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== 5.3#h(0.6em)Linear Inequalities in One Variable with Applications


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)


  
    
=== Learning Objectives


    #strong[After completing this section, you should be able to:]


    + Graph inequalities in one variable.
+ Solve linear inequalities in one variable.
+ Construct a linear inequality to solve applications.


    In this section, we will study linear inequalities in one variable. Inequalities can be used when the possible values (answers) in a certain situation are numerous, not just a few, or when the exact value (answer) is not known but it is known to be within a range of possible values. There are many real-world scenarios that can be represented by linear inequalities. For example, consider the survey of the mayoral election in  Surveys and polls are usually conducted with only a small group of people. The margin of error indicates a range of how the actual group of voters would vote given the results of the survey. This range can be expressed using inequalities.


    Another example involves college tuition. Say a local community college charges \$113 per credit hour. You budget \$1,500 for tuition this fall semester. What are the number of credit hours that you could take this fall? Since this answer could be many different values, it can be expressed as an inequality.


  
  
    
=== Graphing Inequalities on the Number Line


    In Algebraic Expressions, we introduced equality and the #math.equation(block: false, alt: "equals")[$=$] symbol. In this section, we look at inequality and the symbols #math.equation(block: false, alt: "less than")[$<$], #math.equation(block: false, alt: "greater than")[$>$], #math.equation(block: false, alt: "less than or equal to")[$≤$], and #math.equation(block: false, alt: "greater than or equal to")[$≥$]. The table below summarizes the symbols and their meaning.


    #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Symbol], [Meaning]),
  [#math.equation(block: false, alt: "less than")[$<$]], [less than],
  [#math.equation(block: false, alt: "greater than")[$>$]], [greater than],
  [#math.equation(block: false, alt: "less than or equal to")[$≤$]], [less than or equal to],
  [#math.equation(block: false, alt: "greater than or equal to")[$≥$]], [greater than or equal to],
))


    Suppose you had the inequality statement #math.equation(block: false, alt: "x greater than 3")[$x > 3$]. What possible number or numbers would make the inequality #math.equation(block: false, alt: "x greater than 3")[$x > 3$] true? If you are thinking, "#math.equation(block: false, alt: "x")[$x$] could be 4," that's correct, but #math.equation(block: false, alt: "x")[$x$] could also be 5, 6, 37, 1 million, or even 3.001. The number of solutions is infinite; any number greater than 3 is a solution to the inequality #math.equation(block: false, alt: "x greater than 3")[$x > 3$].


    Rather than trying to list all possible solutions, we show all the solutions to the inequality #math.equation(block: false, alt: "x greater than 3")[$x > 3$] on the number line. All the numbers to the right of 3 on the number line are shaded, to show that all numbers greater than 3 are solutions. At the number 3 itself, an open parenthesis is drawn, since the number 3 is not part of the solutions of #math.equation(block: false, alt: "x greater than 3")[$x > 3$].


    We can also represent inequalities using interval notation. There is no upper end to the solution to this inequality. In interval notation, we express #math.equation(block: false, alt: "x greater than 3")[$x > 3$] as #math.equation(block: false, alt: "open parenthesis 3 , infinity close parenthesis")[$( 3 , ∞ )$]. The symbol #math.equation(block: false, alt: "infinity")[$∞$] is read as "infinity." Infinity is not an actual number.  shows both the number line and the interval notation for #math.equation(block: false, alt: "x greater than 3")[$x > 3$].


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)


    We used the left parenthesis symbol to show that the endpoint of the inequality is not included. Parentheses are used when the endpoints are not included as a possible answer to the inequality. The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals.


    The inequality #math.equation(block: false, alt: "x less than or equal to 1")[$x ≤ 1$] means all numbers less than or equal to 1. To illustrate that solution on a number line, we first put a bracket at #math.equation(block: false, alt: "x equals 1")[$x = 1$]; brackets are used when the endpoint is included. We then shade in all the numbers to the left of 1, to show that all numbers less than one are solutions. There is no lower end to those numbers. We write #math.equation(block: false, alt: "x less than or equal to 1")[$x ≤ 1$] in interval notation as #math.equation(block: false, alt: "open parenthesis minus infinity , 1 ]")[$( − ∞ , 1 ]$]. The symbol #math.equation(block: false, alt: "minus infinity")[$− ∞$] is read as "negative infinity."  shows both the number line and interval notation for #math.equation(block: false, alt: "x equals 1")[$x = 1$].


