#set document(title: "8.4 Linear 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")) == 8.4#h(0.6em)Linear Inequalities In this section, we study linear inequalities in two variables and how they arise in applications. === Graphs of Inequalities in Two Variables Ivana is investing in the hotel business. She has bought two hotels, and she will expand her investments when her total profit from the two hotels exceeds #math.equation(block: false, alt: "$ 10 , 000")[$\$ 10 , 000$]. If we let #math.equation(block: false, alt: "x")[$x$] represent the profit from one hotel and let #math.equation(block: false, alt: "y")[$y$] represent the profit from the other, then Ivana will expand her investments when #math.equation(block: true, alt: "x plus y greater than or equal to 10 , 000")[$x + y ≥ 10 , 000$] Notice that the equation #math.equation(block: false, alt: "x plus y equals 10 , 000")[$x + y = 10 , 000$] is not appropriate to model our situation, since Ivana will be delighted if her profits are not exactly equal to #math.equation(block: false, alt: "$ 10 , 000")[$\$ 10 , 000$] but actually exceed that amount. A #strong[solution] to an inequality in two variables is an ordered pair of numbers that satisfies the inequality. The graph of the inequality must show all the points whose coordinates are solutions. As an example, let us graph the inequality above, #math.equation(block: false, alt: "x plus y greater than or equal to 10 , 000")[$x + y ≥ 10 , 000$]. Rewrite the inequality by subtracting #math.equation(block: false, alt: "x")[$x$] from both sides to get #math.equation(block: true, alt: "y greater than or equal to minus x plus 10 , 000")[$y ≥ − x + 10 , 000$] This inequality says that for each #math.equation(block: false, alt: "x")[$x$]-value, we must choose points with #math.equation(block: false, alt: "y")[$y$]-values greater than or equal to #math.equation(block: false, alt: "minus x plus 10 , 000")[$− x + 10 , 000$]. For example, when #math.equation(block: false, alt: "x equals 2000")[$x = 2000$], we must choose points with #math.equation(block: false, alt: "y")[$y$]-values greater than or equal to #math.equation(block: false, alt: "8000")[$8000$]. Solutions for several choices of #math.equation(block: false, alt: "x")[$x$] are shown in figure (a). #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) A more efficient way to find all the solutions of the inequality is to start with the graph of the corresponding equation #math.equation(block: true, alt: "y equals minus x plus 10 , 000")[$y = − x + 10 , 000$] The graph is a straight line, as illustrated in figure (b). Observe that any point above this line has a y-coordinate greater than #math.equation(block: false, alt: "minus x plus 10 , 000")[$− x + 10 , 000$] and hence satisfies the inequality. Thus, the graph of the inequality includes all the points on or above the line #math.equation(block: false, alt: "y equals minus x plus 10 , 000")[$y = − x + 10 , 000$], as shown by the shaded region in figure (b). You can check that the shaded points are also solutions to the original inequality, #math.equation(block: false, alt: "x plus y greater than or equal to 10 , 000")[$x + y ≥ 10 , 000$]. Consider the point #math.equation(block: false, alt: "open parenthesis minus 1000 , 12 , 000 close parenthesis")[$( − 1000 , " " 12 , 000 )$], which lies in the shaded region above the line. This pair does satisfy #math.equation(block: false, alt: "x plus y greater than or equal to 10 , 000")[$x + y ≥ 10 , 000$], because #math.equation(block: true, alt: "minus 1000 plus 12 , 000 greater than or equal to 10 , 000")[$− 1000 + 12 , 000 ≥ 10 , 000$] (Ivana will expand her investments if her first hotel loses \$#math.equation(block: false, alt: "1000")[$1000$] and her second has a profit of \$#math.equation(block: false, alt: "12 , 000")[$12 , 000$].) On the other hand, the point #math.equation(block: false, alt: "open parenthesis 5000 , 4000 close parenthesis")[$( 5000 , 4000 )$] does not lie in the graph of #math.equation(block: false, alt: "x plus y greater than or equal to 10 , 000")[$x + y ≥ 10 , 000$], because the coordinates do not satisfy the inequality. === Linear Inequalities A #strong[linear inequality] can be written in the form #math.equation(block: true, alt: "a x plus b y plus c less than or equal to 0 or a x plus b y plus c greater than or equal to 0")[$a x + b y + c ≤ 0 " " " " " " " or " " " " " " " a x + b y + c ≥ 0$] The solutions consist of the line #math.equation(block: false, alt: "a x plus b y plus c equals 0")[$a x + b y + c = 0$] and a #strong[half-plane] on one side of that line. We shade the half-plane to show that all its points are included in the solution set. If the inequality is strict, then the graph includes only the half-plane and not the line. In that case, we use a dashed line for the graph of the equation #math.equation(block: false, alt: "a x plus b y plus c equals 0")[$a x + b y + c = 0$] to show that it is not part of the solution. To decide which side of the line to shade, we can solve the inequality for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. If we