | < 9 .$] #solutionbox[ #figure(figph[The solution is negative 9 is less than x which is less than 9. The number line shows open circles at negative 9 and 9 with shading in between the circles. The interval notation is negative 9 to 9 within parentheses.], alt: "The solution is negative 9 is less than x which is less than 9. The number line shows open circles at negative 9 and 9 with shading in between the circles. The interval notation is negative 9 to 9 within parentheses.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "| x | less than 1 .")[$| x | < 1 .$] #solutionbox[ #figure(figph[The solution is negative 1 is less than x which is less than 1. The number line shows open circles at negative 1 and 1 with shading in between the circles. The interval notation is negative 1 to 1 within parentheses.], alt: "The solution is negative 1 is less than x which is less than 1. The number line shows open circles at negative 1 and 1 with shading in between the circles. The interval notation is negative 1 to 1 within parentheses.", caption: none) ] ] #examplebox("Example 7")[][ Solve #math.equation(block: false, alt: "| 5 x minus 6 | less than or equal to 4 .")[$| 5 x − 6 | ≤ 4 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1.] Isolate the absolute value expression. #linebreak() It is isolated.], [#math.equation(block: false, alt: "| 5 x minus 6 | less than or equal to 4")[$| 5 x − 6 | ≤ 4$]]), [#strong[Step 2.] Write the equivalent compound inequality.], [#math.equation(block: false, alt: "−4 less than or equal to 5 x minus 6 less than or equal to 4")[$−4 ≤ 5 x − 6 ≤ 4$]], [#strong[Step 3.] Solve the compound inequality.], [#math.equation(block: false, alt: "2 less than or equal to 5 x less than or equal to 10")[$2 ≤ 5 x ≤ 10$] #linebreak() #math.equation(block: false, alt: "the fraction 2 over 5 less than or equal to x less than or equal to 2")[$frac(2, 5) ≤ x ≤ 2$]], [#strong[Step 4.] Graph the solution.], [#figure(figph[A number line displaying the closed interval from 2/5 to 2.0, inclusive, with additional markers at 0 and 1.0.], alt: "A number line displaying the closed interval from 2/5 to 2.0, inclusive, with additional markers at 0 and 1.0.", caption: none)], [#strong[Step 5.] Write the solution using interval notation.], [#math.equation(block: false, alt: "[ the fraction 2 over 5 , 2 ]")[$[ frac(2, 5) , 2 ]$]], [Check: #linebreak() The check is left to you.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "| 2 x minus 1 | less than or equal to 5 .")[$| 2 x − 1 | ≤ 5 .$] Graph the solution and write the solution in interval notation: #solutionbox[ #figure(figph[The solution is negative 2 is less than or equal to x which is less than or equal to 3. The number line shows closed circles at negative 2 and 3 with shading between the circles. The interval notation is negative 2 to 3 within brackets.], alt: "The solution is negative 2 is less than or equal to x which is less than or equal to 3. The number line shows closed circles at negative 2 and 3 with shading between the circles. The interval notation is negative 2 to 3 within brackets.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "| 4 x minus 5 | less than or equal to 3 .")[$| 4 x − 5 | ≤ 3 .$] Graph the solution and write the solution in interval notation: #solutionbox[ #figure(figph[The solution is one-half is less than or equal to x which is less than or equal to 2. The number line shows closed circles at one-half and 2 with shading between the circles. The interval notation is one-half to 2 within brackets.], alt: "The solution is one-half is less than or equal to x which is less than or equal to 2. The number line shows closed circles at one-half and 2 with shading between the circles. The interval notation is one-half to 2 within brackets.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve absolute value inequalities with \< or ≤.] + Isolate the absolute value expression. + Write the equivalent compound inequality. #linebreak() #math.equation(block: true, alt: "| u | less than a, is equivalent to, − a less than u less than a; | u | less than or equal to a, is equivalent to, − a less than or equal to u less than or equal to a")[$| u | < a & & & & & "is equivalent to" & & & & & "−" a < u < a \ | u | ≤ a & & & & & "is equivalent to" & & & & & "−" a ≤ u ≤ a$] + Solve the compound inequality. + Graph the solution + Write the solution using interval notation. ] === Solve Absolute Value Inequalities with “Greater Than” What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. We started with the inequality #math.equation(block: false, alt: "| x | less than or equal to 5 .")[$| x | ≤ 5 .$] We saw that the numbers whose distance is less than or equal to five from zero on the number line were #math.equation(block: false, alt: "−5")[$−5$] and 5 and all the numbers between #math.equation(block: false, alt: "−5")[$−5$] and 5. #figure(figph[The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.], alt: "The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.", caption: none) Now we want to look at the inequality #math.equation(block: false, alt: "| x | greater than or equal to 5 .")[$| x | ≥ 5 .$] Where are the numbers whose distance from zero is greater than or equal to five? Again both #math.equation(block: false, alt: "−5")[$−5$] and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than #math.equation(block: false, alt: "−5")[$−5$] and greater than 5 on the number line. #figure(figph[The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its left and a left bracket at 5 with shading to its right. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is greater than or equal to 5, then x is less than or equal to negative 5 or x is greater than or equal to 5.], alt: "The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its left and a left bracket at 5 with shading to its right. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is greater than or equal to 5, then x is less than or equal to negative 5 or x is greater than or equal to 5.", caption: none) In a more general way, we can see that if #math.equation(block: false, alt: "| u | greater than or equal to a ,")[$| u | ≥ a ,$] then #math.equation(block: false, alt: "u less than or equal to − a")[$u ≤ "−" a$] or #math.equation(block: false, alt: "u greater than or equal to a .")[$u ≥ a .$] #figure(figph[The figure is a number line with negative a, 0, and a displayed. There is a right bracket at negative a that has shading to its left and a left bracket at a with shading to its right. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is greater than or equal to a, then u is less than or equal to negative a or u is greater than or equal to a.], alt: "The figure is a number line with negative a, 0, and a displayed. There is a right bracket at negative a that has shading to its left and a left bracket at a with shading to its right. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is greater than or equal to a, then u is less than or equal to negative a or u is greater than or equal to a.", caption: none) This result is summarized here. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Absolute Value Inequalities with \> or ≥] For any algebraic expression, #emph[u], and any positive real number, #emph[a], #math.equation(block: true, alt: "if, | u | greater than a ,, then u less than minus a or u greater than a; if, | u | greater than or equal to a ,, then u less than or equal to − a or u greater than or equal to a")[$"if" & & & & & | u | > a , & & & & & "then" #h(0.2em) u < − a #h(0.5em) "or" #h(0.5em) u > a \ "if" & & & & & | u | ≥ a , & & & & & "then" #h(0.2em) u ≤ "−" a #h(0.5em) "or" #h(0.5em) u ≥ a$] ] #examplebox("Example 8")[][ Solve #math.equation(block: false, alt: "| x | greater than 4 .")[$| x | > 4 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "| x | greater than 4")[$| x | > 4$]]), [Write the equivalent inequality.], [#math.equation(block: false, alt: "x less than −4 or x greater than 4")[$x < −4 #h(0.5em) "or" #h(0.5em) x > 4$]], [Graph the solution.], [#figure(figph[A number line illustrates two inequalities: x \< -4, shown in dark teal from -10 to -4 (exclusive), and x \> 4, shown in light blue from 4 (exclusive) to 10. Arrows indicate extension to negative and positive infinity.], alt: "A number line illustrates two inequalities: x < -4, shown in dark teal from -10 to -4 (exclusive), and x > 4, shown in light blue from 4 (exclusive) to 10. Arrows indicate extension to negative and positive infinity.", caption: none)], [Write the solution using interval notation.], [#math.equation(block: false, alt: "open parenthesis − ∞ , −4 close parenthesis union open parenthesis 4 , ∞ close parenthesis")[$( "−" ∞ , −4 ) ∪ ( 4 , ∞ )$]], [Check:], [], )) To verify, check a value in each section of the number line showing the solution. Choose numbers such as #math.equation(block: false, alt: "−6 ,")[$−6 ,$] 0, and 7. #figure(figph[The figure is a number line with a right parenthesis at negative 4 with shading to its left and a left parenthesis at 4 shading to its right. The values negative 6, 0, and 7 are marked with points. The absolute value of negative 6 is greater than negative 4 is true. It does not satisfy the absolute value of x is greater than 4. The absolute value of 0 is greater than 4 is false. It does not satisfy the absolute value of x is greater than 4. The absolute value of 7 is less than 4 is true. It does satisfy the absolute value of x is greater than 4.], alt: "The figure is a number line with a right parenthesis at negative 4 with shading to its left and a left parenthesis at 4 shading to its right. The values negative 6, 0, and 7 are marked with points. The absolute value of negative 6 is greater than negative 4 is true. It does not satisfy the absolute value of x is greater than 4. The absolute value of 0 is greater than 4 is false. It does not satisfy the absolute value of x is greater than 4. The absolute value of 7 is less than 4 is true. It does satisfy the absolute value of x is greater than 4.