#set document(title: "2.6 Solve Compound 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")) == 2.6#h(0.6em)Solve Compound Inequalities #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Simplify: #math.equation(block: false, alt: "the fraction 2 over 5 open parenthesis x plus 10 close parenthesis .")[$frac(2, 5) #h(0.2em) ( x + 10 ) .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "the fraction 2 over 5 x plus 4")[$frac(2, 5) x + 4$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify: #math.equation(block: false, alt: "− open parenthesis x minus 4 close parenthesis .")[$"−" ( x − 4 ) .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "minus x plus 4")[$− x + 4$] ] ] === Solve Compound Inequalities with “and” Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. A #strong[compound inequality] is made up of two inequalities connected by the word “and” or the word “or.” For example, the following are compound inequalities. #math.equation(block: true, alt: "x plus 3 greater than −4 and 4 x minus 5 less than or equal to 3")[$x + 3 > −4 #h(1em) "and" #h(1em) 4 x − 5 ≤ 3$]#math.equation(block: true, alt: "2 open parenthesis y plus 1 close parenthesis less than 0 or y minus 5 greater than or equal to −2")[$2 ( y + 1 ) < 0 #h(1em) "or" #h(1em) y − 5 ≥ −2$]#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Inequality] A #strong[compound inequality] is made up of two inequalities connected by the word “and” or the word “or.” ] To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions. To solve a compound inequality with the word “and,” we look for all numbers that make #emph[both] inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make #emph[either] inequality true. Let’s start with the compound inequalities with “and.” Our solution will be the numbers that are solutions to #emph[both] inequalities known as the intersection of the two inequalities. Consider the intersection of two streets—the part where the streets overlap—belongs to both streets. #figure(figph[The figure is an illustration of two streets with their intersection shaded], alt: "The figure is an illustration of two streets with their intersection shaded", caption: none) To find the solution of the compound inequality, we look at the graphs of each inequality and then find the numbers that belong to both graphs—where the graphs overlap. For the compound inequality #math.equation(block: false, alt: "x greater than −3")[$x > −3$] and #math.equation(block: false, alt: "x less than or equal to 2 ,")[$x ≤ 2 ,$] we graph each inequality. We then look for where the graphs “overlap”. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. #figure(figph[The figure shows the graph of x is greater than negative 3 with a left parenthesis at negative 3 and shading to its right, the graph of x is less than or equal to 2 with a bracket at 2 and shading to its left, and the graph of x is greater than negative 3 and x is less than or equal to 2 with a left parenthesis at negative 3 and a right parenthesis at 2 and shading between negative 3 and 2. Negative 3 and 2 are marked by lines on each number line.], alt: "The figure shows the graph of x is greater than negative 3 with a left parenthesis at negative 3 and shading to its right, the graph of x is less than or equal to 2 with a bracket at 2 and shading to its left, and the graph of x is greater than negative 3 and x is less than or equal to 2 with a left parenthesis at negative 3 and a right parenthesis at 2 and shading between negative 3 and 2. Negative 3 and 2 are marked by lines on each number line.", caption: none) We can see that the numbers between #math.equation(block: false, alt: "−3")[$−3$] and #math.equation(block: false, alt: "2")[$2$] are shaded on both of the first two graphs. They will then be shaded on the solution graph. The number #math.equation(block: false, alt: "−3")[$−3$] is not shaded on the first graph and so since it is not shaded on both graphs, it is not included on the solution graph. The number two is shaded on both the first and second graphs. Therefore, it is be shaded on the solution graph. This is how we will show our solution in the next examples. #examplebox("Example 1")[][ Solve #math.equation(block: false, alt: "6 x minus 3 less than 9")[$6 x − 3 < 9$] and #math.equation(block: false, alt: "2 x plus 9 greater than or equal to 3 .")[$2 x + 9 ≥ 3 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 4, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "6 x minus 3 less than 9")[$6 x − 3 < 9 #h(0.55em)$]], [and], [#math.equation(block: false, alt: "2 x plus 9 greater than or equal to 3")[$2 x + 9 ≥ 3 #h(0.65em)$]], [#strong[Step 1.] Solve each #linebreak() inequality.], [#math.equation(block: false, alt: "6 x minus 3 less than 9")[$6 x − 3 < 9 #h(0.55em)$]], [], [#math.equation(block: false, alt: "2 x plus 9 greater than or equal to 3")[$2 x + 9 ≥ 3 #h(0.65em)$]], [], [#math.equation(block: false, alt: "6 x less than 12")[$6 x < 12$]], [], [#math.equation(block: false, alt: "2 x greater than or equal to −6")[$2 x ≥ −6$]], [], [#math.equation(block: false, alt: "x less than 2")[$x < 2 #h(0.55em)$]], [and], [#math.equation(block: false, alt: "x greater than or equal to −3")[$x ≥ −3$]], [#strong[Step 2.] Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on #emph[both] of the first two graphs.], [#figure(figph[This image displays three number lines. The first shows x \< 2, the second shows x \>= -3, and the third shows their intersection, which is -3 \<= x \< 2, illustrating how to combine two inequalities.], alt: "This image displays three number lines. The first shows x < 2, the second shows x >= -3, and the third shows their intersection, which is -3 <= x < 2, illustrating how to combine two inequalities.", caption: none)], [], [], [#strong[Step 3.] Write the solution in interval notation.], [#math.equation(block: false, alt: "[ −3 , 2 close parenthesis")[$[ −3 , 2 )$]], [], [], [All the numbers that make both inequalities true are the solution to the compound inequality.], [], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "4 x minus 7 less than 9")[$4 x − 7 < 9$] and #math.equation(block: false, alt: "5 x plus 8 greater than or equal to 3 .")[$5 x + 8 ≥ 3 .$] #solutionbox[ #figure(figph[The solution is negative 1 is less than or equal to x which is less than 4. On a number line it is shown with a closed circle at negative 1 and an open circle at 4 with shading in between the closed and open circles. Its interval notation is negative 1 to 4 within a bracket and a parenthesis.], alt: "The solution is negative 1 is less than or equal to x which is less than 4. On a number line it is shown with a closed circle at negative 1 and an open circle at 4 with shading in between the closed and open circles. Its interval notation is negative 1 to 4 within a bracket and a parenthesis.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "3 x minus 4 less than 5")[$3 x − 4 < 5$] and #math.equation(block: false, alt: "4 x plus 9 greater than or equal to 1 .")[$4 x + 9 ≥ 1 .