obtain #math.equation(block: true, alt: "y greater than or equal to m x plus b open parenthesis or y greater than m x plus b close parenthesis")[$y ≥ m x + b " " " " " " ( "or " " " " " " " y > m x + b )$] then we shade the half-plane #emph[above] the line. If the inequality is equivalent to #math.equation(block: true, alt: "y less than or equal to m x plus b open parenthesis or y less than m x plus b close parenthesis")[$y ≤ m x + b " " " " " " ( "or " " " " " " " y < m x + b )$] then we shade the half-plane #emph[below] the line. #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Be careful when isolating #math.equation(block: false, alt: "y")[$y$]: We must remember to reverse the direction of the inequality whenever we multiply or divide by a negative number. (See Algebra Skills Refresher Linear Equations and Inequalities if you would like to review solving inequalities.) ] #examplebox("Example 1")[][ Graph #math.equation(block: false, alt: "4 x minus 3 y greater than or equal to 12")[$" " " " 4 x − 3 y ≥ 12$] #solutionbox[ We solve the inequality for #math.equation(block: false, alt: "y")[$y$]. #math.equation(block: true, alt: "4 x minus 3 y, greater than or equal to 12, Subtract 4 x from both sides.; minus 3 y, greater than or equal to minus 4 x plus 12, Divide both sides by minus 3.; y, less than or equal to the fraction 4 over 3 x minus 4")[$4 x − 3 y & ≥ 12 & & "Subtract" " " 4 bold(italic(x)) " " "from both sides." \ − 3 y & ≥ − 4 x + 12 & & "Divide both sides by" " " − 3. \ y & ≤ frac(4, 3) x − 4$] We graph the corresponding line #math.equation(block: true, alt: "y equals the fraction 4 over 3 x minus 4")[$y = frac(4, 3) x − 4$] Note that the #math.equation(block: false, alt: "y")[$y$]-intercept is #math.equation(block: false, alt: "minus 4")[$− 4$] and the slope is #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$]. (See Linear Functions to review the slope-intercept method of graphing.) Finally, we shade the half-plane below the line. The completed graph is shown below. #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) ] ] + Find one #math.equation(block: false, alt: "y")[$y$]-value that satisfies the inequality #math.equation(block: false, alt: "y minus 3 x less than 6")[$y − 3 x < 6$] for each of the #math.equation(block: false, alt: "x")[$x$]-values in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]]), [#math.equation(block: false, alt: "y")[$y$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) + Graph the line #math.equation(block: false, alt: "y minus 3 x equals 6")[$y − 3 x = 6$]. Then plot your solutions from part (a) on the same grid. + Graph the solutions of the inequality #math.equation(block: false, alt: "y minus 3 x less than 6")[$y − 3 x < 6$]. #solutionbox[ + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], )) #linebreak() (Many answers are possible.) + A graph is below. + See graph. ] Graph for (b) and (c): #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) + Find one #math.equation(block: false, alt: "y")[$y$]-value that satisfies the inequality #math.equation(block: false, alt: "y minus 3 x less than 6")[$y − 3 x < 6$] for each of the #math.equation(block: false, alt: "x")[$x$]-values in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]]), [#math.equation(block: false, alt: "y")[$y$]], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], )) + Graph the line #math.equation(block: false, alt: "y minus 3 x equals 6")[$y − 3 x = 6$]. Then plot your solutions from part (a) on the same grid. + Graph the solutions of the inequality #math.equation(block: false, alt: "y minus 3 x less than 6")[$y − 3 x < 6$]. #solutionbox[ + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], )) + #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) + See graph. ] The solutions of #math.equation(block: false, alt: "x greater than minus 4")[$x > − 4$] all lie to which side of the boundary line? \_\_\_\_\_ #solutionbox[ To the right ] The solutions of #math.equation(block: false, alt: "x greater than minus 4")[$x > − 4$] all lie to which side of the boundary line? + Above + Below + To the left + To the right === Using a Test Point A second method for graphing inequalities does not require us to solve for #math.equation(block: false, alt: "y")[$y$]. Once we have graphed the boundary line, we can decide which half-plane to shade by using a test point. The test point can be any point that is not on the boundary line itself. The example's inequality 3x − 2y \< 6, with a movable test point. The shaded half-plane is the solution set, and the boundary line 3x − 2y = 6 is dashed because the inequality is strict — points on the line itself are not solutions. The two sliders move the red test point: drag it to (0, 0), substitute, and 0 \< 6 is true, so the point sits inside the shading; drag it to (5, 0) — the second test point the discussion tries — and 15 \< 6 is false, so the point sits outside. That is the whole method: one point off the line decides which side to shade. If you park the point exactly on the dashed line, at (2, 0) say, the substitution gives 6 \< 6, false — which is why a test point must never be chosen on the boundary.