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    summarizes the general representations in both number line form and interval notation of solutions for #math.equation(block: false, alt: "x greater than a")[$x > a$], #math.equation(block: false, alt: "x less than a")[$x < a$], #math.equation(block: false, alt: "x greater than or equal to a")[$x ≥ a$], and #math.equation(block: false, alt: "x less than or equal to a")[$x ≤ a$].


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  content((0.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-infinity, a)])
})],
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  import cetz.draw: *
  content((0.00pt, 24.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[x ≤ a])
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  content((280.00pt, -9.0pt), anchor: "north", text(size: 8pt)[10])
  content((0.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[\]])
  circle((0.00pt, 0.0pt), radius: 3.00pt, fill: black, stroke: 0.7pt + black)
  content((0.00pt, -14pt), anchor: "north", text(size: 9pt)[a])
  content((0.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-infinity, a\]])
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)]),
  alt: "Four number lines. The first has an open parenthesis at a. The region to the right of the parenthesis is shaded. Text reads, x is greater than a, (a, infinity). Both have a left parenthesis. The second has an open square bracket at a. The region to the right of the bracket is shaded. Text reads, x is greater than or equal to a, (a, infinity). Both have a left bracket. The third has a close parenthesis at a. The region to the left of the parenthesis is shaded. Text reads, x is less than a, (negative infinity, a). Both have a right parenthesis. The fourth has a close square bracket at a. The region to the left of the bracket is shaded. Text reads, x is less than or equal to a, (negative infinity, a). Both have a right bracket.", caption: [Summary of representations in number line form and interval notation.],
)


    #examplebox("Example 1")[Graphing an Inequality][
Graph the inequality #math.equation(block: false, alt: "x greater than or equal to minus 3")[$x ≥ − 3$] and write the solution in interval notation.


      
      #solutionbox[
Shade to the right of #math.equation(block: false, alt: "minus 3")[$− 3$] to show all the numbers greater than #math.equation(block: false, alt: "minus 3")[$− 3$], and put a bracket at #math.equation(block: false, alt: "minus 3")[$− 3$] to show that the numbers are greater than or equal to #math.equation(block: false, alt: "minus 3")[$− 3$] 


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-84.00pt, 0.0pt), (-16.80pt, 0.0pt), stroke: (paint: rgb("#D62128"), thickness: 2.5pt))
  line((-123.20pt, 0.0pt), (-16.80pt, 0.0pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, 3.60pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, -3.60pt), stroke: 0.7pt + black)
  line((-16.80pt, 0.0pt), (-22.80pt, 3.60pt), stroke: 0.7pt + black)
  line((-16.80pt, 0.0pt), (-22.80pt, -3.60pt), stroke: 0.7pt + black)
  line((-112.00pt, -4.0pt), (-112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−4])
  line((-84.00pt, -4.0pt), (-84.00pt, 4.0pt), stroke: 0.6pt + black)
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  line((-56.00pt, -4.0pt), (-56.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-56.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−2])
  line((-28.00pt, -4.0pt), (-28.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-28.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−1])
  content((-84.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[\[])
  content((-70.00pt, -28.00pt), anchor: "north", text(size: 9pt)[\[-3, infinity)])
})]),
  alt: "A number line ranges from negative 4 to negative 1, in increments of 1. An open square bracket is marked at negative 3. The region to the right of the square bracket is shaded on the number line.", caption: none,
)


        Write in interval notation starting at #math.equation(block: false, alt: "minus 3")[$− 3$] with a bracket to show that #math.equation(block: false, alt: "minus 3")[$− 3$] is included in the solution and then infinity because the solution includes all the numbers greater than or equal to #math.equation(block: false, alt: "minus 3")[$− 3$]:


        #math.equation(block: true, alt: "[ minus 3 , infinity close parenthesis")[$[ − 3 , ∞ )$]
]
]


    
    #examplebox("Example 2")[Graphing a Compound Inequality][
Graph the inequality #math.equation(block: false, alt: "x greater than minus 3")[$x > − 3$] and #math.equation(block: false, alt: "x less than 4")[$x < 4$] and write the solution in interval notation.