$] #solutionbox[ #figure(figph[The solution is negative 2 is less than or equal to x which is less than 3. On a number line it is shown with a closed circle at negative 2 and an open circle at 3 with shading in between the closed and open circles. Its interval notation is negative 2 to 3 within a bracket and a parenthesis.], alt: "The solution is negative 2 is less than or equal to x which is less than 3. On a number line it is shown with a closed circle at negative 2 and an open circle at 3 with shading in between the closed and open circles. Its interval notation is negative 2 to 3 within a bracket and a parenthesis.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve a compound inequality with “and.”] + Solve each inequality. + Graph each solution. Then graph the numbers that make #emph[both] inequalities true. #linebreak() This graph shows the solution to the compound inequality. + Write the solution in interval notation. ] #examplebox("Example 2")[][ Solve #math.equation(block: false, alt: "3 open parenthesis 2 x plus 5 close parenthesis less than or equal to 18")[$3 ( 2 x + 5 ) ≤ 18$] and #math.equation(block: false, alt: "2 open parenthesis x minus 7 close parenthesis less than −6 .")[$2 ( x − 7 ) < −6 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 4, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "3 open parenthesis 2 x plus 5 close parenthesis less than or equal to 18")[$3 ( 2 x + 5 ) ≤ 18$]], [and], [#math.equation(block: false, alt: "2 open parenthesis x minus 7 close parenthesis less than −6")[$2 ( x − 7 ) < −6$]], [Solve each #linebreak() inequality.], [#math.equation(block: false, alt: "6 x plus 15 less than or equal to 18")[$6 x + 15 ≤ 18$]], [], [#math.equation(block: false, alt: "2 x minus 14 less than −6")[$2 x − 14 < −6$]], [], [#math.equation(block: false, alt: "6 x less than or equal to 3")[$6 x ≤ 3 #h(0.5em)$]], [], [#math.equation(block: false, alt: "2 x less than 8")[$2 x < 8 #h(0.65em)$]], [], [#math.equation(block: false, alt: "x less than or equal to the fraction 1 over 2")[$x ≤ frac(1, 2) #h(0.4em)$]], [and], [#math.equation(block: false, alt: "x less than 4")[$x < 4 #h(0.65em)$]], [Graph each #linebreak() solution.], [#figure(figph[Graph of two inequalities on number lines. The first shows x less than or equal to 1/2. The second shows x less than 4.], alt: "Graph of two inequalities on number lines. The first shows x less than or equal to 1/2. The second shows x less than 4.", caption: none)], [], [], [Graph the numbers #linebreak() that make both #linebreak() inequalities true.], [#figure(figph[A number line shows the inequality 'x \< 1/2 and x \< 4'. A dashed red line marks 1/2 with an open parenthesis facing left, and a thick dark arrow extends left from there, illustrating the solution x \< 1/2.], alt: "A number line shows the inequality 'x < 1/2 and x < 4'. A dashed red line marks 1/2 with an open parenthesis facing left, and a thick dark arrow extends left from there, illustrating the solution x < 1/2.", caption: none)], [], [], [Write the solution #linebreak() in interval notation.], [#math.equation(block: false, alt: "open parenthesis − ∞ , the fraction 1 over 2 ]")[$( "−" ∞ , frac(1, 2) ]$]], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "2 open parenthesis 3 x plus 1 close parenthesis less than or equal to 20")[$2 ( 3 x + 1 ) ≤ 20$] and #math.equation(block: false, alt: "4 open parenthesis x minus 1 close parenthesis less than 2 .")[$4 ( x − 1 ) < 2 .$] #solutionbox[ #figure(figph[The solution is x is less than three-halves. On a number line it is shown with an open circle at three-halves with shading to its left. Its interval notation is negative infinity to three-halves within a parentheses.], alt: "The solution is x is less than three-halves. On a number line it is shown with an open circle at three-halves with shading to its left. Its interval notation is negative infinity to three-halves within a parentheses.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "5 open parenthesis 3 x minus 1 close parenthesis less than or equal to 10")[$5 ( 3 x − 1 ) ≤ 10$] and #math.equation(block: false, alt: "4 open parenthesis x plus 3 close parenthesis less than 8 .")[$4 ( x + 3 ) < 8 .$] #solutionbox[ #figure(figph[The solution is x is less than negative 1. On a number line it is shown with an open circle at 1 with shading to its left. Its interval notation is negative infinity to negative 1 within parentheses.], alt: "The solution is x is less than negative 1. On a number line it is shown with an open circle at 1 with shading to its left. Its interval notation is negative infinity to negative 1 within parentheses.", caption: none) ] ] #examplebox("Example 3")[][ Solve #math.equation(block: false, alt: "the fraction 1 over 3 x minus 4 greater than or equal to −2")[$frac(1, 3) x − 4 ≥ −2$] and #math.equation(block: false, alt: "−2 open parenthesis x minus 3 close parenthesis greater than or equal to 4 .")[$−2 ( x − 3 ) ≥ 4 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 4, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "the fraction 1 over 3 x minus 4 greater than or equal to −2")[$frac(1, 3) x − 4 ≥ −2$]], [and], [#math.equation(block: false, alt: "minus 2 open parenthesis x minus 3 close parenthesis greater than or equal to 4")[$− 2 ( x − 3 ) ≥ 4 #h(0.75em)$]], [Solve each inequality.], [#math.equation(block: false, alt: "the fraction 1 over 3 x minus 4 greater than or equal to −2")[$frac(1, 3) x − 4 ≥ −2$]], [], [#math.equation(block: false, alt: "−2 x plus 6 greater than or equal to 4")[$−2 x + 6 ≥ 4 #h(0.75em)$]], [], [#math.equation(block: false, alt: "the fraction 1 over 3 x greater than or equal to 2")[$frac(1, 3) x ≥ 2 #h(0.65em)$]], [], [#math.equation(block: false, alt: "−2 x greater than or equal to −2")[$−2 x ≥ −2$]], [], [#math.equation(block: false, alt: "x greater than or equal to 6")[$x ≥ 6 #h(0.65em)$]], [and], [#math.equation(block: false, alt: "x less than or equal to 1")[$x ≤ 1 #h(0.75em)$]], [Graph each solution.], [#figure(figph[Two number lines illustrate inequalities. The top line shows x '\>= ' 6, with a light blue arrow starting at 6 and pointing right. The bottom line shows x '\<=' 1, with a dark blue arrow starting at 1 and pointing left.], alt: "Two number lines illustrate inequalities. The top line shows x '>= ' 6, with a light blue arrow starting at 6 and pointing right. The bottom line shows x '<=' 1, with a dark blue arrow starting at 1 and pointing left.", caption: none)], [], [], [Graph the numbers that #linebreak() make both inequalities #linebreak() true.], [#figure(figph[A number line from -1 to 9 illustrates the contradictory inequalities 'x greater than or equal to 6 and x less than or equal to 1'], alt: "A number line from -1 to 9 illustrates the contradictory inequalities 'x greater than or equal to 6 and x less than or equal to 1'", caption: none)], [], [], [], [There are no numbers that make both inequalities true. #linebreak() #linebreak() This is a contradiction so there is no solution.], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 1 over 4 x minus 3 greater than or equal to −1")[$frac(1, 4) x − 3 ≥ −1$] and #math.equation(block: false, alt: "−3 open parenthesis x minus 2 close parenthesis greater than or equal to 2 .")