{"functions":\[{"color":"\#1f4e79","domain":\[-8,10\],"expression":"(3\*x - 6)/2","expression2":"","id":"halfplane","inequality":"gt","lineDash":\[\],"lineWidth":2.5,"mode":"inequality","tMax":0,"tMin":0,"variable":"x","visible":true},{"color":"\#b8442a","expression":"p + 0.15\*cos(t)","expression2":"q + 0.15\*sin(t)","id":"test-point","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-p","label":"Test point x: p","max":6,"min":-4,"name":"p","step":1,"value":5},{"id":"p-q","label":"Test point y: q","max":4,"min":-4,"name":"q","step":1,"value":0}\],"title":"3x \\u2212 2y \< 6 and a movable test point (p, q)","version":1,"viewport":{"centerX":1,"centerY":0,"scale":44}}#examplebox("Example 2")[][ Graph the solutions of the inequality #math.equation(block: false, alt: "3 x minus 2 y less than 6")[$" " " " 3 x − 2 y < 6$] #solutionbox[ First, we graph the line #math.equation(block: false, alt: "3 x minus 2 y equals 6")[$3 x − 2 y = 6$], as shown below. We will use the intercept method. The intercepts are #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 0 , minus 3 close parenthesis")[$( 0 , − 3 )$], so we sketch the boundary line through those points. Next, we choose a test point. Because #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] does not lie on the line, we choose it as our test point. We substitute the coordinates of the test point into the inequality to obtain #math.equation(block: true, alt: "3 open parenthesis 0 close parenthesis minus 2 open parenthesis 0 close parenthesis less than 6")[$3 ( 0 ) − 2 ( 0 ) < 6$] Because this is a true statement, #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] is a solution of the inequality. Since all the solutions lie on the same side of the boundary line, we shade the half-plane that contains the test point. In this example, the boundary line is a dashed line because the original inequality was strict. #figure(figph[strict linear inequality in two variables], alt: "strict linear inequality in two variables", caption: none) ] ] We can choose #emph[any] point for the test point as long as it does not lie on the boundary line. We chose #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] in Example because the coordinates are easy to substitute into the inequality. If the test point #emph[is] a solution to the inequality, then the half-plane including that point should be shaded. If the test point is #emph[not] a solution to the inequality, then the #emph[other] half-plane should be shaded. For example, suppose we had chosen #math.equation(block: false, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$] as the test point in Example. When we substitute its coordinates into the inequality, we find #math.equation(block: true, alt: "3 open parenthesis 5 close parenthesis minus 2 open parenthesis 0 close parenthesis less than 6")[$3 ( 5 ) − 2 ( 0 ) < 6$] which is a #emph[false] statement. This tells us that #math.equation(block: false, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$] is not a solution to the inequality, so the solutions must lie on the other side of the boundary line. Using #math.equation(block: false, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$] as the test point gives us the same solutions we found in Example. Which point would not be a good test point for the inequality #math.equation(block: false, alt: "y less than or equal to the fraction 2 x over 3")[$y ≤ display(frac(2 x, 3))$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] would not be a good test point because it lies on the boundary line. ] Which point would not be a good test point for the inequality #math.equation(block: false, alt: "y less than or equal to the fraction 2 x over 3")[$y ≤ display(frac(2 x, 3))$] ? + #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] + #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$] + #math.equation(block: false, alt: "open parenthesis 2 , 3 close parenthesis")[$( 2 , 3 )$] + #math.equation(block: false, alt: "open parenthesis minus 1 , minus 1 close parenthesis")[$( − 1 , − 1 )$] Here is a summary of our test point method for graphing inequalities. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Graph an Inequality Using a Test Point:] + Graph the corresponding equation to obtain the boundary line. + Choose a test point that does not lie on the boundary line. + Substitute the coordinates of the test point into the inequality. + If the resulting statement is true, shade the half-plane that includes the test point. + If the resulting statement is false, shade the half-plane that does not include the test point. + If the inequality is strict, make the boundary line a dashed line. ] Graph the solutions of the inequality #math.equation(block: false, alt: "y greater than the fraction minus 3 over 2 x")[$y > display(frac(− 3, 2)) x$] + Graph the line #math.equation(block: false, alt: "y equals the fraction minus 3 over 2 x")[$y = display(frac(− 3, 2)) x$]. (Use the slope-intercept method.) + Choose a test point. (Do not choose #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]!) + Decide which side of