      
      #solutionbox[
#strong[Step 1:] Graph #math.equation(block: false, alt: "x greater than minus 3")[$x > − 3$] .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
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  line((-123.20pt, 0.0pt), (-117.20pt, 3.60pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, -3.60pt), stroke: 0.7pt + black)
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  line((56.00pt, -4.0pt), (56.00pt, 4.0pt), stroke: 0.6pt + black)
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  line((112.00pt, -4.0pt), (112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[4])
  line((140.00pt, -4.0pt), (140.00pt, 4.0pt), stroke: 0.6pt + black)
  content((140.00pt, -9.0pt), anchor: "north", text(size: 8pt)[5])
  content((-84.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[(])
  content((112.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[)])
  content((14.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-3, 4)])
})]),
  alt: "A number line ranges from negative 4 to 5, in increments of 1. An open parenthesis is marked at negative 3. The region to the right of the parenthesis is shaded on the number line. Text reads, negative 3 is less than x is less than 4.", caption: none,
)


        #strong[Step 2:] Graph #math.equation(block: false, alt: "x less than 4")[$x < 4$] .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-123.20pt, 0.0pt), (112.00pt, 0.0pt), stroke: (paint: rgb("#D62128"), thickness: 2.5pt))
  line((-123.20pt, 0.0pt), (151.20pt, 0.0pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, 3.60pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, -3.60pt), stroke: 0.7pt + black)
  line((151.20pt, 0.0pt), (145.20pt, 3.60pt), stroke: 0.7pt + black)
  line((151.20pt, 0.0pt), (145.20pt, -3.60pt), stroke: 0.7pt + black)
  line((-112.00pt, -4.0pt), (-112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−4])
  line((-84.00pt, -4.0pt), (-84.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-84.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−3])
  line((-56.00pt, -4.0pt), (-56.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-56.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−2])
  line((-28.00pt, -4.0pt), (-28.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-28.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−1])
  line((0.00pt, -4.0pt), (0.00pt, 4.0pt), stroke: 0.6pt + black)
  content((0.00pt, -9.0pt), anchor: "north", text(size: 8pt)[0])
  line((28.00pt, -4.0pt), (28.00pt, 4.0pt), stroke: 0.6pt + black)
  content((28.00pt, -9.0pt), anchor: "north", text(size: 8pt)[1])
  line((56.00pt, -4.0pt), (56.00pt, 4.0pt), stroke: 0.6pt + black)
  content((56.00pt, -9.0pt), anchor: "north", text(size: 8pt)[2])
  line((84.00pt, -4.0pt), (84.00pt, 4.0pt), stroke: 0.6pt + black)
  content((84.00pt, -9.0pt), anchor: "north", text(size: 8pt)[3])
  line((112.00pt, -4.0pt), (112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[4])
  line((140.00pt, -4.0pt), (140.00pt, 4.0pt), stroke: 0.6pt + black)
  content((140.00pt, -9.0pt), anchor: "north", text(size: 8pt)[5])
  content((112.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[)])
  content((14.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-infinity, 4)])
})]),
  alt: "A number line ranges from negative 4 to 5, in increments of 1. A close parenthesis is marked at 4. The region to the left of the parenthesis is shaded on the number line.", caption: none,
)


        #strong[Step 3:] Graph both on the same number line and think of where the solutions are to BOTH inequalities . This will be where BOTH are shaded.