[$−3 ( x − 2 ) ≥ 2 .$] #solutionbox[ #figure(figph[The inequality is a contradiction. So, there is no solution. As a result, there is no graph of the number line or interval notation.], alt: "The inequality is a contradiction. So, there is no solution. As a result, there is no graph of the number line or interval notation.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 1 over 5 x minus 5 greater than or equal to −3")[$frac(1, 5) x − 5 ≥ −3$] and #math.equation(block: false, alt: "−4 open parenthesis x minus 1 close parenthesis greater than or equal to −2 .")[$−4 ( x − 1 ) ≥ −2 .$] #solutionbox[ #figure(figph[The inequality is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.], alt: "The inequality is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.", caption: none) ] ] Sometimes we have a compound inequality that can be written more concisely. For example, #math.equation(block: false, alt: "a less than x")[$a < x$] and #math.equation(block: false, alt: "x less than b")[$x < b$] can be written simply as #math.equation(block: false, alt: "a less than x less than b")[$a < x < b$] and then we call it a #strong[double inequality]. The two forms are equivalent. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Double Inequality] A double inequality is a compound inequality such as #math.equation(block: false, alt: "a less than x less than b .")[$a < x < b .$] It is equivalent to #math.equation(block: false, alt: "a less than x")[$a < x$] and #math.equation(block: false, alt: "x less than b .")[$x < b .$] #math.equation(block: true, alt: "Other forms:, a less than x less than b, is equivalent to, a less than x, and, x less than b; a less than or equal to x less than or equal to b, is equivalent to, a less than or equal to x, and, x less than or equal to b; a greater than x greater than b, is equivalent to, a greater than x, and, x greater than b; a greater than or equal to x greater than or equal to b, is equivalent to, a greater than or equal to x, and, x greater than or equal to b")[$"Other forms:" & & & a < x < b & & & "is equivalent to" & & & a < x & & & "and" & & & x < b \ a ≤ x ≤ b & & & "is equivalent to" & & & a ≤ x & & & "and" & & & x ≤ b \ a > x > b & & & "is equivalent to" & & & a > x & & & "and" & & & x > b \ a ≥ x ≥ b & & & "is equivalent to" & & & a ≥ x & & & "and" & & & x ≥ b$] ] To solve a double inequality we perform the same operation on all three “parts” of the double inequality with the goal of isolating the variable in the center. #examplebox("Example 4")[][ Solve #math.equation(block: false, alt: "−4 less than or equal to 3 x minus 7 less than 8 .")[$−4 ≤ 3 x − 7 < 8 .$] 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 displayed: -4 is less than or equal to 3x minus 7, which is less than 8.], alt: "A mathematical inequality is displayed: -4 is less than or equal to 3x minus 7, which is less than 8.", caption: none)]), [Add 7 to all three parts.], [#figure(figph[An algebraic inequality '-4 \<= 3x - 7 \< 8' is shown with '+7' being added to all three parts, resulting in '-4 + 7 \<= 3x - 7 + 7 \< 8 + 7'.], alt: "An algebraic inequality '-4 <= 3x - 7 < 8' is shown with '+7' being added to all three parts, resulting in '-4 + 7 <= 3x - 7 + 7 < 8 + 7'.", caption: none)], [Simplify.], [#figure(figph[A mathematical inequality is shown, reading '3 ×\<= 3x ×\< 15' against a white background.], alt: "A mathematical inequality is shown, reading '3 ×<= 3x ×< 15' against a white background.", caption: none)], [Divide each part by three.], [#figure(figph[A mathematical inequality shown, with 3/3 less than or equal to 3x/3, which is less than 15/3. The denominators are highlighted in red, indicating a division step in solving the inequality.], alt: "A mathematical inequality shown, with 3/3 less than or equal to 3x/3, which is less than 15/3. The denominators are highlighted in red, indicating a division step in solving the inequality.", caption: none)], [Simplify.], [#figure(figph[An indistinct mathematical inequality 1 \<= x \< 5.], alt: "An indistinct mathematical inequality 1 <= x < 5.", caption: none)], [Graph the solution.], [#figure(figph[A number line shows an interval starting at 1 (inclusive) and ending at 5 (exclusive). The interval is represented by a solid teal line between a square bracket at 1 and a parenthesis at 5, on a number line ranging from -2 to 8.], alt: "A number line shows an interval starting at 1 (inclusive) and ending at 5 (exclusive). The interval is represented by a solid teal line between a square bracket at 1 and a parenthesis at 5, on a number line ranging from -2 to 8.", caption: none)], [Write the solution in interval notation.], [#figure(figph[The range 1 inclusive to 5 exclusive is displayed.], alt: "The range 1 inclusive to 5 exclusive is displayed.", caption: none)], )) ] ] When written as a double inequality, #math.equation(block: false, alt: "1 less than or equal to x less than 5 ,")[$1 ≤ x < 5 ,$] it is easy to see that the solutions are the numbers caught between one and five, including one, but not five. We can then graph the solution immediately as we did above. Another way to graph the solution of #math.equation(block: false, alt: "1 less than or equal to x less than 5")[$1 ≤ x < 5$] is to graph both the solution of #math.equation(block: false, alt: "x greater than or equal to 1")[$x ≥ 1$] and the solution of #math.equation(block: false, alt: "x less than 5 .")[$x < 5 .$] We would then find the numbers that make both inequalities true as we did in previous examples. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "−5 less than or equal to 4 x minus 1 less than 7 .")[$−5 ≤ 4 x − 1 < 7 .$] #solutionbox[ #figure(figph[The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 1 to 2 within a bracket and a parenthesis.], alt: "The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 1 to 2 within a bracket and a parenthesis.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "−3 less than 2 x minus 5 less than or equal to 1 .")[$−3 < 2 x − 5 ≤ 1 .$] #solutionbox[ #figure(figph[The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 with shading between the closed and open circles. Its interval notation is negative 1 to 3 within a parenthesis and a bracket.], alt: "The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 with shading between the closed and open circles. Its interval notation is negative 1 to 3 within a parenthesis and a bracket.", caption: none) ] ] === Solve Compound Inequalities with “or” To solve a #strong[compound inequality] with “or”, we start out just as we did with the compound inequalities with “and”—we solve the two inequalities. Then we find all the numbers that make #emph[either] inequality true. Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together. To write the solution in #strong[interval notation], we will often use the #strong[union symbol], #math.equation(block: false, alt: "union")[$∪$] to show the union of the solutions shown in the graphs. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve a compound inequality with “or.”] + Solve each inequality. + Graph each solution. Then graph the numbers that make either inequality true. + Write the solution in interval notation. ] #examplebox("Example 5")[][ Solve #math.equation(block: false, alt: "5 minus 3 x less than or equal to −1")[$5 − 3 x ≤ −1$] or #math.equation(block: false, alt: "8 plus 2 x less than or equal to 5 .")[$8 + 2 x ≤ 5 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 4, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "5 minus 3 x less than or equal to −1")[$5 − 3 x ≤ −1$]], [or], [#math.equation(block: false, alt: "8 plus 2 x less than or equal to 5")[$8 + 2 x ≤ 5 #h(1.25em)$]], [Solve each inequality.], [#math.equation(block: false, alt: "5 minus 3 x less than or equal to −1")[$5 − 3 x ≤ −1$]], [], [#math.equation(block: false, alt: "8 plus 2 x less than or equal to 5")[$8 + 2 x ≤ 5 #h(1.25em)$]], [], [#math.equation(block: false, alt: "−3 x less than or equal to −6")[$−3 x ≤ −6$]], [], [#math.equation(block: false, alt: "2 x less than or equal to minus 3")[$#h(3em) 2 x ≤ − 3 #h(0.1em)$]], [], [#math.equation(block: false, alt: "x greater than or equal to 2")[$x ≥ 2 #h(0.65em)$]], [or], [#math.equation(block: false, alt: "x less than or equal to minus the fraction 3 over 2")[$#h(3.2em) x ≤ − frac(3, 2)$]], [Graph each solution.], [#figure(figph[Two number lines display inequalities. The top line shows x \>= 2, shaded light blue to the right. The bottom line shows x \<= -3/2, shaded dark blue to the left.], alt: "Two number lines display inequalities. The top line shows x >= 2, shaded light blue to the right. The bottom line shows x <= -3/2, shaded dark blue to the left.", caption: none)], [], [], [Graph numbers that #linebreak() make either inequality #linebreak() true.], [#figure(figph[A number line illustrating the solution set for the inequality x '\>= 2 or x \<= -3/2, with a dark blue ray extending left from -3/2 (inclusive) and a light blue ray extending right from 2 (inclusive).], alt: "A number line illustrating the solution set for the inequality x '>= 2 or x <= -3/2, with a dark blue ray extending left from -3/2 (inclusive) and a light blue ray extending right from 2 (inclusive).", caption: none)], [], [], [], [#math.equation(block: false, alt: "open parenthesis − ∞ , minus the fraction 3 over 2 ] union [ 2 , ∞ close parenthesis")[$( "−" ∞ , − frac(3, 2) ] ∪ [ 2 , ∞ )$]], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "1 minus 2 x less than or equal to −3")[$1 − 2 x ≤ −3$] or #math.equation(block: false, alt: "7 plus 3 x less than or equal to 4 .")[$7 + 3 x ≤ 4 .$] #solutionbox[ #figure(figph[The solution is x is greater than or equal to 2 or x is less than or equal to 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 2 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 2 and infinity within a bracket and a parenthesis.], alt: "The solution is x is greater than or equal to 2 or x is less than or equal to 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 2 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 2 and infinity within a bracket and a parenthesis.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "2 minus 5 x less than or equal to −3")[$2 − 5 x ≤ −3$] or #math.equation(block: false, alt: "5 plus 2 x less than or equal to 3 .")[$5 + 2 x ≤ 3 .$] #solutionbox[ #figure(figph[The solution is x is greater than or equal to 1 or x is less than or equal to negative 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 1 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 1 and infinity within a bracket and a parenthesis.], alt: "The solution is x is greater than or equal to 1 or x is less than or equal to negative 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 1 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 1 and infinity within a bracket and a parenthesis.", caption: none) ] ] #examplebox("Example 6")[][ Solve #math.equation(block: false, alt: "the fraction 2 over 3 x minus 4 less than or equal to 3")[$frac(2, 3) x − 4 ≤ 3$] or #math.equation(block: false, alt: "the fraction 1 over 4 open parenthesis x plus 8 close parenthesis greater than or equal to −1 .")[$frac(1, 4) ( x + 8 ) ≥ −1 .$] Graph the solution and write the solution in interval notation. #solutionbox[ #figure(table( columns: 4, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "the fraction 2 over 3 x minus 4 less than or equal to 3")[$frac(2, 3) x − 4 ≤ 3 #h(1.2em)$]], [or], [#math.equation(block: false, alt: "the fraction 1 over 4 open parenthesis x plus 8 close parenthesis greater than or equal to −1")[$frac(1, 4) ( x + 8 ) ≥ −1 #h(1.8em)$]], [Solve each #linebreak() inequality.], [#math.equation(block: false, alt: "3 open parenthesis the fraction 2 over 3 x minus 4 close parenthesis less than or equal to 3 open parenthesis 3 close parenthesis")[$#h(0.4em) 3 ( frac(2, 3) x − 4 ) ≤ 3 ( 3 )$]], [], [#math.equation(block: false, alt: "4 times the fraction 1 over 4 open parenthesis x plus 8 close parenthesis greater than or equal to 4 times open parenthesis −1 close parenthesis")[$4 · frac(1, 4) ( x + 8 ) ≥ 4 · ( −1 )$]], [], [#math.equation(block: false, alt: "2 x minus 12 less than or equal to 9")[$2 x − 12 ≤ 9 #h(1.2em)$]], [], [#math.equation(block: false, alt: "x plus 8 greater than or equal to −4")[$x + 8 ≥ −4 #h(1.75em)$]], [], [#math.equation(block: false, alt: "2 x less than or equal to 21")[$2 x ≤ 21 #h(0.7em)$]], [], [#math.equation(block: false, alt: "x greater than or equal to −12")[$x ≥ −12 #h(1.25em)$]], [], [#math.equation(block: false, alt: "x less than or equal to the fraction 21 over 2")[$x ≤ frac(21, 2) #h(0.55em)$]], [], [], [], [#math.equation(block: false, alt: "x less than or equal to the fraction 21 over 2")[$x ≤ frac(21, 2) #h(0.55em)$]], [or], [#math.equation(block: false, alt: "x greater than or equal to −12")[$x ≥ −12 #h(1.25em)$]], [Graph each #linebreak() solution.], [#figure(figph[Two number lines display inequalities. The top line graphs x \<= 21/2 (10.5), shaded left from a closed bracket at 10.5. The bottom line graphs x \>= -12, shaded right from a closed bracket at -12.], alt: "Two number lines display inequalities. The top line graphs x <= 21/2 (10.5), shaded left from a closed bracket at 10.5. The bottom line graphs x >= -12, shaded right from a closed bracket at -12.", caption: none)], [], [], [Graph numbers #linebreak() that make either #linebreak() inequality true.], [#figure(figph[A number line graph representing the solution to the inequality 'x ×\<= 21/2 or x ×\>= -12'. The entire number line is highlighted in blue, indicating that all real numbers are solutions.], alt: "A number line graph representing the solution to the inequality 'x ×<= 21/2 or x ×>= -12'. The entire number line is highlighted in blue, indicating that all real numbers are solutions.", caption: none)], [], [], [], [The solution covers all real numbers.], [], [], [], [#math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]], [], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 3 over 5 x minus 7 less than or equal to −1")[$frac(3, 5) x − 7 ≤ −1$] or #math.equation(block: false, alt: "the fraction 1 over 3 open parenthesis x plus 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) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the compound inequality. Graph the solution and write the solution in interval notation: #math.equation(block: false, alt: "the fraction 3 over 4 x minus 3 less than or equal to 3")[$frac(3, 4) x − 3 ≤ 3$] or #math.equation(block: false, alt: "the fraction 2 over 5 open parenthesis x plus 10 close parenthesis greater than or equal to 0 .")