the line to shade. \_\_\_\_\_ + Should the boundary line be dashed or solid? \_\_\_\_\_ #solutionbox[ + A graph is below. + Choose any point not on the boundary line #math.equation(block: false, alt: "y equals the fraction minus 3 over 2 x")[$y = display(frac(− 3, 2)) x$]. + Shade above the boundary line. + Use a dashed boundary line because the inequality is strict. ] Graph of inequality: #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) Graph the solutions of the inequality #math.equation(block: false, alt: "y greater than the fraction minus 3 over 2 x")[$y > display(frac(− 3, 2)) x$] + Graph the line #math.equation(block: false, alt: "y equals the fraction minus 3 over 2 x")[$y = display(frac(− 3, 2)) x$]. (Use the slope-intercept method.) + Choose a test point. (Do not choose #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]!) + Decide which side of the line to shade. + Should the boundary line be dashed or solid? #solutionbox[ #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) ] Recall that the equation of a vertical line has the form #math.equation(block: true, alt: "x equals k")[$x = k$] where #math.equation(block: false, alt: "k")[$k$] is a constant, and a horizontal line has an equation of the form #math.equation(block: true, alt: "y equals k")[$y = k$] Similarly, the inequality #math.equation(block: false, alt: "x greater than or equal to k")[$x ≥ k$] may represent the inequality in two variables #math.equation(block: true, alt: "x plus 0 y greater than or equal to k")[$x + 0 y ≥ k$] Its graph is then a region in the plane. #examplebox("Example 3")[][ Graph #math.equation(block: false, alt: "x greater than or equal to 2")[$" " " " x ≥ 2 " " " "$] in the plane. #solutionbox[ First, we graph the equation #math.equation(block: false, alt: "x equals 2")[$x = 2$]; its graph is a vertical line. Because the origin does not lie on this line, we can use it as a test point. Substitute #math.equation(block: false, alt: "0")[$0$] for #math.equation(block: false, alt: "x")[$x$] (there is no #math.equation(block: false, alt: "y")[$y$]) into the inequality to obtain #math.equation(block: true, alt: "0 greater than or equal to 2")[$0 ≥ 2$] Since this is a false statement, we shade the half-plane that does not contain the origin. We see in the figure below that the graph of the inequality contains all points whose #math.equation(block: false, alt: "x")[$x$]-coordinates are greater than or equal to #math.equation(block: false, alt: "2")[$2$]. #figure(figph[graph of x greater or equal to 2 in the plane], alt: "graph of x greater or equal to 2 in the plane", caption: none) ] ] Graph #math.equation(block: false, alt: "minus 2 less than or equal to y less than 3")[$" " " " − 2 ≤ y < 3 " " " "$] in the plane. #solutionbox[ #figure(figph[linear compound inequality in two variables], alt: "linear compound inequality in two variables", caption: none) ] Graph #math.equation(block: false, alt: "minus 2 less than or equal to y less than 3")[$" " " " − 2 ≤ y < 3 " " " "$] in the plane. #solutionbox[ #figure(figph[linear compound inequality in two variables], alt: "linear compound inequality in two variables", caption: none) ] === Systems of Inequalities Some applications are best described by a system of two or more inequalities. The solutions to a system of inequalities include all points that are solutions to each inequality in the system. The graph of the system is the intersection of the shaded regions for each inequality in the system. For example, the figure at right shows the solutions of the system #math.equation(block: true, alt: "y greater than x and y greater than 2")[$y > x " " " " " " " and " " " " " " " y > 2$] #figure(figph[system of two inequalities], alt: "system of two inequalities", caption: none) #examplebox("Example 4")[][ Laura is a finicky eater, and she dislikes most foods that are high in calcium. Her morning cereal satisfies some of her calcium requirements, but she needs an additional 500 milligrams of calcium, which she will get from a combination of broccoli, at 160 milligrams per serving, and zucchini, at 30 milligrams per serving. Draw a graph representing the possible combinations of broccoli and zucchini that fulfill Laura’s calcium requirements. #solutionbox[ + Number of servings of broccoli: #math.equation(block: false, alt: "x")[$" " " " x$] #linebreak() Number of servings of zucchini: #math.equation(block: false, alt: "y")[$" " " " y$] + To consume at least #math.equation(block: false, alt: "500")[$500$] milligrams of calcium, Laura must choose #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] so that #math.equation(block: true, alt: "160 x plus 30 y greater than or equal to 500")[$160 x + 30 y ≥ 500$] It makes no sense to consider negative values of #math.equation(block: false, alt: "x")[$x$] or of #math.equation(block: false, alt: "y")[$y$], since Laura cannot eat a negative number of servings. Thus, we have two more inequalities to satisfy: #math.equation(block: true, alt: "x greater than or equal to 0 and y greater than or equal to 0")[$x ≥ 0 " " " " " " " and " " " " " " " y ≥ 0$] + We graph all three