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "| x | greater than 2 .")[$| x | > 2 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(figph[The solution is x is less than negative 2 or x is greater than 2. The number line shows an open circle at negative 2 with shading to its left and an open circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 2 to infinity within parentheses.], alt: "The solution is x is less than negative 2 or x is greater than 2. The number line shows an open circle at negative 2 with shading to its left and an open circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 2 to infinity within parentheses.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "| x | greater than 1 .")[$| x | > 1 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(figph[The solution is x is less than negative 1 or x is greater than 1. The number line shows an open circle at negative 1 with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to negative 1 within parentheses and 1 to infinity within parentheses.], alt: "The solution is x is less than negative 1 or x is greater than 1. The number line shows an open circle at negative 1 with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to negative 1 within parentheses and 1 to infinity within parentheses.", caption: none) ] ] #examplebox("Example 9")[][ Solve #math.equation(block: false, alt: "| 2 x minus 3 | greater than or equal to 5 .")[$| 2 x − 3 | ≥ 5 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "| 2 x minus 3 | greater than or equal to 5")[$| 2 x − 3 | ≥ 5$]]), [#strong[Step 1.] Isolate the absolute value expression. It is isolated.], [], [#strong[Step 2.] Write the equivalent compound inequality.], [#math.equation(block: false, alt: "2 x minus 3 less than or equal to −5 or 2 x minus 3 greater than or equal to 5")[$2 x − 3 ≤ −5 #h(0.5em) "or" #h(0.5em) 2 x − 3 ≥ 5$]], [#strong[Step 3.] Solve the compound inequality.], [#math.equation(block: false, alt: "2 x less than or equal to − 2 or 2 x greater than or equal to 8")[$2 x ≤ "−" 2 #h(0.5em) "or" #h(0.5em) 2 x ≥ 8$] #linebreak() #math.equation(block: false, alt: "x less than or equal to − 1 or x greater than or equal to 4")[$x ≤ "−" 1 #h(0.5em) "or" #h(0.5em) x ≥ 4$]], [#strong[Step 4.] Graph the solution.], [#figure(figph[A number line illustrates two separate intervals: numbers less than -3 (dark blue arrow extending left) and numbers greater than 5 (light blue arrow extending right), with labels at 0, 5, and 10.], alt: "A number line illustrates two separate intervals: numbers less than -3 (dark blue arrow extending left) and numbers greater than 5 (light blue arrow extending right), with labels at 0, 5, and 10.", caption: none)], [#strong[Step 5.] Write the solution using interval notation.], [#math.equation(block: false, alt: "open parenthesis − ∞ , −1 ] union [ 4 , ∞ close parenthesis")[$( "−" ∞ , −1 ] ∪ [ 4 , ∞ )$]], [Check: #linebreak() The check is left to you.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "| 4 x minus 3 | greater than or equal to 5 .")[$| 4 x − 3 | ≥ 5 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(figph[The solution is x is less than or equal to negative one-half or x is greater than or equal 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis], alt: "The solution is x is less than or equal to negative one-half or x is greater than or equal 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "| 3 x minus 4 | greater than or equal to 2 .")[$| 3 x − 4 | ≥ 2 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(figph[The solution is x is less than or equal to two-thirds or x is greater than or equal 2. The number line shows a closed circle at two-thirds with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to two-thirds within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.], alt: "The solution is x is less than or equal to two-thirds or x is greater than or equal 2. The number line shows a closed circle at two-thirds with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to two-thirds within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve absolute value inequalities with \> or ≥.] + Isolate the absolute value expression. + Write the equivalent compound inequality. #linebreak() #math.equation(block: true, alt: "| u | greater than a is equivalent to u less than minus a or u greater than a; | u | greater than or equal to a is equivalent to u less than or equal to − a or u greater than or equal to a")[$| u | > a #h(2em) "is equivalent to" #h(2em) u < − a #h(0.5em) "or" #h(0.5em) u > a \ | u | ≥ a #h(2em) "is equivalent to" #h(2em) u ≤ "−" a #h(0.5em) "or" #h(0.5em) u ≥ a$] + Solve the compound inequality. + Graph the solution + Write the solution using interval notation. ] === Solve Applications with Absolute Value Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain #emph[tolerance] of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected. #math.equation(block: true, alt: "| actual-ideal | less than or equal to tolerance")[$| "actual-ideal" | ≤ "tolerance"$]#examplebox("Example 10")[][ The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by #math.equation(block: false, alt: "0.075")[$0.075$] mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected? #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [Let #emph[x] = the actual measurement.]), [Use an absolute value inequality to express this situation.], [#math.equation(block: false, alt: "| actual-ideal | less than or equal to tolerance")[$| "actual-ideal" | ≤ "tolerance"$]], [], [#math.equation(block: false, alt: "| x minus 60 | less than or equal to 0.075")[$| x − 60 | ≤ 0.075$]], [Rewrite as a compound inequality.], [#math.equation(block: false, alt: "− 0.075 less than or equal to x minus 60 less than or equal to 0.075")[$"−" 0.075 ≤ x − 60 ≤ 0.075$]], [Solve the inequality.], [#math.equation(block: false, alt: "59.925 less than or equal to x less than or equal to 60.075")[$59.925 ≤ x ≤ 60.075$]], [Answer the question.], [The diameter of the rod can be between 59.925 mm and 60.075 mm.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The ideal diameter of a rod needed for a machine is 80 mm. The actual diameter can vary from the ideal diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected? #solutionbox[ The diameter of the rod can be between 79.991 and 80.009 mm. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The ideal diameter of a rod needed for a machine is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected? #solutionbox[ The diameter of the rod can be between 74.95 and 75.05 mm. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access this online resource for additional instruction and practice with solving linear absolute value equations and inequalities. - #link("https://openstax.org/l/37solvlinabsol")[Solving Linear Absolute Value Equations and Inequalities] ] === Key Concepts - #strong[Absolute Value] #linebreak() The absolute value of a number is its distance from 0 on the number line. #linebreak() The absolute value of a number #emph[n] is written as #math.equation(block: false, alt: "| n |")[$| n |$] and #math.equation(block: false, alt: "| n | greater than or equal to 0")[$| n | ≥ 0$] for all numbers. #linebreak() Absolute values are always greater than or equal to zero. - #strong[Absolute Value Equations] #linebreak() For any algebraic expression, #emph[u], and any positive real number, #emph[a], #linebreak() #math.equation(block: false, alt: "if, | u | equals a; then, u equals − a or u equals a")[$"if" & & & & | u | = a \ "then" & & & & #h(0.3em) u = "−" a #h(0.5em) "or" #h(0.5em) u = a$] #linebreak() Remember that an absolute value cannot be a negative number. - #strong[How to Solve Absolute Value Equations] + Isolate the absolute value expression. + Write the equivalent equations. + Solve each equation. + Check each solution. - #strong[Equations with Two Absolute Values] #linebreak() For any algebraic expressions, #emph[u] and #emph[v], #linebreak() #math.equation(block: false, alt: "if, | u | equals | v |; then, u equals − v or u equals v")[$"if" & & & & | u | = | v | \ "then" & & & & #h(0.3em) u = "−" v #h(0.5em) "or" #h(0.5em) u = v$] - #strong[Absolute Value Inequalities with] #math.equation(block: false, alt: "less than")[$<$] or #math.equation(block: false, alt: "less than or equal to")[$≤$] #linebreak() For any algebraic expression, #emph[u], and any positive real number, #emph[a], #linebreak() #math.equation(block: false, alt: "if, | u | less than a ,, then − a less than u less than a; if, | u | less than or equal to a ,, then − a less than or equal to u less than or equal to a")[$"if" & & & & & | u | < a , & & & & & "then" #h(0.5em) "−" a < u < a \ "if" & & & & & | u | ≤ a , & & & & & "then" #h(0.5em) "−" a ≤ u ≤ a$] - #strong[How To Solve Absolute Value Inequalities with] #math.equation(block: false, alt: "less than")[$<$] or #math.equation(block: false, alt: "less than or equal to")[$≤$] + Isolate the absolute value expression. + Write the equivalent compound inequality. #linebreak() #math.equation(block: false, alt: "| u | less than a, is equivalent to, − a less than u less than a; | u | less than or equal to a, is equivalent to, − a less than or equal to u less than or equal to a")[$| u | < a & & & & & "is equivalent to" & & & & & "−" a < u < a \ | u | ≤ a & & & & & "is equivalent to" & & & & & "−" a ≤ u ≤ a$] + Solve the compound inequality. + Graph the solution + Write the solution using interval notation - #strong[Absolute Value Inequalities with] #math.equation(block: false, alt: "greater than")[$>$] or #math.equation(block: false, alt: "greater than or equal to")[$≥$] #linebreak() For