[$frac(2, 5) ( x + 10 ) ≥ 0 .$] #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) ] ] === Solve Applications with Compound Inequalities Situations in the real world also involve compound inequalities. We will use the same problem solving strategy that we used to solve linear equation and inequality applications. Recall the problem solving strategies are to first read the problem and make sure all the words are understood. Then, identify what we are looking for and assign a variable to represent it. Next, restate the problem in one sentence to make it easy to translate into a #strong[compound inequality]. Last, we will solve the compound inequality. #examplebox("Example 7")[][ Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses. During the summer, a property owner will pay \$24.72 plus \$1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to \$57.06 and \$171.02. How many hcf can the owner use if he wants his usage to stay in the normal range? #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([Identify what we are looking for.], [The number of hcf he can use and stay in the “normal usage” billing range.]), [Name what we are looking for.], [Let #math.equation(block: false, alt: "x equals")[$x =$] the number of hcf he can use.], [Translate to an inequality.], [Bill is \$24.72 plus \$1.54 times the number of hcf he uses or #math.equation(block: false, alt: "24.72 plus 1.54 x .")[$24.72 + 1.54 x .$]], [], [#figure(figph[An image displays the text 'His bill will be between or equal to \$57.06 and \$171.02' with a light blue brace underneath it. Below the brace, the mathematical inequality '57.06 ', followed by the less than or equal to symbol, ' 24.74 + 1.54x ', then the less than or equal to symbol again, and finally ' 171.02' is shown.], alt: "An image displays the text 'His bill will be between or equal to $57.06 and $171.02' with a light blue brace underneath it. Below the brace, the mathematical inequality '57.06 ', followed by the less than or equal to symbol, ' 24.74 + 1.54x ', then the less than or equal to symbol again, and finally ' 171.02' is shown.", caption: none)], [Solve the inequality.], [#figure(figph[The image illustrates the step-by-step solution of a compound inequality, beginning with 57.06 A 24.74 + 1.54x A 171.02, subtracting 24.72 from all parts, and then dividing by 1.54 to isolate x, resulting in 21 A x A 95.], alt: "The image illustrates the step-by-step solution of a compound inequality, beginning with 57.06 A 24.74 + 1.54x A 171.02, subtracting 24.72 from all parts, and then dividing by 1.54 to isolate x, resulting in 21 A x A 95.", caption: none)], [Answer the question.], [The property owner can use 21–95 hcf and still fall within the “normal usage” billing range.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Due to the drought in California, many communities now have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses. During the summer, a property owner will pay \$24.72 plus \$1.32 per hcf for Conservation Usage. The bill for Conservation Usage would be between or equal to \$31.32 and \$52.12. How many hcf can the owner use if she wants her usage to stay in the conservation range? #solutionbox[ The homeowner can use 5–20 hcf and still fall within the “conservation usage” billing range. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses. During the winter, a property owner will pay \$24.72 plus \$1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to \$49.36 and \$86.32. How many hcf will he be allowed to use if he wants his usage to stay in the normal range? #solutionbox[ The homeowner can use 16–40 hcf and still fall within the “normal usage” billing range. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access this online resource for additional instruction and practice with solving compound inequalities. - #link("https://openstax.org/l/37compinequalit")[Compound inequalities] ] === Key Concepts - #strong[How to solve a compound inequality with “and”] + Solve each inequality. + Graph each solution. Then graph the numbers that make #emph[both] inequalities true. This graph shows the solution to the compound inequality. + Write the solution in interval notation. - #strong[Double Inequality] - A #strong[double inequality] is a compound inequality such as #math.equation(block: false, alt: "a less than x less than b")[$a < x < b$]. It is equivalent to #math.equation(block: false, alt: "a less than x")[$a < x$] and #math.equation(block: false, alt: "x less than b .")[$x < b .$] #linebreak() #math.equation(block: false, alt: "Other forms:, a less than x less than b, is equivalent to, a less than x, and, x less than b; a less than or equal to x less than or equal to b, is equivalent to, a less than or equal to x, and, x less than or equal to b; a greater than x greater than b, is equivalent to, a greater than x, and, x greater than b; a greater than or equal to x greater than or equal to b, is equivalent to, a greater than or equal to x, and, x greater than or equal to b")[$"Other forms:" & & & a < x < b & & & "is equivalent to" & & & a < x & & & "and" & & & x < b \ a ≤ x ≤ b & & & "is equivalent to" & & & a ≤ x & & & "and" & & & x ≤ b \ a > x > b & & & "is equivalent to" & & & a > x & & & "and" & & & x > b \ a ≥ x ≥ b & & & "is equivalent to" & & & a ≥ x & & & "and" & & & x ≥ b$] - #strong[How to solve a compound inequality with “or”] + Solve each inequality. + Graph each solution. Then graph the numbers that make either inequality true. + Write the solution in interval notation. ==== Practice Makes Perfect #strong[Solve Compound Inequalities with “and”] In the following exercises, solve each inequality, graph the solution, and write the solution in interval notation. #math.equation(block: false, alt: "x less than 3")[$x < 3$] and #math.equation(block: false, alt: "x greater than or equal to 1")[$x ≥ 1$] #math.equation(block: false, alt: "x less than or equal to 4")[$x ≤ 4$] and #math.equation(block: false, alt: "x greater than −2")[$x > −2$] #solutionbox[ #figure(figph[The solution is negative 2 is less than x which is less than or equal to 4. Its graph has an open circle at 1negative 2 and a closed circle at 4 with shading between the open and closed circles. Its interval notation is negative 2 to 4 within a parenthesis and a bracket.], alt: "The solution is negative 2 is less than x which is less than or equal to 4. Its graph has an open circle at 1negative 2 and a closed circle at 4 with shading between the open and closed circles. Its interval notation is negative 2 to 4 within a parenthesis and a bracket.", caption: none) ] #math.equation(block: false, alt: "x greater than or equal to −4")[$x ≥ −4$] and #math.equation(block: false, alt: "x less than or equal to −1")[$x ≤ −1$] #math.equation(block: false, alt: "x greater than −6")[$x > −6$] and #math.equation(block: false, alt: "x less than −3")[$x < −3$] #solutionbox[ #figure(figph[The solution is negative 6 is less than x which is less than negative 3. Its graph has an open circle at negative 6 and an open circle at negative 3 with shading between open circles. Its interval notation is negative 6 to negative 3 within parentheses.], alt: "The solution is negative 6 is less than x which is less than negative 3. Its graph has an open circle at negative 6 and an open circle at negative 3 with shading between open circles. Its interval notation is negative 6 to negative 3 within parentheses.", caption: none) ] #math.equation(block: false, alt: "5 x minus 2 less than 8")[$5 x − 2 < 8$] and #math.equation(block: false, alt: "6 x plus 9 greater than or equal to 3")[$6 x + 9 ≥ 3$] #math.equation(block: false, alt: "4 x minus 1 less than 7")[$4 x − 1 < 7$] and #math.equation(block: false, alt: "2 x plus 8 greater than or equal to 4")[$2 x + 8 ≥ 4$] #solutionbox[ #figure(figph[The solution is negative 2 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 2 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 2 to 2 within a bracket and a parenthesis.], alt: "The solution is negative 2 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 2 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 2 to 2 within a bracket and a parenthesis.", caption: none) ] #math.equation(block: false, alt: "4 x plus 6 less than or equal to 2")[$4 x + 6 ≤ 2$] and #linebreak() #math.equation(block: false, alt: "2 x plus 1 greater than or equal to −5")[$2 x + 1 ≥ −5$] #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")[$7 x − 1 > −8$] #solutionbox[ #figure(figph[The solution is negative 1 is less than x which is less than or equal to three-halves. Its graph has an open circle at negative 1 and a closed circle at three-halves with shading between the open and closed circles. Its interval notation is negative 1 to three-halves within a parenthesis and a bracket.], alt: "The solution is negative 1 is less than x which is less than or equal to three-halves. Its graph has an open circle at negative 1 and a closed circle at three-halves with shading between the open and closed circles. Its interval notation is negative 1 to three-halves within a parenthesis and a bracket.", caption: none) ] #math.equation(block: false, alt: "2 x minus 11 less than 5")[$2 x − 11 < 5$] and #linebreak() #math.equation(block: false, alt: "3 x minus 8 greater than −5")[$3 x − 8 > −5$] #math.equation(block: false, alt: "7 x minus 8 less than 6")[$7 x − 8 < 6$] and #linebreak() #math.equation(block: false, alt: "5 x plus 7 greater than −3")[$5 x + 7 > −3$] #solutionbox[ #figure(figph[The solution is negative 2 is less than x which is less than 2. Its graph has an open circle at negative 2 and an open circle at 2 with shading between the open circles. Its interval notation is negative 2 to 2 within parentheses.], alt: "The solution is negative 2 is less than x which is less than 2. Its graph has an open circle at negative 2 and an open circle at 2 with shading between the open circles. Its interval notation is negative 2 to 2 within parentheses.", caption: none) ] #math.equation(block: false, alt: "4 open parenthesis 2 x minus 1 close parenthesis less than or equal to 12")[$4 ( 2 x − 1 ) ≤ 12$] and #linebreak() #math.equation(block: false, alt: "2 open parenthesis x plus 1 close parenthesis less than 4")[$2 ( x + 1 ) < 4$] #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: "3 open parenthesis x plus 3 close parenthesis less than 3")[$3 ( x + 3 ) < 3$] #solutionbox[ #figure(figph[The solution is x is less than negative 2. Its graph has an open circle at negative 2 and is shaded to the left. Its interval notation is negative infinity to negative 2 within parentheses.], alt: "The solution is x is less than negative 2. Its graph has an open circle at negative 2 and is shaded to the left. Its interval notation is negative infinity to negative 2 within parentheses.", caption: none) ] #math.equation(block: false, alt: "3 open parenthesis 2 x minus 3 close parenthesis greater than 3")[$3 ( 2 x − 3 ) > 3$] and #linebreak() #math.equation(block: false, alt: "4 open parenthesis x plus 5 close parenthesis greater than or equal to 4")[$4 ( x + 5 ) ≥ 4$] #math.equation(block: false, alt: "−3 open parenthesis x plus 4 close parenthesis less than 0")[$−3 ( x + 4 ) < 0$] and #linebreak() #math.equation(block: false, alt: "−1 open parenthesis 3 x minus 1 close parenthesis less than or equal to 7")[$−1 ( 3 x − 1 ) ≤ 7$] #solutionbox[ #figure(figph[The solution is x is greater than or equal to negative 2. Its graph has a closed circle at negative 2 and is shaded to the right. Its interval notation is negative 2 to infinity within a bracket and a parenthesis.], alt: "The solution is x is greater than or equal to negative 2. Its graph has a closed circle at negative 2 and is shaded to the right. Its interval notation is negative 2 to infinity within a bracket and a parenthesis.", caption: none) ] #math.equation(block: false, alt: "the fraction 1 over 2 open parenthesis 3 x minus 4 close parenthesis less than or equal to 1")[$frac(1, 2) ( 3 x − 4 ) ≤ 1$] and #linebreak() #math.equation(block: false, alt: "the fraction 1 over 3 open parenthesis x plus 6 close parenthesis less than or equal to 4")[$frac(1, 3) ( x + 6 ) ≤ 4$] #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")[$frac(1, 5) ( x − 5 ) ≤ 3$] #solutionbox[ #figure(figph[The solution is x is less than or equal to 12. Its graph has a closed circle at 12 and is shaded to the left. Its interval notation is negative infinity to 12 within a parenthesis and a bracket.], alt: "The solution is x is less than or equal to 12. Its graph has a closed circle at 12 and is shaded to the left. Its interval notation is negative infinity to 12 within a parenthesis and a bracket.", caption: none) ] #math.equation(block: false, alt: "5 x minus 2 less than or equal to 3 x plus 4")[$5 x − 2 ≤ 3 x + 4$] and #linebreak() #math.equation(block: false, alt: "3 x minus 4 greater than or equal to 2 x plus 1")[$3 x − 4 ≥ 2 x + 1$] #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 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: false, alt: "the fraction 2 over 3 x minus 6 greater than or equal to −4")[$frac(2, 3) x − 6 ≥ −4$] and #linebreak() #math.equation(block: false, alt: "−4 open parenthesis x plus 2 close parenthesis greater than or equal to 0")[$−4 ( x + 2 ) ≥ 0$] #math.equation(block: false, alt: "the fraction 1 over 2 open parenthesis x minus 6 close parenthesis plus 2 less than −5")[$frac(1, 2) ( x − 6 ) + 2 < −5$] and #linebreak() #math.equation(block: false, alt: "4 minus the fraction 2 over 3 x less than 6")[$4 − frac(2, 3) x < 6$] #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: "−5 less than or equal to 4 x minus 1 less than 7")[$−5 ≤ 4 x − 1 < 7$] #math.equation(block: true, alt: "−3 less than 2 x minus 5 less than or equal to 1")[$−3 < 2 x − 5 ≤ 1$] #solutionbox[ #figure(figph[The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 and is shaded