inequalities on the same axes. The inequalities #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$] and #math.equation(block: false, alt: "y greater than or equal to 0")[$y ≥ 0$] restrict the solutions to lie in the first quadrant. The solutions common to all three inequalities are shown below. #figure(figph[system of three inequalities], alt: "system of three inequalities", caption: none) + Laura can choose any combination of broccoli and zucchini represented by points in the shaded region. For example, the point #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] is a solution to the system of inequalities, so Laura could choose to eat #math.equation(block: false, alt: "3")[$3$] servings of broccoli and #math.equation(block: false, alt: "1")[$1$] serving of zucchini. ] ] Use the following steps to graph the solutions of the system #math.equation(block: true, alt: "x plus y, less than or equal to 12; 3 x minus 4 y, less than or equal to 8")[$x + y & ≤ 12 \ 3 x − 4 y & ≤ 8$] + Graph the boundary line #math.equation(block: false, alt: "x plus y equals 12")[$x + y = 12$]. + Lightly shade the solutions of the inequality #math.equation(block: false, alt: "x plus y less than or equal to 12")[$x + y ≤ 12$]. + Graph the boundary line #math.equation(block: false, alt: "3 x minus 4 y equals 8")[$3 x − 4 y = 8$]. + Lightly shade the solutions of #math.equation(block: false, alt: "3 x minus 4 y less than or equal to 8")[$3 x − 4 y ≤ 8$]. + Shade the intersection of the two solutions sets. #solutionbox[ #figure(figph[linear system of inequalities in two variables], alt: "linear system of inequalities in two variables", caption: none) ] Use the following steps to graph the solutions of the system #math.equation(block: true, alt: "x plus y, less than or equal to 12; 3 x minus 4 y, less than or equal to 8")[$x + y & ≤ 12 \ 3 x − 4 y & ≤ 8$] + Graph the boundary line #math.equation(block: false, alt: "x plus y equals 12")[$x + y = 12$]. + Lightly shade the solutions of the inequality #math.equation(block: false, alt: "x plus y less than or equal to 12")[$x + y ≤ 12$]. + Graph the boundary line #math.equation(block: false, alt: "3 x minus 4 y equals 8")[$3 x − 4 y = 8$]. + Lightly shade the solutions of #math.equation(block: false, alt: "3 x minus 4 y less than or equal to 8")[$3 x − 4 y ≤ 8$]. + Shade the intersection of the two solutions sets. #solutionbox[ #figure(figph[linear system of inequalities in two variables], alt: "linear system of inequalities in two variables", caption: none) ] Why do we shade the intersection of the solution sets when solving a system of inequalities? \_\_\_\_\_ Why do we shade the intersection of the solution sets when solving a system of inequalities? To describe the solutions of a system of inequalities, it is useful to locate the #strong[vertices], or corner points, of the boundary. #examplebox("Example 5")[][ Graph the solution set of the system below and find the coordinates of its vertices. #math.equation(block: true, alt: "x minus y minus 2, less than or equal to 0; x plus 2 y minus 6, less than or equal to 0; x greater than or equal to 0 , y, less than or equal to 0")[$x − y − 2 & ≤ 0 \ x + 2 y − 6 & ≤ 0 \ x ≥ 0 , " " " " " " " " " " y & ≤ 0$] #solutionbox[ The last two inequalities, #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$] and #math.equation(block: false, alt: "y greater than or equal to 0")[$y ≥ 0$], restrict the solutions to the first quadrant. First, we graph the line #math.equation(block: false, alt: "x minus y minus 2 equals 0")[$x − y − 2 = 0$] and use the test point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] to shade the half-plane, including the origin. Then we graph the line #math.equation(block: false, alt: "x minus 2 y minus 6 equals 0")[$x − 2 y − 6 = 0$] and again use the test point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] to shade the half-plane below the line. The intersection of the shaded regions is shown below. #figure(figph[system of four inequalities], alt: "system of four inequalities", caption: none) To find the coordinates of the vertices #math.equation(block: false, alt: "A")[$A$], #math.equation(block: false, alt: "B")[$B$], #math.equation(block: false, alt: "C")[$C$], and #math.equation(block: false, alt: "D")[$D$], we solve simultaneously the equations of the two lines that intersect at each vertex. Thus, for #math.equation(block: false, alt: "A")[$A$], we solve the system #math.equation(block: true, alt: "x, equals 0; y, equals 0")[$x & = 0 \ y & = 0$] to find #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] for #math.equation(block: false, alt: "B")[$B$], we solve the system #math.equation(block: true, alt: "x, equals 0; x plus 2 y, equals 6")[$x & = 0 \ x + 2 y & = 6$] to find #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$] for #math.equation(block: false, alt: "C")[$C$], we solve the system #math.equation(block: true, alt: "x plus 2 y, equals 6; x minus y, equals 2")[$x + 2 y & = 6 \ x − y & = 2$] to find #math.equation(block: false, alt: "open parenthesis the fraction 10 over 3 , the fraction 4 over 3 close parenthesis")[$( display(frac(10, 