any algebraic expression, #emph[u], and any positive real number, #emph[a], #linebreak() #math.equation(block: false, alt: "if, | u | greater than a ,, then u less than − a or u greater than a; if, | u | greater than or equal to a ,, then u less than or equal to − a or u greater than or equal to a")[$"if" & & & & & | u | > a , & & & & & "then" #h(0.2em) u < "−" a #h(0.5em) "or" #h(0.5em) u > a \ "if" & & & & & | u | ≥ a , & & & & & "then" #h(0.2em) u ≤ "−" a #h(0.5em) "or" #h(0.5em) u ≥ a$] - #strong[How To Solve Absolute Value Inequalities with] #math.equation(block: false, alt: "greater than")[$>$] or #math.equation(block: false, alt: "greater than or equal to")[$≥$] + Isolate the absolute value expression. + Write the equivalent compound inequality. #linebreak() #math.equation(block: false, alt: "| u | greater than a, is equivalent to, u less than − a or u greater than a; | u | greater than or equal to a, is equivalent to, u less than or equal to − a or u greater than or equal to a")[$| u | > a & & & & "is equivalent to" & & & & u < "−" a #h(0.5em) "or" #h(0.5em) u > a \ | u | ≥ a & & & & "is equivalent to" & & & & u ≤ "−" a #h(0.5em) "or" #h(0.5em) u ≥ a$] + Solve the compound inequality. + Graph the solution + Write the solution using interval notation === Section Exercises ==== Practice Makes Perfect #strong[Solve Absolute Value Equations] In the following exercises, solve. ⓐ #math.equation(block: false, alt: "| x | equals 6")[$| x | = 6$] ⓑ #math.equation(block: false, alt: "| y | equals −3")[$| y | = −3$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] ⓐ #math.equation(block: false, alt: "| x | equals 4")[$| x | = 4$] ⓑ #math.equation(block: false, alt: "| y | equals −5")[$| y | = −5$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals 4 , x equals −4")[$x = 4 , x = −4$] ⓑ no solution ⓒ #math.equation(block: false, alt: "z equals 0")[$z = 0$] ] ⓐ #math.equation(block: false, alt: "| x | equals 7")[$| x | = 7$] ⓑ #math.equation(block: false, alt: "| y | equals −11")[$| y | = −11$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] ⓐ #math.equation(block: false, alt: "| x | equals 3")[$| x | = 3$] ⓑ #math.equation(block: false, alt: "| y | equals −1")[$| y | = −1$] ⓒ #math.equation(block: false, alt: "| z | equals 0")[$| z | = 0$] #solutionbox[ ⓐ #math.equation(block: false, alt: "x equals 3 , x equals −3")[$x = 3 , x = −3$] ⓑ no solution ⓒ #math.equation(block: false, alt: "z equals 0")[$z = 0$] ] #math.equation(block: true, alt: "| 2 x minus 3 | minus 4 equals 1")[$| 2 x − 3 | − 4 = 1$] #math.equation(block: true, alt: "| 4 x minus 1 | minus 3 equals 0")[$| 4 x − 1 | − 3 = 0$] #solutionbox[ #math.equation(block: true, alt: "x equals 1 , x equals minus the fraction 1 over 2")[$x = 1 , x = − frac(1, 2)$] ] #math.equation(block: true, alt: "| 3 x minus 4 | plus 5 equals 7")[$| 3 x − 4 | + 5 = 7$] #math.equation(block: true, alt: "| 4 x plus 7 | plus 2 equals 5")[$| 4 x + 7 | + 2 = 5$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals minus the fraction 5 over 2")[$x = −1 , x = − frac(5, 2)$] ] #math.equation(block: true, alt: "4 | x minus 1 | plus 2 equals 10")[$4 | x − 1 | + 2 = 10$] #math.equation(block: true, alt: "3 | x minus 4 | plus 2 equals 11")[$3 | x − 4 | + 2 = 11$] #solutionbox[ #math.equation(block: true, alt: "x equals 7 , x equals 1")[$x = 7 , x = 1$] ] #math.equation(block: true, alt: "3 | 4 x minus 5 | minus 4 equals 11")[$3 | 4 x − 5 | − 4 = 11$] #math.equation(block: true, alt: "3 | x plus 2 | minus 5 equals 4")[$3 | x + 2 | − 5 = 4$] #solutionbox[ #math.equation(block: true, alt: "x equals 1 , x equals −5")[$x = 1 , x = −5$] ] #math.equation(block: true, alt: "−2 | x minus 3 | plus 8 equals −4")[$−2 | x − 3 | + 8 = −4$] #math.equation(block: true, alt: "−3 | x minus 4 | plus 4 equals −5")[$−3 | x − 4 | + 4 = −5$] #solutionbox[ #math.equation(block: true, alt: "x equals 7 , x equals 1")[$x = 7 , x = 1$] ] #math.equation(block: true, alt: "| the fraction 3 over 4 x minus 3 | plus 7 equals 2")[$| frac(3, 4) x − 3 | + 7 = 2$] #math.equation(block: true, alt: "| the fraction 3 over 5 x minus 2 | plus 5 equals 2")[$| frac(3, 5) x − 2 | + 5 = 2$] #solutionbox[ no solution ] #math.equation(block: true, alt: "| the fraction 1 over 2 x plus 5 | plus 4 equals 1")[$| frac(1, 2) x + 5 | + 4 = 1$] #math.equation(block: true, alt: "| the fraction 1 over 4 x plus 3 | plus 3 equals 1")[$| frac(1, 4) x + 3 | + 3 = 1$] #solutionbox[ no solution ] #math.equation(block: true, alt: "| 3 x minus 2 | equals | 2 x minus 3 |")[$| 3 x − 2 | = | 2 x − 3 |$] #math.equation(block: true, alt: "| 4 x plus 3 | equals | 2 x plus 1 |")[$| 4 x + 3 | = | 2 x + 1 |$] #solutionbox[ #math.equation(block: true, alt: "x equals −1 , x equals minus the fraction 2 over 3")[$x = −1 , x = − frac(2, 3)$] ] #math.equation(block: true, alt: "| 6 x minus 5 | equals | 2 x plus 3 |")[$| 6 x − 5 | = | 2 x + 3 |$] #math.equation(block: true, alt: "| 6 minus x | equals | 3 minus 2 x |")[$| 6 − x | = | 3 − 2 x |$] #solutionbox[ #math.equation(block: true, alt: "x equals −3 , x equals 3")[$x = −3 , x = 3$] ] #strong[Solve Absolute Value Inequalities with “less than”] In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation. #math.equation(block: true, alt: "| x | less than 5")[$| x | < 5$] #math.equation(block: true, alt: "| x | less than 1")[$| x | < 1$] #solutionbox[ #figure(figph[The solution is negative 1 is less than x which is less than 1. The number line shows an open circle at negative 1, an open circle at 1, and shading between the circles. The interval notation is negative 1 to 1 within parentheses.], alt: "The solution is negative 1 is less than x which is less than 1. The number line shows an open circle at negative 1, an open circle at 1, and shading between the circles. The interval notation is negative 1 to 1 within parentheses.", caption: none) ] #math.equation(block: true, alt: "| x | less than or equal to 8")[$| x | ≤ 8$] #math.equation(block: true, alt: "| x | less than or equal to 3")[$| x | ≤ 3$] #solutionbox[ #figure(figph[The solution is negative 3 is less than or equal to x which is less than or equal to 3. The number line shows a closed circle at negative 3, a closed circle at 3, and shading between the circles. The interval notation is negative 3 to 3 within brackets.], alt: "The solution is negative 3 is less than or equal to x which is less than or equal to 3. The number line shows a closed circle at negative 3, a closed circle at 3, and shading between the circles. The interval notation is negative 3 to 3 within brackets.", caption: none) ] #math.equation(block: true, alt: "| 3 x minus 3 | less than or equal to 6")[$| 3 x − 3 | ≤ 6$] #math.equation(block: true, alt: "| 2 x minus 5 | less than or equal to 3")[$| 2 x − 5 | ≤ 3$] #solutionbox[ #figure(figph[The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading between the circles. The interval notation is 1 to 4 within brackets.], alt: "The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading between the circles. The interval notation is 1 to 4 within brackets.", caption: none) ] #math.equation(block: true, alt: "| 2 x plus 3 | plus 5 less than 4")[$| 2 x + 3 | + 5 < 4$] #math.equation(block: true, alt: "| 3 x minus 7 | plus 3 less than 1")[$| 3 x − 7 | + 3 < 1$] #solutionbox[ #figure(figph[The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.], alt: "The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.", caption: none) ] #math.equation(block: true, alt: "| 4 x minus 3 | less than 1")[$| 4 x − 3 | < 1$] #math.equation(block: true, alt: "| 6 x minus 5 | less than 7")[$| 6 x − 5 | < 7$] #solutionbox[ #figure(figph[The solution is negative one-third is less than x which is less than 2. The number line shows an open circle at negative one-half, an open circle at 2, and shading between the circles. The interval notation is negative one-third to 2 within parentheses.], alt: "The solution is negative one-third is less than x which is less than 2. The number line shows an open circle at negative one-half, an open circle at 2, and shading between the circles. The interval notation is negative one-third to 2 within parentheses.", caption: none) ] #math.equation(block: true, alt: "| x minus 4 | less than or equal to −1")[$| x − 4 | ≤ −1$] #math.equation(block: true, alt: "| 5 x plus 1 | less than or equal to −2")[$| 5 x + 1 | ≤ −2$] #solutionbox[ #figure(figph[The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.], alt: "The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.", caption: none) ] #strong[Solve Absolute Value Inequalities with “greater than”] In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation. #math.equation(block: true, alt: "| x | greater than 3")[$| x | > 3$] #math.equation(block: true, alt: "| x | greater than 6")[$| x | > 6$] #solutionbox[ #figure(figph[The solution is x is less than negative 6 or x is greater than 6. The number line shows an open circle at negative 6 with shading to its left and an open circle at 6 with shading to its right. The interval notation is the union of negative infinity to negative 6 within parentheses and 6 to infinity within parentheses], alt: "The solution is x is less than negative 6 or x is greater than 6. The number line shows an open circle at negative 6 with shading to its left and an open circle at 6 with shading to its right. The interval notation is the union of negative infinity to negative 6 within parentheses and 6 to infinity within parentheses", caption: none) ] #math.equation(block: true, alt: "| x | greater than or equal to 2")[$| x | ≥ 2$] #math.equation(block: true, alt: "| x | greater than or equal to 5")[$| x | ≥ 5$] #solutionbox[ #figure(figph[The solution is x is less then or equal to negative 5 or x is greater than or equal to 5. The number line shows an open circle at negative 5 with shading to its left and an