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-84.00pt, 0.0pt), (112.00pt, 0.0pt), stroke: (paint: rgb("#D62128"), thickness: 2.5pt))
  line((-123.20pt, 0.0pt), (151.20pt, 0.0pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, 3.60pt), stroke: 0.7pt + black)
  line((-123.20pt, 0.0pt), (-117.20pt, -3.60pt), stroke: 0.7pt + black)
  line((151.20pt, 0.0pt), (145.20pt, 3.60pt), stroke: 0.7pt + black)
  line((151.20pt, 0.0pt), (145.20pt, -3.60pt), stroke: 0.7pt + black)
  line((-112.00pt, -4.0pt), (-112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−4])
  line((-84.00pt, -4.0pt), (-84.00pt, 4.0pt), stroke: 0.6pt + black)
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  line((-56.00pt, -4.0pt), (-56.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-56.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−2])
  line((-28.00pt, -4.0pt), (-28.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-28.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−1])
  line((0.00pt, -4.0pt), (0.00pt, 4.0pt), stroke: 0.6pt + black)
  content((0.00pt, -9.0pt), anchor: "north", text(size: 8pt)[0])
  line((28.00pt, -4.0pt), (28.00pt, 4.0pt), stroke: 0.6pt + black)
  content((28.00pt, -9.0pt), anchor: "north", text(size: 8pt)[1])
  line((56.00pt, -4.0pt), (56.00pt, 4.0pt), stroke: 0.6pt + black)
  content((56.00pt, -9.0pt), anchor: "north", text(size: 8pt)[2])
  line((84.00pt, -4.0pt), (84.00pt, 4.0pt), stroke: 0.6pt + black)
  content((84.00pt, -9.0pt), anchor: "north", text(size: 8pt)[3])
  line((112.00pt, -4.0pt), (112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[4])
  line((140.00pt, -4.0pt), (140.00pt, 4.0pt), stroke: 0.6pt + black)
  content((140.00pt, -9.0pt), anchor: "north", text(size: 8pt)[5])
  content((-84.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[(])
  content((112.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[)])
  content((14.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-3, 4)])
})]),
  alt: "A number line ranges from negative 4 to 5, in increments of 1. An open parenthesis is marked at negative 3 and a close parenthesis is marked at 4. The region within the parentheses is shaded on the number line.", caption: none,
)


        #strong[Step 4:] Write the solution in interval notation:


        #math.equation(block: true, alt: "open parenthesis minus 3 , 4 close parenthesis")[$( − 3 , 4 )$]
]
]


    
    #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
#emph[Where Did the Inequality Symbols Come From?]


      The first use of the #math.equation(block: false, alt: "less than")[$<$] symbol to represent "less than" and #math.equation(block: false, alt: "greater than")[$>$] to represent "greater than" appeared in a mathematics book written by Englishman Thomas Harriot that was published in 1631. However, Harriot did not invent the symbols…the editor of the book did! Harriot used triangular symbols to represent less than and greater than; the editor, for reasons unknown, changed to symbols that are similar to the ones we use today. The symbols used to represent less than or equal to, and greater than or equal to (#math.equation(block: false, alt: "less than or equal to")[$≤$] and #math.equation(block: false, alt: "greater than or equal to")[$≥$]) were first used in 1731 by French hydrologist and surveyor Pierre Bouguer. Interestingly, English mathematician John Wallis had used similar symbols as early as 1670, but he put the bar above the less than and greater than symbols instead of below them.
]


  
  
    
=== Solving Linear Inequalities


    A #strong[linear inequality] is much like a linear equation—but the equal sign is replaced with an inequality sign. A linear inequality is an inequality in one variable that can be written in one of the forms #math.equation(block: false, alt: "a x plus b less than c , a x plus b less than or equal to c , a x plus b greater than or equal to c ,")[$a x + b < c , a x + b ≤ c , a x + b ≥ c ,$] or #math.equation(block: false, alt: "a x plus b greater than c ,")[$a x + b > c ,$] where #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] are all real numbers.


    When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities. We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example, we know that 2 is less than 4, i.e., #math.equation(block: false, alt: "2 less than 4")[$2 < 4$]. If we add 6 to both sides of this inequality, we still have a true statement:


    #math.equation(block: true, alt: "2 plus 6, less than, 4 plus 6; 8, less than, 10")[$2 + 6 & < & 4 + 6 \ 8 & < & 10$]
    The same would happen if we subtracted 6 from both sides of the inequality; the statement would stay true:


    #math.equation(block: true, alt: "2 minus 6, less than, 4 minus 6; minus 4, less than, minus 2")[$2 − 6 & < & 4 − 6 \ − 4 & < & − 2$]
    Notice that the inequality signs stayed the same. This leads us to the Addition and Subtraction Properties of Inequality.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
For any numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$], if #math.equation(block: false, alt: "a less than b")[$a < b$], then #math.equation(block: false, alt: "a plus c less than b plus c")[$a + c < b + c$] and #math.equation(block: false, alt: "a minus c less than b minus c")[$a − c < b − c$].