3)) , display(frac(4, 3)) )$] for #math.equation(block: false, alt: "D")[$D$], we solve the system #math.equation(block: true, alt: "y, equals 0; x minus y, equals 2")[$y & = 0 \ x − y & = 2$] to find #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] The vertices are the points #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$], #math.equation(block: false, alt: "open parenthesis the fraction 10 over 3 , the fraction 4 over 3 close parenthesis")[$( display(frac(10, 3)) , display(frac(4, 3)) )$], and #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$]. ] ] The first quadrant includes all the solutions of \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "x greater than 0 , y greater than 0")[$x > 0 , y > 0$] ] The first quadrant includes all the solutions of + #math.equation(block: false, alt: "x plus y greater than 0")[$x + y > 0$] + #math.equation(block: false, alt: "x minus y greater than 0")[$x − y > 0$] + #math.equation(block: false, alt: "x y greater than 0")[$x y > 0$] + #math.equation(block: false, alt: "x greater than 0 , y greater than 0")[$x > 0 , y > 0$] + Graph the system of inequalities #math.equation(block: true, alt: "5 x plus 4 y, less than 40; minus 3 x plus 4 y, less than 12; x less than 6 , y, greater than 2")[$5 x + 4 y & < 40 \ − 3 x + 4 y & < 12 \ x < 6 , #hide($00$) y & > 2$] + Find the coordinates of the vertices of the solution set. #linebreak() \_\_\_\_\_ Separate different ordered pairs with a comma. #solutionbox[ + See graph below. + #math.equation(block: false, alt: "A open parenthesis the fraction minus 4 over 3 , 2 close parenthesis , B open parenthesis the fraction 7 over 2 , the fraction 45 over 8 close parenthesis , C open parenthesis 6 , the fraction 5 over 2 close parenthesis , D open parenthesis 6 , 2 close parenthesis")[$A ( display(frac(− 4, 3)) , 2 ) , B ( display(frac(7, 2)) , display(frac(45, 8)) ) , C ( 6 , display(frac(5, 2)) ) , D ( 6 , 2 )$] ] #figure(figph[system of linear inequalities in two variables], alt: "system of linear inequalities in two variables", caption: none) + Graph the system of inequalities #math.equation(block: true, alt: "5 x plus 4 y, less than 40; minus 3 x plus 4 y, less than 12; x less than 6 , y, greater than 2")[$5 x + 4 y & < 40 \ − 3 x + 4 y & < 12 \ x < 6 , #hide($00$) y & > 2$] + Find the coordinates of the vertices of the solution set. #solutionbox[ + #figure(figph[system of linear inequalities in two variables], alt: "system of linear inequalities in two variables", caption: none) + #math.equation(block: false, alt: "A open parenthesis the fraction minus 4 over 3 , 2 close parenthesis , B open parenthesis the fraction 7 over 2 , the fraction 45 over 8 close parenthesis , C open parenthesis 6 , the fraction 5 over 2 close parenthesis , D open parenthesis 6 , 2 close parenthesis")[$A ( display(frac(− 4, 3)) , 2 ) , B ( display(frac(7, 2)) , display(frac(45, 8)) ) , C ( 6 , display(frac(5, 2)) ) , D ( 6 , 2 )$] ] To find the vertices of the solution set for a system of inequalities, we \_\_\_\_\_ #solutionbox[ solve a system of equations. ] To find the vertices of the solution set for a system of inequalities, we + use a test point. + use the vertex formula. + solve a system of equations. + find the #math.equation(block: false, alt: "x")[$x$]-intercepts. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Half-plane - Test point - Vertices ==== CONCEPTS + The solutions of a linear inequality in two variables consist of a #strong[half-plane] on one side of the line. The line itself is not included if the inequality is strict. + Once we have graphed the boundary line, we can decide which half-plane to shade by using a #strong[test point]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Graph an Inequality Using a Test Point:] + Graph the corresponding equation to obtain the boundary line. + Choose a test point that does not lie on the boundary line. + Substitute the coordinates of the test point into the inequality. + If the resulting statement is true, shade the half-plane that includes the test point. + If the resulting statement is false, shade the half-plane that does not include the test point. + If the inequality is strict, make the boundary line a dashed line. ] + The solutions to a system of inequalities include all points that are solutions to each inequality in the system. The graph of the system is the intersection of the shaded regions for each inequality in the system. + To describe the solutions of a system of inequalities, it is useful to locate the #strong[vertices], or corner points, of the boundary. ==== STUDY QUESTIONS + The solutions of a linear inequality in two variables form what sort of set? + How can you find the boundary of the solution set? + If your test point is not a solution of the inequality, which side of the line should you shade? + How can you find the vertices of the solution set of a system of inequalities? ==== SKILLS Practice each skill in the Homework problems listed. + Graph the solutions of a linear inequality in two variables: \#1–16 + Graph the solutions of a system of inequalities: \#17–36 + Solve problems using inequalities: \#37–42 === Homework 8.4 For Problems 1–16, graph the inequality. #math.equation(block: true, alt: "y greater than 2 x plus 4")[$y > 2 x + 4$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "y less than 9 minus 3 x")[$y < 9 − 3 x$] #math.equation(block: true, alt: "3 x minus 2 y less than or equal to 12")[$3 x − 2 y ≤ 12$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "2 x plus 5 y greater than or equal to 10")[$2 x + 5 y ≥ 10$] #math.equation(block: true, alt: "x plus 4 y greater than or equal to minus 6")[$x + 4 y ≥ − 6$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "3 x minus y less than or equal to minus 2")[$3 x − y ≤ − 2$] #math.equation(block: true, alt: "x greater than minus 3 y plus 1")[$x > − 3 y + 1$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "x greater than 2 y minus 5")[$x > 2 y − 5$] #math.equation(block: true, alt: "x greater than or equal to minus 3")[$x ≥ − 3$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "y less than 4")[$y < 4$] #math.equation(block: true, alt: "y less than the fraction 1 over 2 x")[$y < display(frac(1, 2)) x$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "y greater than the fraction 4 over 3 x")[$y > display(frac(4, 3)) x$] #math.equation(block: true, alt: "0 greater than or equal to x minus y")[$0 ≥ x − y$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "0 greater than or equal to x plus 3 y")[$0 ≥ x + 3 y$] #math.equation(block: true, alt: "minus 1 less than y less than or equal to 4")[$− 1 < y ≤ 4$] #figure(figph[linear inequality in two variables], alt: "linear inequality in two variables", caption: none) #math.equation(block: true, alt: "minus 2 less than or equal to y less than 0")[$− 2 ≤ y < 0$] For Problems 17–6, graph the system of inequalities. #math.equation(block: true, alt: "y, greater than 2; x, greater than or equal to minus 2")[$y & > 2 \ x & ≥ − 2$] #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) #math.equation(block: true, alt: "y, less than or equal to minus 1; x, greater than 2")[$y & ≤ − 1 \ x & > 2$] #math.equation(block: true, alt: "y, less than x; y, greater than or equal to minus 3")[$y & < x \ y & ≥ − 3$] #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) #math.equation(block: true, alt: "y, greater than or equal to minus x; y, less than 2")[$y & ≥ − x \ y & < 2$] #math.equation(block: true, alt: "x plus y, less than or equal to 6; x plus y, greater than or equal to 4")[$x + y & ≤ 6 \ x + y & ≥ 4$] #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) #math.equation(block: true, alt: "x minus y, less than 3; x minus y, greater than minus 2")[$x − y & < 3 \ x − y & > − 2$] #math.equation(block: true, alt: "2 x minus y, less than or equal to 4; x plus 2 y, greater than 6")[$2 x − y & ≤ 4 \ x + 2 y & > 6$] #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) #math.equation(block: true, alt: "2 y minus x, less than 2; x plus y, less than or equal to 4")[$2 y − x & < 2 \ x + y & ≤ 4$] #math.equation(block: true, alt: "3 y minus 2 x less than 2; y greater than x minus 1")[$3 y − 2 x < 2 \ y > x − 1$] #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) #math.equation(block: true, alt: "2 x plus y less than 4; y less than or equal to 1 minus x")[$2 x + y < 4 \ y ≤ 1 − x$] For Problems 27–36, graph the system of inequalities and find the coordinates of the vertices. #math.equation(block: true, alt: "2 x plus 3 y minus 6 less than 0; x greater than or equal to 0 , y greater than or equal to 0")[$2 x + 3 y − 6 < 0 \ x ≥ 0 , " " y ≥ 0$] #figure(figph[system of inequalities, with vertices], alt: "system of inequalities, with vertices", caption: none) #math.equation(block: true, alt: "3 x plus 2 y less than 6; x greater than or equal to 0 , y greater than or equal to 0")[$3 x + 2 y < 6 \ x ≥ 0 , " " y ≥ 0$] #math.equation(block: true, alt: "5 y minus 3 x less than or equal to 15; x plus y less than or equal to 11; x greater than or equal to 0 , y greater than or equal to 0")[$5 y − 3 x ≤ 15 \ x + y ≤ 11 \ x ≥ 0 , " " y ≥ 0$] #figure(figph[system of inequalities, with vertices], alt: "system of inequalities, with vertices", caption: none) #math.equation(block: true, alt: "y minus 2 x greater than or equal to minus 4; x plus y less than or equal to 5; x greater than or equal to 0 , y greater than or equal to 0")[$y − 2 x ≥ − 4 \ x + y ≤ 5 \ x ≥ 0 , " " y ≥ 0$] #math.equation(block: true, alt: "2 y less than or equal to x; 2 x less than or equal to y plus 12; x greater than or equal to 0 , y greater than or equal to 0")[$2 y ≤ x #hide($0000$) \ 2 x ≤ y + 12 \ x ≥ 0 , " " y ≥ 0$] #figure(figph[system of inequalities, with vertices], alt: "system of inequalities, with vertices", caption: none) #math.equation(block: true, alt: "y greater than or equal to 3 x; 2 y plus x less than or equal to 14; x greater than or equal to 0 , y greater than or equal to 0")[$y ≥ 3 x #hide($0000$) \ 2 y + x ≤ 14 \ x ≥ 0 , " " y ≥ 0$] #math.equation(block: true, alt: "x plus y greater than or equal to 3; 2 y less than or equal to x plus 8; 2 y plus 3 x less than or equal to 24; x greater than or equal to 0 , y greater than or