open circle at 5 with shading to its right. The interval notation is the union of negative infinity to negative 5 within parentheses and 5 to infinity within parentheses.], alt: "The solution is x is less then or equal to negative 5 or x is greater than or equal to 5. The number line shows an open circle at negative 5 with shading to its left and an open circle at 5 with shading to its right. The interval notation is the union of negative infinity to negative 5 within parentheses and 5 to infinity within parentheses.", caption: none) ] #math.equation(block: true, alt: "| 3 x minus 8 | greater than − 1")[$| 3 x − 8 | > "−" 1$] #math.equation(block: true, alt: "| x minus 5 | greater than − 2")[$| x − 5 | > "−" 2$] #solutionbox[ #figure(figph[The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.], alt: "The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.", caption: none) ] #math.equation(block: true, alt: "| 3 x minus 2 | greater than 4")[$| 3 x − 2 | > 4$] #math.equation(block: true, alt: "| 2 x minus 1 | greater than 5")[$| 2 x − 1 | > 5$] #solutionbox[ #figure(figph[The solution is x is less than negative 2 or x is greater than 3. The number line shows an open circle at negative 2 with shading to its left and an open circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 3 to infinity within parentheses.], alt: "The solution is x is less than negative 2 or x is greater than 3. The number line shows an open circle at negative 2 with shading to its left and an open circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 3 to infinity within parentheses.", caption: none) ] #math.equation(block: true, alt: "| x plus 3 | greater than or equal to 5")[$| x + 3 | ≥ 5$] #math.equation(block: true, alt: "| x minus 7 | greater than or equal to 1")[$| x − 7 | ≥ 1$] #solutionbox[ #figure(figph[The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to 6 within parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.], alt: "The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to 6 within parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: true, alt: "3 | x | plus 4 greater than or equal to 1")[$3 | x | + 4 ≥ 1$] #math.equation(block: true, alt: "5 | x | plus 6 greater than or equal to 1")[$5 | x | + 6 ≥ 1$] #solutionbox[ #figure(figph[The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.], alt: "The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.", caption: none) ] In the following exercises, solve. For each inequality, also graph the solution and write the solution in interval notation. #math.equation(block: true, alt: "2 | x plus 6 | plus 4 equals 8")[$2 | x + 6 | + 4 = 8$] #math.equation(block: true, alt: "| 6 x minus 5 | equals | 2 x plus 3 |")[$| 6 x − 5 | = | 2 x + 3 |$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals the fraction 1 over 4")[$x = 2 , x = frac(1, 4)$] ] #math.equation(block: true, alt: "| 3 x minus 4 | greater than or equal to 2")[$| 3 x − 4 | ≥ 2$] #math.equation(block: true, alt: "| 2 x minus 5 | plus 2 equals 3")[$| 2 x − 5 | + 2 = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 3 , x equals 2")[$x = 3 , x = 2$] ] #math.equation(block: true, alt: "| 4 x minus 3 | less than 5")[$| 4 x − 3 | < 5$] #math.equation(block: true, alt: "| 3 x plus 1 | minus 3 equals 7")[$| 3 x + 1 | − 3 = 7$] #solutionbox[ #math.equation(block: true, alt: "x equals 3 , x equals minus the fraction 11 over 3")[$x = 3 , x = − frac(11, 3)$] ] #math.equation(block: true, alt: "| 7 x plus 2 | plus 8 less than 4")[$| 7 x + 2 | + 8 < 4$] #math.equation(block: true, alt: "5 | 2 x minus 1 | minus 3 equals 7")[$5 | 2 x − 1 | − 3 = 7$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction 3 over 2 , x equals minus the fraction 1 over 2")[$x = frac(3, 2) , x = − frac(1, 2)$] ] #math.equation(block: true, alt: "| 8 minus x | equals | 4 minus 3 x |")[$| 8 − x | = | 4 − 3 x |$] #math.equation(block: true, alt: "| x minus 7 | greater than − 3")[$| x − 7 | > "−" 3$] #solutionbox[ #figure(figph[The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.], alt: "The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.", caption: none) ] #strong[Solve Applications with Absolute Value] In the following exercises, solve. A chicken farm ideally produces 200,000 eggs per day. But this total can vary by as much as 25,000 eggs. What is the maximum and minimum expected production at the farm? An organic juice bottler ideally produces 215,000 bottle per day. But this total can vary by as much as 7,500 bottles. What is the maximum and minimum expected production at the bottling company? #solutionbox[ The minimum to maximum expected production is 207,500 to 2,225,000 bottles ] In order to insure compliance with the law, Miguel routinely overshoots the weight of his tortillas by 0.5 gram. He just received a report that told him that he could be losing as much as \$100,000 per year using this practice. He now plans to buy new equipment that guarantees the thickness of the tortilla within 0.005 inches. If the ideal thickness of the tortilla is 0.04 inches, what thickness of tortillas will be guaranteed? At Lilly’s Bakery, the ideal weight of a loaf of bread is 24 ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined? #solutionbox[ The acceptable weight is 22.5 to 25.5 ounces. ] ==== Writing Exercises Write a graphical description of the absolute value of a number. In your own words, explain how to solve the absolute value inequality, #math.equation(block: false, alt: "| 3 x minus 2 | greater than or equal to 4 .")[$| 3 x − 2 | ≥ 4 .$] #solutionbox[ Answers will vary. ] ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve absolute value equations. In row 3, the I can was solve absolute value inequalities with “less than.” In row 4, the I can was solve absolute value inequalities with “greater than.” In row 5, the I can was solve applications with absolute value.], alt: "This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve absolute value equations. In row 3, the I can was solve absolute value inequalities with “less than.” In row 4, the I can was solve absolute value inequalities with “greater than.” In row 5, the I can was solve applications with absolute value.", caption: none) ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve? === Chapter Review Exercises ==== Use a General Strategy to Solve Linear Equations #strong[Solve Equations Using the General Strategy for Solving Linear Equations] In the following exercises, determine whether each number is a solution to the equation. #math.equation(block: true, alt: "10 x minus 1 equals 5 x , x equals the fraction 1 over 5")[$10 x − 1 = 5 x , x = frac(1, 5)$] #math.equation(block: true, alt: "−12 n plus 5 equals 8 n , n equals minus the fraction 5 over 4")[$−12 n + 5 = 8 n , n = − frac(5, 4)$] #solutionbox[ no ] In the following exercises, solve each linear equation. #math.equation(block: true, alt: "6 open parenthesis x plus 6 close parenthesis equals 24")[$6 ( x + 6 ) = 24$] #math.equation(block: true, alt: "− open parenthesis s plus 4 close parenthesis equals 18")[$"−" ( s + 4 ) = 18$] #solutionbox[ #math.equation(block: true, alt: "s equals −22")[$s = −22$] ] #math.equation(block: true, alt: "23 minus 3 open parenthesis y minus 7 close parenthesis equals 8")[$23 − 3 ( y − 7 ) = 8$] #math.equation(block: true, alt: "the fraction 1 over 3 open parenthesis 6 m plus 21 close parenthesis equals m minus 7")[$frac(1, 3) ( 6 m + 21 ) = m − 7$] #solutionbox[ #math.equation(block: true, alt: "m equals −14")[$m = −14$] ] #math.equation(block: true, alt: "4 open parenthesis 3.5 y plus 0.25 close parenthesis equals 365")[$4 ( 3.5 y + 0.25 ) = 365$] #math.equation(block: true, alt: "0.25 open parenthesis q minus 8 close parenthesis equals 0.1 open parenthesis q plus 7 close parenthesis")[$0.25 ( q − 8 ) = 0.1 ( q + 7 )$] #solutionbox[ #math.equation(block: true, alt: "q equals 18")[$q = 18$] ] #math.equation(block: true, alt: "8 open parenthesis r minus 2 close parenthesis equals 6 open parenthesis r plus 10 close parenthesis")[$8 ( r − 2 ) = 6 ( r + 10 )$] #math.equation(block: true, alt: "5 plus 7 open parenthesis 2 minus 5 x close parenthesis equals 2 open parenthesis 9 x plus 1 close parenthesis minus open parenthesis 13 x minus 57 close parenthesis")[$5 + 7 ( 2 − 5 x ) = 2 ( 9 x + 1 ) − ( 13 x − 57 )$] #solutionbox[ #math.equation(block: true, alt: "x equals −1")[$x = −1$] ] #math.equation(block: true, alt: "open parenthesis 9 n plus 5 close parenthesis minus open parenthesis 3 n minus 7 close parenthesis equals 20 minus open parenthesis 4 n minus 2 close parenthesis")[$( 9 n + 5 ) − ( 3 n − 7 ) = 20 − ( 4 n − 2 )$] #math.equation(block: true, alt: "2 [ −16 plus 5 open parenthesis 8 k minus 6 close parenthesis ] equals 8 open parenthesis 3 minus 4 k close parenthesis minus 32")[$2 [ −16 + 5 ( 8 k − 6 ) ] = 8 ( 3 − 4 k ) − 32$] #solutionbox[ #math.equation(block: true, alt: "k equals the fraction 3 over 4")[$k = frac(3, 4)$] ] #strong[Classify Equations] In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution. #math.equation(block: true, alt: "17 y minus 3 open parenthesis 4 minus 2 y close parenthesis equals 11 open parenthesis y minus 1 close parenthesis plus 12 y minus 1")[$17 y − 3 ( 4 − 2 y ) = 11 ( y − 1 ) + 12 y − 1$] #math.equation(block: true, alt: "9 u plus 32 equals 15 open parenthesis u minus 4 close parenthesis minus 3 open parenthesis 2 u plus 21 close parenthesis")[$9 u + 32 = 15 ( u − 4 ) − 3 ( 2 u + 21 )$] #solutionbox[ contradiction; no solution ] #math.equation(block: true, alt: "−8 open parenthesis 7 m plus 4 close parenthesis equals −6 open parenthesis 8 m plus 9 close parenthesis")[$−8 ( 7 m + 4 ) = −6 ( 8 m + 9 )$] #strong[Solve Equations with Fraction or Decimal Coefficients] In the following exercises, solve each equation. #math.equation(block: true, alt: "the fraction 2 over 5 n minus the