      For any numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$], if #math.equation(block: false, alt: "a greater than b")[$a > b$], then #math.equation(block: false, alt: "a plus c greater than b plus c")[$a + c > b + c$] and #math.equation(block: false, alt: "a minus c greater than b minus c")[$a − c > b − c$].
]


    We can add or subtract the same quantity from both sides of an inequality and still keep the inequality the same. But what happens to an inequality when we divide or multiply both sides by a number? Let's first multiply and divide both sides by a positive number, starting with an inequality we know is true, #math.equation(block: false, alt: "10 less than 15")[$10 < 15$]. We will multiply and divide this inequality by 5:


    #math.equation(block: true, alt: "10, less than, 15, 10, less than, 15; 10 open parenthesis 5 close parenthesis, ?, 15 open parenthesis 5 close parenthesis, the fraction 10 over 5, ?, the fraction 15 over 5; 50, ?, 75, 2, ?, 3; 50, less than, 75 open parenthesis true close parenthesis, 2, less than, 3 open parenthesis true close parenthesis")[$10 & < & 15 & 10 & < & 15 \ 10 ( 5 ) & ? & 15 ( 5 ) & frac(10, 5) & ? & frac(15, 5) \ 50 & ? & 75 & 2 & ? & 3 \ 50 & < & 75 ( "true" ) & 2 & < & 3 ( "true" )$]
    The inequality signs stayed the same. Does the inequality stay the same when we divide or multiply by a negative number? Let's use our inequality #math.equation(block: false, alt: "10 less than 15")[$10 < 15$] to find out, multiplying it and dividing it by #math.equation(block: false, alt: "minus 5")[$− 5$]:


    #math.equation(block: true, alt: "10, less than, 15, 10, less than, 15; 10 open parenthesis minus 5 close parenthesis, ?, 15 open parenthesis minus 5 close parenthesis, the fraction 10 over minus 5, ?, the fraction 15 over minus 5; minus 50, ?, minus 75, minus 2, ?, minus 3; minus 50, greater than, minus 75 open parenthesis true close parenthesis, minus 2, greater than, minus 3 open parenthesis true close parenthesis")[$10 & < & 15 & 10 & < & 15 \ 10 ( − 5 ) & ? & 15 ( − 5 ) & frac(10, − 5) & ? & frac(15, − 5) \ − 50 & ? & − 75 & − 2 & ? & − 3 \ − 50 & > & − 75 ( "true" ) & − 2 & > & − 3 ( "true" )$]
    Notice that when we filled in the inequality signs, the inequality signs reversed their direction in order to make it true! To summarize, when we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses. This gives us the Multiplication and Division Property of Inequality.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
For any numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$],


      #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([multiply or divide by a positive:], [if #math.equation(block: false, alt: "a less than b")[$a < b$]
     and #math.equation(block: false, alt: "c greater than 0")[$c > 0$] , then #math.equation(block: false, alt: "a c less than b c")[$a c < b c$] and #math.equation(block: false, alt: "the fraction a over c less than the fraction b over c")[$frac(a, c) < frac(b, c)$]]),
  [], [if #math.equation(block: false, alt: "a greater than b")[$a > b$] and #math.equation(block: false, alt: "c greater than 0")[$c > 0$] , then #math.equation(block: false, alt: "a c greater than b c")[$a c > b c$] and #math.equation(block: false, alt: "the fraction a over c greater than the fraction b over c")[$frac(a, c) > frac(b, c)$]],
  [multiply or divide by a negative:], [if #math.equation(block: false, alt: "a less than b")[$a < b$] and #math.equation(block: false, alt: "c less than 0")[$c < 0$] , then #math.equation(block: false, alt: "a c greater than b c")[$a c > b c$] and #math.equation(block: false, alt: "the fraction a over c greater than the fraction b over c")[$frac(a, c) > frac(b, c)$]],
  [], [if #math.equation(block: false, alt: "a greater than b")[$a > b$] and #math.equation(block: false, alt: "c less than 0")[$c < 0$] , then #math.equation(block: false, alt: "a c less than b c")[$a c < b c$] and #math.equation(block: false, alt: "the fraction a over c less than the fraction b over c")[$frac(a, c) < frac(b, c)$]],
))
]


    To summarize, when we divide or multiply an inequality by:


    - a positive number, the inequality sign stays the same.
- a negative number, the inequality sign reverses.