equal to 0")[$x + y ≥ 3 #hide($000$) \ 2 y ≤ x + 8 #hide($00$) \ 2 y + 3 x ≤ 24 \ x ≥ 0 , " " y ≥ 0$] #figure(figph[system of inequalities, with vertices], alt: "system of inequalities, with vertices", caption: none) #math.equation(block: true, alt: "2 y plus 3 x greater than or equal to 6; 2 y plus x less than or equal to 10; y greater than or equal to 3 x minus 9; x greater than or equal to 0 , y greater than or equal to 0")[$2 y + 3 x ≥ 6 \ 2 y + x ≤ 10 \ y ≥ 3 x − 9 \ x ≥ 0 , " " y ≥ 0$] #math.equation(block: true, alt: "3 y minus x greater than or equal to 3; y minus 4 x greater than or equal to minus 10; y minus 2 less than or equal to x; x greater than or equal to 0 , y greater than or equal to 0")[$3 y − x ≥ 3 #hide($00$) \ y − 4 x ≥ − 10 \ y − 2 ≤ x #hide($000$) \ x ≥ 0 , " " y ≥ 0$] #figure(figph[system of inequalities, with vertices], alt: "system of inequalities, with vertices", caption: none) #math.equation(block: true, alt: "2 y plus x less than or equal to 6; 4 y less than or equal to 2 x plus 8; x less than or equal to 4 y plus 4; x greater than or equal to 0 , y greater than or equal to 0")[$2 y + x ≤ 6 \ 4 y ≤ 2 x + 8 \ x ≤ 4 y + 4 " " \ x ≥ 0 , " " y ≥ 0$] For Problems 37–42, graph the set of solutions to the problem. Two of the inequalities in each system are #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$] and #math.equation(block: false, alt: "y greater than or equal to 0")[$y ≥ 0$]. The math club is selling tickets for a show by a mathemagician. Student tickets will cost \$#math.equation(block: false, alt: "1")[$1$] and faculty tickets will cost \$#math.equation(block: false, alt: "2")[$2$]. The ticket receipts must be at least \$#math.equation(block: false, alt: "250")[$250$] to cover the fee for the performer. Write a system of three inequalities for the number of student tickets and the number of faculty tickets that must be sold, and graph the solutions. #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) The math department is having a book sale of old textbooks to raise at least \$#math.equation(block: false, alt: "300")[$300$] for scholarships. Paperback textbooks will cost \$#math.equation(block: false, alt: "2")[$2$] and the hardcover textbooks will cost \$#math.equation(block: false, alt: "5")[$5$]. Write a system of three inequalities for the number of paperback and hardback textbooks that must be sold, and graph the solutions. Vassilis plans to invest at most \$#math.equation(block: false, alt: "10 , 000")[$10 , 000$] in two banks. One bank pays #math.equation(block: false, alt: "6 %")[$6 upright(%)$] annual interest and the other pays #math.equation(block: false, alt: "5 %")[$5 upright(%)$] annual interest. Vassilis wants at least \$#math.equation(block: false, alt: "540")[$540$] total annual interest from his two investments. Write a system of four inequalities for the amount Vassilis can invest in the two accounts, and graph the system. #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) Jeannette has #math.equation(block: false, alt: "180")[$180$] acres of farmland for growing wheat or soy. She can get a profit of \$#math.equation(block: false, alt: "36")[$36$] per acre for wheat and \$#math.equation(block: false, alt: "24")[$24$] per acre for soy. She wants to have a profit of at least \$#math.equation(block: false, alt: "5400")[$5400$] from her crops. Write a system of four inequalities for the number of acres she can use for each crop, and graph the solutions. Gary's pancake recipe includes corn meal and whole wheat flour. Corn meal has #math.equation(block: false, alt: "2.4")[$2.4$] grams of linoleic acid and #math.equation(block: false, alt: "2.5")[$2.5$] milligrams of niacin per cup. Whole wheat flour has #math.equation(block: false, alt: "0.8")[$0.8$] gram of linoleic acid and #math.equation(block: false, alt: "5")[$5$] milligrams of niacin per cup. These two ingredients should not exceed #math.equation(block: false, alt: "3")[$3$] cups total. The mixture should provide at least #math.equation(block: false, alt: "3.2")[$3.2$] grams of linoleic acid and at least #math.equation(block: false, alt: "10")[$10$] milligrams of niacin. Write a system of five inequalities for the amount of corn meal and the amount of whole wheat flour Gary can use, then graph the solutions. #figure(figph[system of inequalities], alt: "system of inequalities", caption: none) Cho and his brother go into business making comic book costumes. They need #math.equation(block: false, alt: "1")[$1$] hour of cutting and #math.equation(block: false, alt: "2")[$2$] hours of sewing to make a Batman costume. They need #math.equation(block: false, alt: "2")[$2$] hours of cutting and #math.equation(block: false, alt: "1")[$1$] hour of sewing to make a Wonder Woman costume. They have available at most #math.equation(block: false, alt: "10")[$10$] hours per day for cutting and at most #math.equation(block: false, alt: "8")[$8$] hours per day for sewing. They must make at least one costume each day to stay in business. Write a system of five inequalities for the number of each type of costume Cho can make, then graph the solutions.