fraction 1 over 10 equals the fraction 7 over 10")[$frac(2, 5) n − frac(1, 10) = frac(7, 10)$] #solutionbox[ #math.equation(block: true, alt: "n equals 2")[$n = 2$] ] #math.equation(block: true, alt: "the fraction 3 over 4 a minus the fraction 1 over 3 equals the fraction 1 over 2 a plus the fraction 5 over 6")[$frac(3, 4) a − frac(1, 3) = frac(1, 2) a + frac(5, 6)$] #math.equation(block: true, alt: "the fraction 1 over 2 open parenthesis k plus 3 close parenthesis equals the fraction 1 over 3 open parenthesis k plus 16 close parenthesis")[$frac(1, 2) ( k + 3 ) = frac(1, 3) ( k + 16 )$] #solutionbox[ #math.equation(block: true, alt: "k equals 23")[$k = 23$] ] #math.equation(block: true, alt: "the fraction 5 y minus 1 over 3 plus 4 equals the fraction −8 y plus 4 over 6")[$frac(5 y − 1, 3) + 4 = frac(−8 y + 4, 6)$] #math.equation(block: true, alt: "0.8 x minus 0.3 equals 0.7 x plus 0.2")[$0.8 x − 0.3 = 0.7 x + 0.2$] #solutionbox[ #math.equation(block: true, alt: "x equals 5")[$x = 5$] ] #math.equation(block: true, alt: "0.10 d plus 0.05 open parenthesis d minus 4 close parenthesis equals 2.05")[$0.10 d + 0.05 ( d − 4 ) = 2.05$] ==== Use a Problem-Solving Strategy #strong[Use a Problem Solving Strategy for Word Problems] In the following exercises, solve using the problem solving strategy for word problems. Three-fourths of the people at a concert are children. If there are 87 children, what is the total number of people at the concert? #solutionbox[ There are 116 people. ] There are nine saxophone players in the band. The number of saxophone players is one less than twice the number of tuba players. Find the number of tuba players. #strong[Solve Number Word Problems] In the following exercises, solve each number word problem. The sum of a number and three is forty-one. Find the number. #solutionbox[ 38 ] One number is nine less than another. Their sum is negative twenty-seven. Find the numbers. One number is two more than four times another. Their sum is negative thirteen. Find the numbers. #solutionbox[ #math.equation(block: true, alt: "−3 , −10")[$−3 , −10$] ] The sum of two consecutive integers is #math.equation(block: false, alt: "−135 .")[$−135 .$] Find the numbers. Find three consecutive even integers whose sum is 234. #solutionbox[ 76, 78, 80 ] Find three consecutive odd integers whose sum is 51. Koji has \$5,502 in his savings account. This is \$30 less than six times the amount in his checking account. How much money does Koji have in his checking account? #solutionbox[ \$922 ] #strong[Solve Percent Applications] In the following exercises, translate and solve. What number is 67% of 250? 12.5% of what number is 20? #solutionbox[ 160 ] What percent of 125 is 150? In the following exercises, solve. The bill for Dino’s lunch was \$19.45. He wanted to leave 20% of the total bill as a tip. How much should the tip be? #solutionbox[ #math.equation(block: true, alt: "$ 3.89")[$"$" 3.89$] ] Dolores bought a crib on sale for \$350. The sale price was 40% of the original price. What was the original price of the crib? Jaden earns \$2,680 per month. He pays \$938 a month for rent. What percent of his monthly pay goes to rent? #solutionbox[ 35% ] Angel received a raise in his annual salary from \$55,400 to \$56,785. Find the percent change. Rowena’s monthly gasoline bill dropped from \$83.75 last month to \$56.95 this month. Find the percent change. #solutionbox[ 32% ] Emmett bought a pair of shoes on sale at 40% off from an original price of \$138. Find ⓐ the amount of discount and ⓑ the sale price. Lacey bought a pair of boots on sale for \$95. The original price of the boots was \$200. Find ⓐ the amount of discount and ⓑ the discount rate. (Round to the nearest tenth of a percent, if needed.) #solutionbox[ ⓐ \$105 ⓑ #math.equation(block: false, alt: "52.5 %")[$52.5 "%"$] ] Nga and Lauren bought a chest at a flea market for \$50. They re-finished it and then added a 350% mark-up. Find ⓐ the amount of the mark-up and ⓑ the list price. #strong[Solve Simple Interest Applications] In the following exercises, solve. Winston deposited \$3,294 in a bank account with interest rate 2.6% How much interest was earned in five years? #solutionbox[ \$428.22 ] Moira borrowed \$4,500 from her grandfather to pay for her first year of college. Three years later, she repaid the \$4,500 plus \$243 interest. What was the rate of interest? Jaime’s refrigerator loan statement said he would pay \$1,026 in interest for a four-year loan at 13.5%. How much did Jaime borrow to buy the refrigerator? #solutionbox[ \$1,900 ] ==== Solve a formula for a Specific Variable #strong[Solve a Formula for a Specific Variable] In the following exercises, solve the formula for the specified variable. Solve the formula #linebreak() #math.equation(block: false, alt: "V equals L W H")[$V = L W H$] for #emph[L]. Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 d sub 1 d sub 2")[$A = frac(1, 2) d_(1) d_(2)$] for #math.equation(block: false, alt: "d sub 2 .")[$d_(2) .$] #solutionbox[ #math.equation(block: true, alt: "d sub 2 equals the fraction 2 A over d sub 1")[$d_(2) = frac(2 A, d_(1))$] ] Solve the formula #linebreak() #math.equation(block: false, alt: "h equals 48 t plus the fraction 1 over 2 a t squared")[$h = 48 t + frac(1, 2) a t^(2)$] for #emph[a]. Solve the formula #linebreak() #math.equation(block: false, alt: "4 x minus 3 y equals 12")[$4 x − 3 y = 12$] for #emph[y]. #solutionbox[ #math.equation(block: true, alt: "y equals the fraction 4 x over 3 minus 4")[$y = frac(4 x, 3) − 4$] ] #strong[Use Formulas to Solve Geometry Applications] In the following exercises, solve using a geometry formula. What is the height of a triangle with area #math.equation(block: false, alt: "67.5")[$67.5$] square meters and base 9 meters? The measure of the smallest angle in a right triangle is #math.equation(block: false, alt: "45 °")[$45 "°"$] less than the measure of the next larger angle. Find the measures of all three angles. #solutionbox[ #math.equation(block: true, alt: "22.5 ° , 67.5 ° , 90 °")[$22.5 "°" , #h(0.2em) 67.5 "°" , 90 "°"$] ] The perimeter of a triangle is 97 feet. One side of the triangle is eleven feet more than the smallest side. The third side is six feet more than twice the smallest side. Find the lengths of all sides. Find the length of the hypotenuse. #figure(figph[The figure is a right triangle with a base of 10 units and a height of 24 units.], alt: "The figure is a right triangle with a base of 10 units and a height of 24 units.", caption: none) #solutionbox[ 26 ] Find the length of the missing side. Round to the nearest tenth, if necessary. #figure(figph[The figure is a right triangle with a height of 15 units and a hypotenuse of 17 units.], alt: "The figure is a right triangle with a height of 15 units and a hypotenuse of 17 units.", caption: none) Sergio needs to attach a wire to hold the antenna to the roof of his house, as shown in the figure. The antenna is eight feet tall and Sergio has 10 feet of wire. How far from the base of the antenna can he attach the wire? Approximate to the nearest tenth, if necessary. #figure(figph[The figure is a right triangle with a height of 8 feet and a hypotenuse of 10 feet.], alt: "The figure is a right triangle with a height of 8 feet and a hypotenuse of 10 feet.", caption: none) #solutionbox[ 6 feet ] Seong is building shelving in his garage. The shelves are 36 inches wide and 15 inches tall. He wants to put a diagonal brace across the back to stabilize the shelves, as shown. How long should the brace be? #figure(figph[The figure illustrates rectangular shelving whose width of 36 inch and height of 15 inches forms a right triangle with a diagonal brace.], alt: "The figure illustrates rectangular shelving whose width of 36 inch and height of 15 inches forms a right triangle with a diagonal brace.", caption: none) The length of a rectangle is 12 cm more than the width. The perimeter is 74 cm. Find the length and the width. #solutionbox[ #math.equation(block: false, alt: "24.5")[$24.5$] cm, #math.equation(block: false, alt: "12.5")[$12.5$] cm ] The width of a rectangle is three more than twice the length. The perimeter is 96 inches. Find the length and the width. The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side. #solutionbox[ 9 ft, 14 ft, 12 ft ] ==== Solve Mixture and Uniform Motion Applications #strong[Solve Coin Word Problems] In the following exercises, solve. Paulette has \$140 in \$5 and \$10 bills. The number of \$10 bills is one less than twice the number of \$5 bills. How many of each does she have? Lenny has \$3.69 in pennies, dimes, and quarters. The number of pennies is three more than the number of dimes. The number of quarters is twice the number of dimes. How many of each coin does he have? #solutionbox[ nine pennies, six dimes, 12 quarters ] #strong[Solve Ticket and Stamp Word Problems] In the following exercises, solve each ticket or stamp word problem. Tickets for a basketball game cost \$2 for students and \$5 for adults. The number of students was three less than 10 times the number of adults. The total amount of money from ticket sales was \$619. How many of each ticket were sold? 125 tickets were sold for the jazz band concert for a total of \$1,022. Student tickets cost \$6 each and general admission tickets cost \$10 each. How many of each kind of ticket were sold? #solutionbox[ 57 students, 68 general admission ] Yumi spent \$34.15 buying stamps. The number of \$0.56 stamps she bought was 10 less than four times the number of \$0.41 stamps. How many of each did she buy? #strong[Solve Mixture Word Problems] In the following exercises, solve. Marquese is making 10 pounds of trail mix from raisins and nuts. Raisins cost \$3.45 per pound and nuts cost \$7.95 per pound. How many pounds of raisins and how many pounds of nuts should