    #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
#emph[Be careful to only reverse the inequality sign when you are multiplying and dividing by a negative. You do NOT reverse the inequality sign when you add or subtract a negative. For example, #math.equation(block: false, alt: "2 x less than minus 4")[$2 x < − 4$] is solved by dividing both sides of the inequality by 2 to get #math.equation(block: false, alt: "x less than minus 2")[$x < − 2$]. You do NOT reverse the inequality sign because there is a negative 4. As another example, #math.equation(block: false, alt: "minus 2 x plus 5 less than 3 x")[$− 2 x + 5 < 3 x$] is solved by adding #math.equation(block: false, alt: "minus 2 x")[$− 2 x$] to both sides to get #math.equation(block: false, alt: "5 less than 5 x")[$5 < 5 x$]. This does not reverse the inequality sign because we were not multiplying or dividing by a negative. We then divide both sides by 5 and get #math.equation(block: false, alt: "1 less than x")[$1 < x$].]
]


    #examplebox("Example 3")[Solving a Linear Inequality Using One Operation][
Solve #math.equation(block: false, alt: "9 y less than 54")[$9 y < 54$], graph the solution on the number line, and write the solution in interval notation.


      
      #solutionbox[
#math.equation(block: true, alt: "9 y, less than, 54; the fraction 9 y over 9, less than, the fraction 54 over 9; y, less than, 6")[$9 y & < & 54 \ frac(9 y, 9) & < & frac(54, 9) \ y & < & 6$]


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((100.80pt, 0.0pt), (168.00pt, 0.0pt), stroke: (paint: rgb("#D62128"), thickness: 2.5pt))
  line((100.80pt, 0.0pt), (207.20pt, 0.0pt), stroke: 0.7pt + black)
  line((100.80pt, 0.0pt), (106.80pt, 3.60pt), stroke: 0.7pt + black)
  line((100.80pt, 0.0pt), (106.80pt, -3.60pt), stroke: 0.7pt + black)
  line((207.20pt, 0.0pt), (201.20pt, 3.60pt), stroke: 0.7pt + black)
  line((207.20pt, 0.0pt), (201.20pt, -3.60pt), stroke: 0.7pt + black)
  line((112.00pt, -4.0pt), (112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[4])
  line((140.00pt, -4.0pt), (140.00pt, 4.0pt), stroke: 0.6pt + black)
  content((140.00pt, -9.0pt), anchor: "north", text(size: 8pt)[5])
  line((168.00pt, -4.0pt), (168.00pt, 4.0pt), stroke: 0.6pt + black)
  content((168.00pt, -9.0pt), anchor: "north", text(size: 8pt)[6])
  line((196.00pt, -4.0pt), (196.00pt, 4.0pt), stroke: 0.6pt + black)
  content((196.00pt, -9.0pt), anchor: "north", text(size: 8pt)[7])
  content((168.00pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[)])
  content((154.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-infinity, 6)])
})]),
  alt: "A number line ranges from 4 to 7, in increments of 1. A close parenthesis is marked at 6. The region to the left of the parenthesis is shaded on the number line. Text reads (negative infinity, 6).", caption: none,
)
]
]


    
    #examplebox("Example 4")[Solving a Linear Inequality Using Multiple Operations][
Solve the inequality #math.equation(block: false, alt: "6 y less than or equal to 11 y plus 17")[$6 y ≤ 11 y + 17$], graph the solution on the number line, and write the solution in interval notation.