between the open and closed circles. Its interval notation is 1 to 3 within a parenthesis and a bracket.], alt: "The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 and is shaded between the open and closed circles. Its interval notation is 1 to 3 within a parenthesis and a bracket.", caption: none) ] #math.equation(block: true, alt: "5 less than 4 x plus 1 less than 9")[$5 < 4 x + 1 < 9$] #math.equation(block: true, alt: "−1 less than 3 x plus 2 less than 8")[$−1 < 3 x + 2 < 8$] #solutionbox[ #figure(figph[The solution is negative 1 is less than x which is less than 2. Its graph has an open circle at negative 1 an open circle at 2 and is shaded between. Its interval notation is negative 1 to 2 within parentheses.], alt: "The solution is negative 1 is less than x which is less than 2. Its graph has an open circle at negative 1 an open circle at 2 and is shaded between. Its interval notation is negative 1 to 2 within parentheses.", caption: none) ] #math.equation(block: true, alt: "−8 less than 5 x plus 2 less than or equal to −3")[$−8 < 5 x + 2 ≤ −3$] #math.equation(block: true, alt: "−6 less than or equal to 4 x minus 2 less than −2")[$−6 ≤ 4 x − 2 < −2$] #solutionbox[ #figure(figph[The solution is negative 1 is less than or equal to x which is less than or 0. Its graph has a closed circle at negative 1 and an open circle at 0 and is shaded between the closed and open circles. Its interval notation is negative 1 to 0 within a bracket and a parenthesis.], alt: "The solution is negative 1 is less than or equal to x which is less than or 0. Its graph has a closed circle at negative 1 and an open circle at 0 and is shaded between the closed and open circles. Its interval notation is negative 1 to 0 within a bracket and a parenthesis.", caption: none) ] #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: "x less than or equal to −2")[$x ≤ −2$] or #math.equation(block: false, alt: "x greater than 3")[$x > 3$] #math.equation(block: false, alt: "x less than or equal to −4")[$x ≤ −4$] or #math.equation(block: false, alt: "x greater than −3")[$x > −3$] #solutionbox[ #figure(figph[The solution is x is less than or equal to negative 4 or x is greater than negative 3. The graph of the solutions on a number line has a closed circle at negative 4 and shading to the left and an open circle at negative 3 with shading to the right. The interval notation is the union of negative infinity to negative 4 within a parenthesis and a bracket and negative 3 and infinity within parentheses.], alt: "The solution is x is less than or equal to negative 4 or x is greater than negative 3. The graph of the solutions on a number line has a closed circle at negative 4 and shading to the left and an open circle at negative 3 with shading to the right. The interval notation is the union of negative infinity to negative 4 within a parenthesis and a bracket and negative 3 and infinity within parentheses.", caption: none) ] #math.equation(block: false, alt: "x less than 2")[$x < 2$] or #math.equation(block: false, alt: "x greater than or equal to 5")[$x ≥ 5$] #math.equation(block: false, alt: "x less than 0")[$x < 0$] or #math.equation(block: false, alt: "x greater than or equal to 4")[$x ≥ 4$] #solutionbox[ #figure(figph[The solution is x is less than 0 or x is greater than or equal to 2. The graph of the solutions on a number line has an open circle at 0 and shading to the left and a closed circle at 4 with shading to the right. The interval notation is the union of negative infinity to 0 within parentheses and 4 to infinity within a bracket and parenthesis.], alt: "The solution is x is less than 0 or x is greater than or equal to 2. The graph of the solutions on a number line has an open circle at 0 and shading to the left and a closed circle at 4 with shading to the right. The interval notation is the union of negative infinity to 0 within parentheses and 4 to infinity within a bracket and parenthesis.", caption: none) ] #math.equation(block: false, alt: "2 plus 3 x less than or equal to 4")[$2 + 3 x ≤ 4$] or #linebreak() #math.equation(block: false, alt: "5 minus 2 x less than or equal to −1")[$5 − 2 x ≤ −1$] #math.equation(block: false, alt: "4 minus 3 x less than or equal to −2")[$4 − 3 x ≤ −2$] or #linebreak() #math.equation(block: false, alt: "2 x minus 1 less than or equal to −5")[$2 x − 1 ≤ −5$] #solutionbox[ #figure(figph[The solution is x is less than or equal to negative 2 or x is greater than or equal to 2. The graph of the solutions on a number line has a closed circle at negative 2 and shading to the left and a closed circle at 2 with shading to the 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 or equal to negative 2 or x is greater than or equal to 2. The graph of the solutions on a number line has a closed circle at negative 2 and shading to the left and a closed circle at 2 with shading to the 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: false, alt: "2 open parenthesis 3 x minus 1 close parenthesis less than 4")[$2 ( 3 x − 1 ) < 4$] or #linebreak() #math.equation(block: false, alt: "3 x minus 5 greater than 1")[$3 x − 5 > 1$] #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 graph of the solutions on a number line has an open circle at two-thirds and shading to the left and an open circle at 1 with shading to the right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 and infinity within parentheses.], alt: "The solution is x is less than two-thirds or x is greater than 1. The graph of the solutions on a number line has an open circle at two-thirds and shading to the left and an open circle at 1 with shading to the right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 and infinity within parentheses.", caption: none) ] #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")[$4 ( 2 − x ) > 0$] #math.equation(block: false, alt: "the fraction 2 over 3 x minus 3 greater than 5")[$frac(2, 3) x − 3 > 5$] or #math.equation(block: false, alt: "3 open parenthesis 5 minus x close parenthesis greater than 6")[$3 ( 5 − x ) > 6$] #solutionbox[ #figure(figph[The solution is x is less than 3 or x is greater than 12. The graph of the solutions on a number line has an open circle at 3 and shading to the left and an open circle at 12 with shading to the right. The interval notation is the union of negative infinity to 3 within parentheses and 12 and infinity within parentheses.], alt: "The solution is x is less than 3 or x is greater than 12. The graph of the solutions on a number line has an open circle at 3 and shading to the left and an open circle at 12 with shading to the right. The interval notation is the union of negative infinity to 3 within parentheses and 12 and infinity within parentheses.", caption: none) ] #math.equation(block: false, alt: "3 x minus 2 greater than 4")[$3 x − 2 > 4$] or #math.equation(block: false, alt: "5 x minus 3 less than or equal to 7")[$5 x − 3 ≤ 7$] #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")[$3 ( x + 4 ) ≤ 6$] #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: 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: false, alt: "the fraction 3 over 4 x plus 2 