Marquese use for the trail mix to cost him \$6.96 per pound? #solutionbox[ #math.equation(block: false, alt: "2.2")[$2.2$] lbs of raisins, #math.equation(block: false, alt: "7.8")[$7.8$] lbs of nuts ] Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tile. She will use basic tiles that cost \$8 per square foot and decorator tiles that cost \$20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be \$10 per square foot? Enrique borrowed \$23,500 to buy a car. He pays his uncle 2% interest on the \$4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total \$23,500? (Round your answer to the nearest tenth of a percent.) #solutionbox[ #math.equation(block: true, alt: "9.7 %")[$9.7 "%"$] ] #strong[Solve Uniform Motion Applications] In the following exercises, solve. When Gabe drives from Sacramento to Redding it takes him 2.2 hours. It takes Elsa two hours to drive the same distance. Elsa’s speed is seven miles per hour faster than Gabe’s speed. Find Gabe’s speed and Elsa’s speed. Louellen and Tracy met at a restaurant on the road between Chicago and Nashville. Louellen had left Chicago and drove 3.2 hours towards Nashville. Tracy had left Nashville and drove 4 hours towards Chicago, at a speed one mile per hour faster than Louellen’s speed. The distance between Chicago and Nashville is 472 miles. Find Louellen’s speed and Tracy’s speed. #solutionbox[ Louellen 65 mph, Tracy 66 mph ] Two busses leave Amarillo at the same time. The Albuquerque bus heads west on the I-40 at a speed of 72 miles per hour, and the Oklahoma City bus heads east on the I-40 at a speed of 78 miles per hour. How many hours will it take them to be 375 miles apart? Kyle rowed his boat upstream for 50 minutes. It took him 30 minutes to row back downstream. His speed going upstream is two miles per hour slower than his speed going downstream. Find Kyle’s upstream and downstream speeds. #solutionbox[ upstream 3 mph, downstream 5 mph ] At 6:30, Devon left her house and rode her bike on the flat road until 7:30. Then she started riding uphill and rode until 8:00. She rode a total of 15 miles. Her speed on the flat road was three miles per hour faster than her speed going uphill. Find Devon’s speed on the flat road and riding uphill. Anthony drove from New York City to Baltimore, which is a distance of 192 miles. He left at 3:45 and had heavy traffic until 5:30. Traffic was light for the rest of the drive, and he arrived at 7:30. His speed in light traffic was four miles per hour more than twice his speed in heavy traffic. Find Anthony’s driving speed in heavy traffic and light traffic. #solutionbox[ heavy traffic 32 mph, light traffic 68 mph ] ==== Solve Linear Inequalities #strong[Graph Inequalities on the Number Line] In the following exercises, graph the inequality on the number line and write in interval notation. #math.equation(block: true, alt: "x less than − 1")[$x < "−" 1$] #math.equation(block: true, alt: "x greater than or equal to −2.5")[$x ≥ −2.5$] #solutionbox[ #figure(figph[The solution is x is greater than or equal to negative 2.5. The number line shows a left bracket at negative 2.5 with shading to its right. The interval notation is negative 2.5 to infinity within a bracket and a parenthesis.], alt: "The solution is x is greater than or equal to negative 2.5. The number line shows a left bracket at negative 2.5 with shading to its right. The interval notation is negative 2.5 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: true, alt: "x less than or equal to the fraction 5 over 4")[$x ≤ frac(5, 4)$] #math.equation(block: true, alt: "x greater than 2")[$x > 2$] #solutionbox[ #figure(figph[The solution is x is greater than 2. The number line shows a left parenthesis at 2 with shading to its right. The interval notation is 2 to infinity within parentheses.], alt: "The solution is x is greater than 2. The number line shows a left parenthesis at 2 with shading to its right. The interval notation is 2 to infinity within parentheses.", caption: none) ] #math.equation(block: true, alt: "−2 less than x less than 0")[$−2 < x < 0$] #math.equation(block: true, alt: "−5 less than or equal to x less than − 3")[$−5 ≤ x < "−" 3$] #solutionbox[ #figure(figph[The solution is negative 5 is less than or equal to x which is less than negative 3. The number line shows a closed circle at negative 5, an open circle at negative 3, and shading between the circles. The interval notation is negative 5 to negative 3 within a bracket and a parenthesis.], alt: "The solution is negative 5 is less than or equal to x which is less than negative 3. The number line shows a closed circle at negative 5, an open circle at negative 3, and shading between the circles. The interval notation is negative 5 to negative 3 within a bracket and a parenthesis.", caption: none) ] #math.equation(block: true, alt: "0 less than or equal to x less than or equal to 3.5")[$0 ≤ x ≤ 3.5$] #strong[Solve Linear Inequalities] In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation. #math.equation(block: true, alt: "n minus 12 less than or equal to 23")[$n − 12 ≤ 23$] #solutionbox[ #figure(figph[The solution is n is less than or equal to 35. The number line shows a a right bracket at 35 with shading to its left. The interval notation is negative infinity to 35 within a parenthesis and a bracket.], alt: "The solution is n is less than or equal to 35. The number line shows a a right bracket at 35 with shading to its left. The interval notation is negative infinity to 35 within a parenthesis and a bracket.", caption: none) ] #math.equation(block: true, alt: "a plus the fraction 2 over 3 greater than or equal to the fraction 7 over 12")[$a + frac(2, 3) ≥ frac(7, 12)$] #math.equation(block: true, alt: "9 x greater than 54")[$9 x > 54$] #solutionbox[ #figure(figph[The solution is x is greater than 6. The number line shows a left parenthesis at 6 with shading to its right. The interval notation is 6 to infinity within parentheses.], alt: "The solution is x is greater than 6. The number line shows a left parenthesis at 6 with shading to its right. The interval notation is 6 to infinity within parentheses.", caption: none) ] #math.equation(block: true, alt: "the fraction q over −2 greater than or equal to −24")[$frac(q, −2) ≥ −24$] #math.equation(block: true, alt: "6 p greater than 15 p minus 30")[$6 p > 15 p − 30$] #solutionbox[ #figure(figph[The solution is p is less than ten-thirds. The number line shows a right parenthesis at ten-thirds with shading to its left. The interval notation is negative infinity to ten-thirds within parentheses.], alt: "The solution is p is less than ten-thirds. The number line shows a right parenthesis at ten-thirds with shading to its left. The interval notation is negative infinity to ten-thirds within parentheses.", caption: none) ] #math.equation(block: true, alt: "9 h minus 7 open parenthesis h minus 1 close parenthesis less than or equal to 4 h minus 23")[$9 h − 7 ( h − 1 ) ≤ 4 h − 23$] #math.equation(block: true, alt: "5 n minus 15 open parenthesis 4 minus n close parenthesis less than or equal to 10 open parenthesis n minus 6 close parenthesis plus 10 n")[$5 n − 15 ( 4 − n ) ≤ 10 ( n − 6 ) + 10 n$] #solutionbox[ #figure(figph[The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.], alt: "The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.", caption: none) ] #math.equation(block: true, alt: "the fraction 3 over 8 a minus the fraction 1 over 12 a greater than the fraction 5 over 12 a plus the fraction 3 over 4")[$frac(3, 8) a − frac(1, 12) a > frac(5, 12) a + frac(3, 4)$] #strong[Translate Words to an Inequality and Solve] In the following exercises, translate and solve. Then write the solution in interval notation and graph on the number line. Five more than #emph[z] is at most 19. #solutionbox[ #figure(figph[The inequality is z plus 5 is less than or equal to 19. Its solution is z is less than or equal to 14. The number line shows a right bracket at 14 with shading to its left. The interval notation is negative infinity to 14 within a parenthesis and a bracket.], alt: "The inequality is z plus 5 is less than or equal to 19. Its solution is z is less than or equal to 14. The number line shows a right bracket at 14 with shading to its left. The interval notation is negative infinity to 14 within a parenthesis and a bracket.", caption: none) ] Three less than #emph[c] is at least 360. Nine times #emph[n] exceeds 42. #solutionbox[ #figure(figph[The inequality is 9 n is greater than 42. Its solution is n is greater than fourteen-thirds. The number line shows a left parentheses at fourteen-thirds with shading to its right. The interval notation is fourteen-thirds to infinity within parentheses.], alt: "The inequality is 9 n is greater than 42. Its solution is n is greater than fourteen-thirds. The number line shows a left parentheses at fourteen-thirds with shading to its right. The interval notation is fourteen-thirds to infinity within parentheses.", caption: none) ] Negative two times #emph[a] is no more than eight. #strong[Solve Applications with Linear Inequalities] In the following exercises, solve. Julianne has a weekly food budget of \$231 for her family. If she plans to budget the same amount for each of the seven days of the week, what is the maximum amount she can spend on food each day? #solutionbox[ \$33 per day ] Rogelio paints watercolors. He got a \$100 gift card to the art supply store and wants to use it to buy 12″ × 16″ canvases. Each canvas costs \$10.99. What is the maximum number of canvases he can buy with his gift card? Briana has been offered a sales job in another city. The offer was for \$42,500 plus 8% of her total sales. In order to make it worth the move, Briana needs to have an annual salary of at least \$66,500. What would her total sales need to be for her to move? #solutionbox[ at least \$300,000 ] Renee’s car costs her \$195 per month plus \$0.09 per mile. How many miles can Renee drive so that her monthly car expenses are no more than \$250? Costa is an accountant. During tax season, he charges \$125 to do a simple tax return. His expenses for buying software, renting an office, and advertising are \$6,000. How many tax returns must he do if he wants to make a profit of at least \$8,000? #solutionbox[ at least 112 jobs ] Jenna is planning a five-day resort vacation with three of her friends. It will cost her \$279 for airfare, \$300 for food and entertainment, and \$65 per day for her share of the hotel. She has \$550 saved towards her vacation and can earn \$25 per hour as an assistant in her uncle’s photography studio. How many hours must she work in order to have enough money for her vacation? ==== Solve Compound Inequalities #strong[Solve Compound Inequalities with “and”] In each of the following exercises, solve each inequality, graph the solution, and write the solution in interval notation. #math.equation(block: false, alt: "x less than or equal to 5")[$x ≤ 5$] and #math.equation(block: false, alt: "x greater than − 3")[$x > "−" 3$] #solutionbox[ #figure(figph[The solution is negative 3 is less than x which is less than or equal to 5. The number line shows an open circle at negative 3 and a closed circle at 5. The interval notation is negative 3 to 5 within a parenthesis and a bracket.], alt: "The solution is negative 3 is less than x which is less than or equal to 5. The number line shows an open circle at negative 3 and a closed circle at 5. The interval notation is negative 3 to 5 within a parenthesis and a bracket.", caption: none) ] #math.equation(block: false, alt: "4 x minus 2 less than or equal to 4")[$4 x − 2 ≤ 4$] and #linebreak() #math.equation(block: false, alt: "7 x minus 1 greater than − 8")[$7 x − 1 > "−" 8$] #math.equation(block: false, alt: "5 open parenthesis 3 x minus 2 close parenthesis less than or equal to 5")[$5 ( 3 x − 2 ) ≤ 5$] and #linebreak() #math.equation(block: false, alt: "4 open parenthesis x plus 2 close parenthesis less than 3")[$4 ( x + 2 ) < 3$] #solutionbox[ #figure(figph[The solution is negative x is less than negative five-fourths. The number line shows an open circle at negative five-fourths with shading to its left. The interval notation is negative infinity to negative five-fourths within parentheses.], alt: "The solution is negative x is less than negative five-fourths. The number line shows an open circle at negative five-fourths with shading to its left. The interval notation is negative infinity to negative five-fourths within parentheses.", caption: none) ] #math.equation(block: false, alt: "the fraction 3 over 4 open parenthesis x minus 8 close parenthesis less than or equal to 3")[$frac(3, 4) ( x − 8 ) ≤ 3$] and #linebreak() #math.equation(block: false, alt: "the fraction 1 over 5 open parenthesis x minus 5 close parenthesis less than or equal to 3")[$frac(1, 5) ( x − 5 ) ≤ 3$] #math.equation(block: false, alt: "the fraction 3 over 4 x minus 5 greater than or equal to −2")[$frac(3, 4) x − 5 ≥ −2$] and #linebreak() #math.equation(block: false, alt: "−3 open parenthesis x plus 1 close parenthesis greater than or equal to 6")[$−3 ( x + 1 ) ≥ 6$] #solutionbox[ #figure(figph[The solution is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation], alt: "The solution is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation", caption: none) ] #math.equation(block: true, alt: "−5 less than or equal to 4 x minus 1 less than 7")[$−5 ≤ 4 x − 1 < 7$] #strong[Solve Compound Inequalities with “or”] In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation. #math.equation(block: false, alt: "5 minus 2 x less than or equal to −1")[$5 − 2 x ≤ −1$] or #linebreak() #math.equation(block: false, alt: "6 plus 3 x less than or equal to 4")[$6 + 3 x ≤ 4$] #solutionbox[ #figure(figph[The solution is x is less than negative two-thirds or x is greater than or equal to 3. The number line shows a closed circle at negative two-thirds with shading to its left and a closed circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative two-thirds within a parenthesis and a bracket and 3 to infinity within a bracket and a parenthesis.], alt: "The solution is x is less than negative two-thirds or x is greater than or equal to 3. The number line shows a closed circle at negative two-thirds with shading to its left and a closed circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative two-thirds within a parenthesis and a bracket and 3 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: false, alt: "3 open parenthesis 2 x minus 3 close parenthesis less than − 5")[$3 ( 2 x − 3 ) < "−" 5$] or #linebreak() #math.equation(block: false, alt: "4 x minus 1 greater than 3")[$4 x − 1 > 3$] #math.equation(block: false, alt: "the fraction 3 over 4 x minus 2 greater than 4")[$frac(3, 4) x − 2 > 4$] or #math.equation(block: false, alt: "4 open parenthesis 2 minus x close parenthesis greater than 0")[$4 ( 2 − x ) > 0$] #solutionbox[ #figure(figph[The solution is x is less than 2 or x is greater than 8. The number line shows an open circle at 2 with shading to its left and an open circle at 8 with shading to its right. The interval notation is the union of negative infinity to 8 within parentheses and 8 to infinity within parentheses.], alt: "The solution is x is less than 2 or x is greater than 8. The number line shows an open circle at 2 with shading to its left and an open circle at 8 with shading to its right. The interval notation is the union of negative infinity to 8 within parentheses and 8 to infinity within parentheses.", caption: none) ] #math.equation(block: false, alt: "2 open parenthesis x plus 3 close parenthesis greater than or equal to 0")[$2 ( x + 3 ) ≥ 0$] or #linebreak() #math.equation(block: false, alt: "3 open parenthesis x plus 4 close parenthesis less than or equal to 6")[$3 ( x + 4 ) ≤ 6$] #math.equation(block: false, alt: "the fraction 1 over 2 x minus 3 less than or equal to 4")[$frac(1, 2) x − 3 ≤ 4$] or #linebreak() #math.equation(block: false, alt: "the fraction 1 over 3 open parenthesis x minus 6 close parenthesis greater than or equal to −2")[$frac(1, 3) ( x − 6 ) ≥ −2$] #solutionbox[ #figure(figph[The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.], alt: "The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.", caption: none) ] #strong[Solve Applications with Compound Inequalities] In the following exercises, solve. Liam is playing a number game with his sister Audry. Liam is thinking of a number and wants Audry to guess it. Five more than three times her number is between 2 and 32. Write a compound inequality that shows the range of numbers that Liam might be thinking of. Elouise is creating a rectangular garden in her back yard. The length of the garden is 12 feet. The perimeter of the garden must be at least 36 feet and no more than 48 feet. Use a compound inequality to find the range of values for the width of the garden. #solutionbox[ #math.equation(block: true, alt: "6 less than or equal to w less than or equal to 12")[$6 ≤ w ≤ 12$] ] ==== Solve Absolute Value Inequalities #strong[Solve Absolute Value Equations] In the following exercises, solve. #math.equation(block: true, alt: "| x | equals 8")[$| x | = 8$] #math.equation(block: true, alt: "| y | equals −14")[$| y | = −14$] #solutionbox[ no solution ] #math.equation(block: true, alt: "| z | equals 0")[$| z | = 0$] #math.equation(block: true, alt: "| 3 x minus 4 | plus 5 equals 7")[$| 3 x − 4 | + 5 = 7$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals the fraction 2 over 3")[$x = 2 , x = frac(2, 3)$] ] #math.equation(block: true, alt: "4 | x minus 1 | plus 2 equals 10")[$4 | x − 1 | + 2 = 10$] #math.equation(block: true, alt: "−2 | x minus 3 | plus 8 equals −4")[$−2 | x − 3 | + 8 = −4$] #solutionbox[ #math.equation(block: true, alt: "x equals 9 , x equals −3")[$x = 9 , x = −3$] ] #math.equation(block: true, alt: "| the fraction 1 over 2 x plus 5 | plus 4 equals 1")[$| frac(1, 2) x + 5 | + 4 = 1$] #math.equation(block: true, alt: "| 6 x minus 5 | equals | 2 x plus 3 |")[$| 6 x − 5 | = | 2 x + 3 |$] #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals the fraction 1 over 4")[$x = 2 , x = frac(1, 4)$] ] #strong[Solve Absolute Value Inequalities with “less than”] In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation. #math.equation(block: true, alt: "| x | less than or equal to 8")[$| x | ≤ 8$] #math.equation(block: true, alt: "| 2 x minus 5 | less than or equal to 3")[$| 2 x − 5 | ≤ 3$] #solutionbox[ #figure(figph[The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading in between the circles. The interval notation is 1 to 4 within brackets.], alt: "The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading in between the circles. The interval notation is 1 to 4 within brackets.", caption: none) ] #math.equation(block: true, alt: "| 6 x minus 5 | less than 7")[$| 6 x − 5 | < 7$] #math.equation(block: true, alt: "| 5 x plus 1 | less than or equal to −2")[$| 5 x + 1 | ≤ −2$] #solutionbox[ #figure(figph[The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.], alt: "The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.", caption: none) ] #strong[Solve Absolute Value Inequalities with “greater than”] In the following exercises, solve. Graph the solution and write the solution in interval notation. #math.equation(block: true, alt: "| x | greater than 6")[$| x | > 6$] #math.equation(block: true, alt: "| x | greater than or equal to 2")[$| x | ≥ 2$] #solutionbox[ #figure(figph[The solution is x is less than negative 2 or x is greater than 6. The number line shows a closed circle at negative 2 with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.], alt: "The solution is x is less than negative 2 or x