      
      #solutionbox[
#math.equation(block: true, alt: "6 y, less than or equal to, 11 y plus 17; 6 y minus 11 y, less than or equal to, 11 y plus 17 minus 11 y; minus 5 y, less than or equal to, 17; the fraction minus 5 y over minus 5, greater than or equal to, minus the fraction 17 over 5; y, greater than or equal to, minus the fraction 17 over 5")[$6 y & ≤ & 11 y + 17 \ 6 y − 11 bold(italic(y)) & ≤ & 11 bold(italic(y)) + 17 − 11 y \ − 5 y & ≤ & 17 \ frac(− 5 y, − 5) & ≥ & − frac(17, 5) \ y & ≥ & − frac(17, 5)$]
        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((-95.20pt, 0.0pt), (-44.80pt, 0.0pt), stroke: (paint: rgb("#D62128"), thickness: 2.5pt))
  line((-151.20pt, 0.0pt), (-44.80pt, 0.0pt), stroke: 0.7pt + black)
  line((-151.20pt, 0.0pt), (-145.20pt, 3.60pt), stroke: 0.7pt + black)
  line((-151.20pt, 0.0pt), (-145.20pt, -3.60pt), stroke: 0.7pt + black)
  line((-44.80pt, 0.0pt), (-50.80pt, 3.60pt), stroke: 0.7pt + black)
  line((-44.80pt, 0.0pt), (-50.80pt, -3.60pt), stroke: 0.7pt + black)
  line((-140.00pt, -4.0pt), (-140.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-140.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−5])
  line((-112.00pt, -4.0pt), (-112.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-112.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−4])
  line((-84.00pt, -4.0pt), (-84.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-84.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−3])
  line((-56.00pt, -4.0pt), (-56.00pt, 4.0pt), stroke: 0.6pt + black)
  content((-56.00pt, -9.0pt), anchor: "north", text(size: 8pt)[−2])
  content((-95.20pt, 0.0pt), text(size: 14pt, fill: rgb("#D62128"))[(])
  content((-98.00pt, -28.00pt), anchor: "north", text(size: 9pt)[(-17/5, infinity)])
})]),
  alt: "A number line ranges from negative 5 to negative 2, in increments of 1. An open square bracket is marked at negative 17 over 5. The region to the right of the square bracket is shaded on the number line. Text reads (negative 17 over 5, infinity).", caption: none,
)
]
]


    
  
  
    
=== Solving Applications with Linear Inequalities


    Many real-life situations require us to solve inequalities. The method we will use to solve applications with #strong[linear inequalities] is very much like the one we used when we solved applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality.


    Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down.


    #examplebox("Example 5")[Constructing a Linear Inequality to Solve an Application with Tablet Computers][
A teacher won a mini grant of \$4,000 to buy tablet computers for their classroom. The tablets they would like to buy cost \$254.12 each, including tax and delivery. What is the maximum number of tablets the teacher can buy?


      
      #solutionbox[
Let #math.equation(block: false, alt: "t equals the number of tablets .")[$t = "the number of tablets" .$]


        #math.equation(block: false, alt: "t")[$t$] times \$254.12 has to be less than \$4,000, so #math.equation(block: false, alt: "254.12 t less than or equal to 4 , 000")[$254.12 t ≤ 4 , 000$].


        Solve for #math.equation(block: false, alt: "t")[$t$]:


        #math.equation(block: true, alt: "the fraction 254.12 t over 254.12, less than or equal to, the fraction 4,000 over 254.12; t, less than or equal to, 15.74")[$frac(254.12 t, 254.12) & ≤ & frac(4","000, 254.12) \ t & ≤ & 15.74$]
        The teacher can buy 15 tablets and stay under \$4,000.
]
]


    
    #examplebox("Example 6")[Constructing a Linear Inequality to Solve a Tuition Application][
The local community college charges \$113 per credit hour. Your budget is \$1,500 for tuition this fall semester. What number of credit hours could you take this fall?


      
      #solutionbox[
Let #math.equation(block: false, alt: "c equals")[$c =$] the number of credit hours you could take.


        #math.equation(block: false, alt: "c")[$c$] times \$113 has to be less than \$1,500, so #math.equation(block: false, alt: "113 c less than or equal to 1 , 500")[$113 c ≤ 1 , 500$].


        Solve for #math.equation(block: false, alt: "c")[$c$]:


        #math.equation(block: true, alt: "the fraction 113 c over 113, less than or equal to, the fraction 1500 over 113; c, less than or equal to, 13.27")[$frac(113 c, 113) & ≤ & frac(1500, 113) \ c & ≤ & 13.27$]
        You can take up to 13 credits and stay under \$1,500.
]
]


    
    #examplebox("Example 7")[Constructing a Linear Inequality to Solve an Application with Travel Costs][
Brenda’s best friend is having a destination wedding and the event will last 3 days and 3 nights. Brenda has \$500 in savings and can earn \$15 an hour babysitting. She expects to pay \$350 for airfare, \$375 for food and entertainment, and \$60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?


      
      #solutionbox[
Let #math.equation(block: false, alt: "b equals")[$b =$] number of babysitting hours.