less than or equal to −1")[$frac(3, 4) x + 2 ≤ −1$] or #linebreak() #math.equation(block: false, alt: "the fraction 1 over 2 open parenthesis x plus 8 close parenthesis greater than or equal to −3")[$frac(1, 2) ( x + 8 ) ≥ −3$] #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[Mixed practice] 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: "3 x plus 7 less than or equal to 1")[$3 x + 7 ≤ 1$] and #linebreak() #math.equation(block: false, alt: "2 x plus 3 greater than or equal to −5")[$2 x + 3 ≥ −5$] #math.equation(block: false, alt: "6 open parenthesis 2 x minus 1 close parenthesis greater than 6")[$6 ( 2 x − 1 ) > 6$] and #linebreak() #math.equation(block: false, alt: "5 open parenthesis x plus 2 close parenthesis greater than or equal to 0")[$5 ( x + 2 ) ≥ 0$] #solutionbox[ #figure(figph[The solution is x is less than 1. Its graph has an open circle at negative 1 is shaded to the right. Its interval notation is 1 to infinity within parentheses.], alt: "The solution is x is less than 1. Its graph has an open circle at negative 1 is shaded to the right. Its interval notation is 1 to infinity within parentheses.", caption: none) ] #math.equation(block: false, alt: "4 minus 7 x greater than or equal to −3")[$4 − 7 x ≥ −3$] or #linebreak() #math.equation(block: false, alt: "5 open parenthesis x minus 3 close parenthesis plus 8 greater than 3")[$5 ( x − 3 ) + 8 > 3$] #math.equation(block: false, alt: "the fraction 1 over 2 x minus 5 less than or equal to 3")[$frac(1, 2) x − 5 ≤ 3$] or #linebreak() #math.equation(block: false, alt: "the fraction 1 over 4 open parenthesis x minus 8 close parenthesis greater than or equal to −3")[$frac(1, 4) ( x − 8 ) ≥ −3$] #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) ] #math.equation(block: true, alt: "−5 less than or equal to 2 x minus 1 less than 7")[$−5 ≤ 2 x − 1 < 7$] #math.equation(block: false, alt: "the fraction 1 over 5 open parenthesis x minus 5 close parenthesis plus 6 less than 4")[$frac(1, 5) ( x − 5 ) + 6 < 4$] and #linebreak() #math.equation(block: false, alt: "3 minus the fraction 2 over 3 x less than 5")[$3 − frac(2, 3) x < 5$] #solutionbox[ #figure(figph[The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.], alt: "The inequality 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: false, alt: "4 x minus 2 greater than 6")[$4 x − 2 > 6$] or #linebreak() #math.equation(block: false, alt: "3 x minus 1 less than or equal to −2")[$3 x − 1 ≤ −2$] #math.equation(block: false, alt: "6 x minus 3 less than or equal to 1")[$6 x − 3 ≤ 1$] and #linebreak() #math.equation(block: false, alt: "5 x minus 1 greater than −6")[$5 x − 1 > −6$] #solutionbox[ #figure(figph[The solution is negative 1 is less than x which is less than or equal to two-thirds. Its graph has an open circle at negative 1 and a closed circle at two-thirds and is shaded between the open and closed circles. Its interval notation is negative 1 to two-thirds within a parenthesis and a bracket.], alt: "The solution is negative 1 is less than x which is less than or equal to two-thirds. Its graph has an open circle at negative 1 and a closed circle at two-thirds and is shaded between the open and closed circles. Its interval notation is negative 1 to two-thirds within a parenthesis and a bracket.", caption: none) ] #math.equation(block: false, alt: "−2 open parenthesis 3 x minus 4 close parenthesis less than or equal to 2")[$−2 ( 3 x − 4 ) ≤ 2$] and #linebreak() #math.equation(block: false, alt: "−4 open parenthesis x minus 1 close parenthesis less than 2")[$−4 ( x − 1 ) < 2$] #math.equation(block: true, alt: "−5 less than or equal to 3 x minus 2 less than or equal to 4")[$−5 ≤ 3 x − 2 ≤ 4$] #solutionbox[ #figure(figph[The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and a closed circle at 2 and is shaded between the closed circles. Its interval notation is negative 1 to 4 within brackets.], alt: "The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and a closed circle at 2 and is shaded between the closed circles. Its interval notation is negative 1 to 4 within brackets.", caption: none) ] #strong[Solve Applications with Compound Inequalities] In the following exercises, solve. Penelope is playing a number game with her sister June. Penelope is thinking of a number and wants June 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 Penelope might be thinking of. Gregory is thinking of a number and he wants his sister Lauren to guess the number. His first clue is that six less than twice his number is between four and forty-two. Write a compound inequality that shows the range of numbers that Gregory might be thinking of. #solutionbox[ #math.equation(block: true, alt: "5 less than or equal to n less than or equal to 24")[$5 ≤ n ≤ 24$] ] Andrew is creating a rectangular dog run in his back yard. The length of the dog run is 18 feet. The perimeter of the dog run must be at least 42 feet and no more than 72 feet. Use a compound inequality to find the range of values for the width of the dog run. 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")[$6 ≤ w ≤ 12$] ] ==== Everyday Math #strong[Blood Pressure] A person’s blood pressure is measured with two numbers. The systolic blood pressure measures the pressure of the blood on the arteries as the heart beats. The diastolic blood pressure measures the pressure while the heart is resting. ⓐ Let #emph[x] be your systolic blood pressure. Research and then write the compound inequality that shows you what a normal systolic blood pressure should be for someone your age. ⓑ Let #emph[y] be your diastolic blood pressure. Research and then write the compound inequality that shows you what a normal diastolic blood pressure should be for someone your age. #strong[Body Mass Index] (BMI) is a measure of body fat is determined using your height and weight. ⓐ Let #emph[x] be your BMI. Research and then write the compound inequality to show the BMI range for you to be considered normal weight. ⓑ Research a BMI calculator and determine your BMI. Is it a solution to the inequality in part (a)? #solutionbox[ ⓐ answers vary ⓑ answers vary ] ==== Writing Exercises In your own words, explain the difference between the properties of equality and the properties of inequality. Explain the steps for solving the compound inequality #math.equation(block: false, alt: "2 minus 7 x greater than or equal to −5")[$2 − 7 x ≥ −5$] or #math.equation(block: false, alt: "4 open parenthesis x minus 3 close parenthesis plus 7 greater than 3 .")[$4 ( x − 3 ) + 7 > 3 .$] #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 four 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 compound inequalities with “and.” In row 3, the I can was solve compound inequalities with “or.” In row 4, the I can was solve applications with compound inequalities.], alt: "This table has four columns and four 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 compound inequalities with “and.” In row 3, the I can was solve compound inequalities with “or.” In row 4, the I can was solve applications with compound inequalities.", caption: none) ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?