is greater than 6. The number line shows a closed circle at negative 2 with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: true, alt: "| x minus 5 | greater than − 2")[$| x − 5 | > "−" 2$] #math.equation(block: true, alt: "| x minus 7 | greater than or equal to 1")[$| x − 7 | ≥ 1$] #solutionbox[ #figure(figph[The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to negative 6 within a parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.], alt: "The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to negative 6 within a parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: true, alt: "3 | x | plus 4 greater than or equal to 1")[$3 | x | + 4 ≥ 1$] #strong[Solve Applications with Absolute Value] In the following exercises, solve. A craft beer brewer needs 215,000 bottle per day. But this total can vary by as much as 5,000 bottles. What is the maximum and minimum expected usage at the bottling company? #solutionbox[ The minimum to maximum expected usage is 210,000 to 220,000 bottles ] At Fancy Grocery, the ideal weight of a loaf of bread is 16 ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined? === Practice Test In the following exercises, solve each equation. #math.equation(block: true, alt: "−5 open parenthesis 2 x plus 1 close parenthesis equals 45")[$−5 ( 2 x + 1 ) = 45$] #solutionbox[ #math.equation(block: true, alt: "x equals −5")[$x = −5$] ] #math.equation(block: true, alt: "the fraction 1 over 4 open parenthesis 12 m plus 28 close parenthesis equals 6 plus 2 open parenthesis 3 m plus 1 close parenthesis")[$frac(1, 4) ( 12 m + 28 ) = 6 + 2 ( 3 m + 1 )$] #math.equation(block: true, alt: "8 open parenthesis 3 a plus 5 close parenthesis minus 7 open parenthesis 4 a minus 3 close parenthesis equals 20 minus 3 a")[$8 ( 3 a + 5 ) − 7 ( 4 a − 3 ) = 20 − 3 a$] #solutionbox[ #math.equation(block: true, alt: "a equals 41")[$a = 41$] ] #math.equation(block: true, alt: "0.1 d plus 0.25 open parenthesis d plus 8 close parenthesis equals 4.1")[$0.1 d + 0.25 ( d + 8 ) = 4.1$] #math.equation(block: true, alt: "14 n minus 3 open parenthesis 4 n plus 5 close parenthesis equals −9 plus 2 open parenthesis n minus 8 close parenthesis")[$14 n − 3 ( 4 n + 5 ) = −9 + 2 ( n − 8 )$] #solutionbox[ contradiction; no solution ] #math.equation(block: true, alt: "3 open parenthesis 3 u plus 2 close parenthesis plus 4 [ 6 minus 8 open parenthesis u minus 1 close parenthesis ] equals 3 open parenthesis u minus 2 close parenthesis")[$3 ( 3 u + 2 ) + 4 [ 6 − 8 ( u − 1 ) ] = 3 ( u − 2 )$] #math.equation(block: true, alt: "the fraction 3 over 4 x minus the fraction 2 over 3 equals the fraction 1 over 2 x plus the fraction 5 over 6")[$frac(3, 4) x − frac(2, 3) = frac(1, 2) x + frac(5, 6)$] #solutionbox[ #math.equation(block: true, alt: "x equals 6")[$x = 6$] ] #math.equation(block: true, alt: "| 3 x minus 4 | equals 8")[$| 3 x − 4 | = 8$] #math.equation(block: true, alt: "| 2 x minus 1 | equals | 4 x plus 3 |")[$| 2 x − 1 | = | 4 x + 3 |$] #solutionbox[ #math.equation(block: true, alt: "x equals −2 , x equals minus the fraction 1 over 3")[$x = −2 , x = − frac(1, 3)$] ] Solve the formula #linebreak() #math.equation(block: false, alt: "x plus 2 y equals 5")[$x + 2 y = 5$] for #emph[y]. In the following exercises, graph the inequality on the number line and write in interval notation. #math.equation(block: true, alt: "x greater than or equal to −3.5")[$x ≥ −3.5$] #solutionbox[ #figure(figph[The inequality is x is greater than or equal to negative 3.5. The number line shows a left bracket at negative 3.5 and shading to the right. The interval notation is negative 3.5 to infinity within a bracket and a parenthesis.], alt: "The inequality is x is greater than or equal to negative 3.5. The number line shows a left bracket at negative 3.5 and shading to the right. The interval notation is negative 3.5 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: true, alt: "x less than the fraction 11 over 4")[$x < frac(11, 4)$] #math.equation(block: true, alt: "−2 less than or equal to x less than 5")[$−2 ≤ x < 5$] #solutionbox[ #figure(figph[The inequality is negative two is less than or equal to x which is less than 5. The number line shows a closed circle at negative 2 and an open circle at 5 with shading between the circles. The interval notation is negative 2 to 5 within a bracket and a parenthesis.], alt: "The inequality is negative two is less than or equal to x which is less than 5. The number line shows a closed circle at negative 2 and an open circle at 5 with shading between the circles. The interval notation is negative 2 to 5 within a bracket and a parenthesis.", caption: none) ] In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation. #math.equation(block: true, alt: "8 k greater than or equal to 5 k minus 120")[$8 k ≥ 5 k − 120$] #math.equation(block: true, alt: "3 c minus 10 open parenthesis c minus 2 close parenthesis less than 5 c plus 16")[$3 c − 10 ( c − 2 ) < 5 c + 16$] #solutionbox[ #figure(figph[The solution is c is greater than one-third. The number line shows a left parenthesis at one-third with shading to its right. The interval notation is one-third to infinity within parentheses.], alt: "The solution is c is greater than one-third. The number line shows a left parenthesis at one-third with shading to its right. The interval notation is one-third to infinity within parentheses.", caption: none) ] #math.equation(block: false, alt: "the fraction 3 over 4 x minus 5 greater than or equal to −2")[$frac(3, 4) x − 5 ≥ −2$] and #linebreak() #math.equation(block: false, alt: "−3 open parenthesis x plus 1 close parenthesis greater than or equal to 6")[$−3 ( x + 1 ) ≥ 6$] #math.equation(block: false, alt: "3 open parenthesis 2 x minus 3 close parenthesis less than − 5")[$3 ( 2 x − 3 ) < "−" 5$] or #linebreak() #math.equation(block: false, alt: "4 x minus 1 greater than 3")[$4 x − 1 > 3$] #solutionbox[ #figure(figph[The solution is x is less than two-thirds or x is greater than 1. The number line shows an open circle at two-thirds with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 to infinity within parentheses.], alt: "The solution is x is less than two-thirds or x is greater than 1. The number line shows an open circle at two-thirds with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 to infinity within parentheses.", caption: none) ] #math.equation(block: false, alt: "the fraction 1 over 2 x minus 3 less than or equal to 4")[$frac(1, 2) x − 3 ≤ 4$] or #linebreak() #math.equation(block: false, alt: "the fraction 1 over 3 open parenthesis x minus 6 close parenthesis greater than or equal to −2")[$frac(1, 3) ( x − 6 ) ≥ −2$] #math.equation(block: true, alt: "| 4 x minus 3 | greater than or equal to 5")[$| 4 x − 3 | ≥ 5$] #solutionbox[ #figure(figph[The solution is x is less than or equal to negative one-half or x is greater than or equal to 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and bracket and 2 to infinity within a bracket and a parenthesis.], alt: "The solution is x is less than or equal to negative one-half or x is greater than or equal to 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and bracket and 2 to infinity within a bracket and a parenthesis.", caption: none) ] In the following exercises, translate to an equation or inequality and solve. Four less than twice #emph[x] is 16. Find the length of the missing side. #figure(figph[The figure is a right triangle with a base of 6 units and a height of 9 units.], alt: "The figure is a right triangle with a base of 6 units and a height of 9 units.", caption: none) #solutionbox[ #math.equation(block: true, alt: "10.8")[$10.8$] ] One number is four more than twice another. Their sum is #math.equation(block: false, alt: "−47 .")[$−47 .$] Find the numbers. The sum of two consecutive odd integers is #math.equation(block: false, alt: "−112 .")[$−112 .$] Find the numbers. #solutionbox[ #math.equation(block: true, alt: "−57 , −55")[$−57 , −55$] ] Marcus bought a television on sale for \$626.50 The original price of the television was \$895. Find ⓐ the amount of discount and ⓑ the discount rate. Bonita has \$2.95 in dimes and quarters in her pocket. If she has five more dimes than quarters, how many of each coin does she have? #solutionbox[ 12 dimes, seven quarters ] Kim is making eight gallons of punch from fruit juice and soda. The fruit juice costs \$6.04 per gallon and the soda costs \$4.28 per gallon. How much fruit juice and how much soda should she use so that the punch costs \$5.71 per gallon? The measure of one angle of a triangle is twice the measure of the smallest angle. The measure of the third angle is three times the measure of the smallest angle. Find the measures of all three angles. #solutionbox[ #math.equation(block: true, alt: "30 ° , 60 ° , 90 °")[$30 "°" , 60 "°" , 90 "°"$] ] The length of a rectangle is five feet more than four times the width. The perimeter is 60 feet. Find the dimensions of the rectangle. Two planes leave Dallas at the same time. One heads east at a speed of 428 miles per hour. The other plane heads west at a speed of 382 miles per hour. How many hours will it take them to be 2,025 miles apart? #solutionbox[ #math.equation(block: false, alt: "2.5")[$2.5$] hours ] Leon drove from his house in Cincinnati to his sister’s house in Cleveland, a distance of 252 miles. It took him #math.equation(block: false, alt: "4 the fraction 1 over 2")[$4 frac(1, 2)$] hours. For the first half hour, he had heavy traffic, and the rest of the time his speed was five miles per hour less than twice his speed in heavy traffic. What was his speed in heavy traffic? Sara has a budget of \$1,000 for costumes for the 18 members of her musical theater group. If all the costumes are the same price, what is the maximum she can spend for each costume? #solutionbox[ At most \$55.56 per costume. ]