        #math.equation(block: false, alt: "b")[$b$] times \$15 plus \$500 has to be more than #math.equation(block: false, alt: "$ 350 plus $ 375 plus $ 60 / night")[$upright(\$) 350 + upright(\$) 375 + upright(\$) 60 / "night"$], so #math.equation(block: false, alt: "15 b plus 500 greater than or equal to 350 plus 375 plus 60 open parenthesis 3 close parenthesis")[$15 b + 500 ≥ 350 + 375 + 60 ( 3 )$].


        Solve for #math.equation(block: false, alt: "b")[$b$]:


        #math.equation(block: true, alt: "15 b plus 500 minus 500, greater than or equal to, 905 minus 500; 15 b, greater than or equal to, 405; the fraction 15 b over 15, greater than or equal to, the fraction 405 over 15; b, greater than or equal to, 27")[$15 b + 500 − 500 & ≥ & 905 − 500 \ 15 b & ≥ & 405 \ frac(15 b, 15) & ≥ & frac(405, 15) \ b & ≥ & 27$]
        Brenda must babysit at least 27 hours.
]
]


    
    #notebox("Tech Check", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
The Desmos activities called #link("https://openstax.org/r/Inequalities_on_a")["Inequalities on a Number Line"] and #link("https://openstax.org/r/Compound_Inequalities")["Compound Inequalities on a Number Line"] are ways for students to develop and deepen their understanding of inequalities. Teachers will need a Desmos account to assign the activity for student use. Once they have assigned the activity to their students, teachers need to share the code for the activity with their students. Students #link("https://openstax.org/r/will_input")[will input the code] to work on the activity.
]


  
  
  
  
    
=== Key Terms


    - linear inequality
- Addition and Subtraction Property of Linear Inequalities
- Multiplication and Division Property of Linear Inequalities


  
  
    
=== Key Concepts


    - Inequalities can be used when the possible values (answers) in a certain situation are numerous, or when the exact value (answer) is not known, but it is known to be within a range of possible values.
- Linear inequalities can be represented using a number line or using interval notation.


  
  
    
=== Formulas


    - For any numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: ",")[$,$] if #math.equation(block: false, alt: "a less than b")[$a < b$], then #math.equation(block: false, alt: "a plus c less than b plus c")[$a + c < b + c$] and #math.equation(block: false, alt: "a minus c less than b minus c")[$a − c < b − c$].
- For any numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$], if #math.equation(block: false, alt: "a greater than b")[$a > b$], then #math.equation(block: false, alt: "a plus c greater than b plus c")[$a + c > b + c$] and #math.equation(block: false, alt: "a minus c greater than b minus c")[$a − c > b − c$].
- For any numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$], #linebreak()  multiply or divide by a positive: #linebreak() 
     if #math.equation(block: false, alt: "a less than b")[$a < b$] and #math.equation(block: false, alt: "c greater than 0")[$c > 0$], then #math.equation(block: false, alt: "a c less than b c")[$a c < b c$] and #math.equation(block: false, alt: "the fraction a over c less than the fraction b over c")[$frac(a, c) < frac(b, c)$] #linebreak() 
     if #math.equation(block: false, alt: "a greater than b")[$a > b$] and #math.equation(block: false, alt: "c greater than 0")[$c > 0$], then #math.equation(block: false, alt: "a c greater than b c")[$a c > b c$] and #math.equation(block: false, alt: "the fraction a over c greater than the fraction b over c")[$frac(a, c) > frac(b, c)$] #linebreak() 
     multiply or divide by a negative:  #linebreak() 
     if #math.equation(block: false, alt: "a less than b")[$a < b$] and #math.equation(block: false, alt: "c less than 0")[$c < 0$], then #math.equation(block: false, alt: "a c greater than b c")[$a c > b c$] and #math.equation(block: false, alt: "the fraction a over c greater than the fraction b over c")[$frac(a, c) > frac(b, c)$] #linebreak() 
     if #math.equation(block: false, alt: "a greater than b")[$a > b$] and #math.equation(block: false, alt: "c less than 0")[$c < 0$], then #math.equation(block: false, alt: "a c less than b c")[$a c < b c$] and #math.equation(block: false, alt: "the fraction a over c less than the fraction b over c")[$frac(a, c) < frac(b, c)$]