#set document(title: "4.7 Graphing Systems of 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")) == 4.7#h(0.6em)Graphing Systems of Linear Inequalities #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Solve the inequality #math.equation(block: false, alt: "2 a less than 5 a plus 12 .")[$2 a < 5 a + 12 .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "minus 4 less than a")[$− 4 < a$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine whether the ordered pair #math.equation(block: false, alt: "open parenthesis 3 , the fraction 1 over 2 close parenthesis")[$( 3 , frac(1, 2) )$] is a solution to the system #math.equation(block: false, alt: "y greater than 2 x plus 3 .")[$y > 2 x + 3 .$] #linebreak() If you missed this problem, review . #solutionbox[ no ] ] === Determine whether an ordered pair is a solution of a system of linear inequalities The definition of a #strong[system of linear inequalities] is very similar to the definition of a system of linear equations. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[System of Linear Inequalities] Two or more linear inequalities grouped together form a system of linear inequalities. ] A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here. #math.equation(block: true, alt: "{ x plus 4 y greater than or equal to 10; 3 x minus 2 y less than 12")[$\{ x + 4 y ≥ 10 \ 3 x − 2 y < 12$]To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] that make both inequalities true. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solutions of a System of Linear Inequalities] Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the #emph[x, y] coordinate system that includes all the points whose ordered pairs make the inequalities true. ] To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system. #examplebox("Example 1")[][ Determine whether the ordered pair is a solution to the system #math.equation(block: false, alt: "{ x plus 4 y greater than or equal to 10; 3 x minus 2 y less than 12 .")[$\{ x + 4 y ≥ 10 \ 3 x − 2 y < 12 .$] ⓐ #math.equation(block: false, alt: "open parenthesis −2 , 4 close parenthesis")[$( −2 , 4 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] #solutionbox[ ⓐ Is the ordered pair #math.equation(block: false, alt: "open parenthesis −2 , 4 close parenthesis")[$( −2 , 4 )$] a solution? #figure(figph[We substitute x equal to negative 2 and y equal to 4 into both inequalities. First inequality is x plus 4 times y greater than or equal to 10. So negative 2 plus 4 open parentheses 4 close parenthesis is greater than or equal to 10 or not. 14 is greater than or equal to 10 is true. Second inequality, 3 times x minus 2 times y is less than 12. Three open parentheses negative 2 close parentheses minus two open parentheses 4 close parentheses is less than 12 or not. Negative 14 is less than 12 is true.], alt: "We substitute x equal to negative 2 and y equal to 4 into both inequalities. First inequality is x plus 4 times y greater than or equal to 10. So negative 2 plus 4 open parentheses 4 close parenthesis is greater than or equal to 10 or not. 14 is greater than or equal to 10 is true. Second inequality, 3 times x minus 2 times y is less than 12. Three open parentheses negative 2 close parentheses minus two open parentheses 4 close parentheses is less than 12 or not. Negative 14 is less than 12 is true.", caption: none) The ordered pair #math.equation(block: false, alt: "open parenthesis −2 , 4 close parenthesis")[$( −2 , 4 )$] made both inequalities true. Therefore #math.equation(block: false, alt: "open parenthesis −2 , 4 close parenthesis")[$( −2 , 4 )$] is a solution to this system. ⓑ Is the ordered pair #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] a solution? #figure(figph[We substitute x equal to three and y equal to one into both inequalities. First inequality is x plus four times y greater than or equal to ten. So three plus four open parentheses one close parenthesis is greater than or equal to ten or not. Seven greater than or equal to ten is false. Second inequality, three times x minus two times y is less than twelve. Three open parentheses three close parentheses minus two open parentheses one close parentheses is less than twelve or not. Seven less than 12 holds true.], alt: "We substitute x equal to three and y equal to one into both inequalities. First inequality is x plus four times y greater than or equal to ten. So three plus four open parentheses one close parenthesis is greater than or equal to ten or not. Seven greater than or equal to ten is false. Second inequality, three times x minus two times y is less than twelve. Three open parentheses three close parentheses minus two open parentheses one close parentheses is less than twelve or not. Seven less than 12 holds true.", caption: none) The ordered pair #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] made one inequality true, but the other one false. Therefore #math.equation(block: false, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] is not a solution to this system. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine whether the ordered pair is a solution to the system: #math.equation(block: false, alt: "{ x minus 5 y greater than 10; 2 x plus 3 y greater than −2 .")[$\{ x − 5 y > 10 \ 2 x + 3 y > −2 .$] ⓐ #math.equation(block: false, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 6 , −3 close parenthesis")[$( 6 , −3 )$] #solutionbox[ ⓐ no ⓑ yes ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine whether the ordered pair is a solution to the system: #math.equation(block: false, alt: "{ y greater than 4 x minus 2; 4 x minus y less than 20 .")[$\{ y > 4 x − 2 \ 4 x − y < 20 .$] ⓐ #math.equation(block: false, alt: "open parenthesis −2 , 1 close parenthesis")[$( −2 , 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 4 , −1 close parenthesis")[$( 4 , −1 )$] #solutionbox[ ⓐ yes ⓑ no ] ] === Solve a System of Linear Inequalities by Graphing The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph. #examplebox("Example 2")[How to Solve a System of Linear Inequalities by Graphing][ Solve the system by graphing: #math.equation(block: false, alt: "{ y greater than or equal to 2 x minus 1; y less than x plus 1 .")[$\{ y ≥ 2 x − 1 \ y < x + 1 .$] #solutionbox[ #figure(figph[Step 1. Graph the first inequality. We graph y less than 2x minus 1. Graph the boundary line y equal to 2x minus 1. It is a solid line because the inequality sign is less than. Shade in the side of the boundary line where the inequality is true. We choose 0, 0 as a test point. It is a solution to the equation, so we shade in above the boundary line.], alt: "Step 1. Graph the first inequality. We graph y less than 2x minus 1. Graph the boundary line y equal to 2x minus 1. It is a solid line because the inequality sign is less than. Shade in the side of the boundary line where the inequality is true. We choose 0, 0 as a test point. It is a solution to the equation, so we shade in above the boundary line.", caption: none) #figure(figph[Step 2. On the same grid, graph the second inequality y less than x plus 1. Graph the boundary line y equal to x plus 1. It is a dashed line because the inequality sign is less than. Shade in the side of that boundary line where the inequality is true. Again, we use 0, 0 as a test point. It is a solution so we shade below the line y equals x plus 1.], alt: "Step 2. On the same grid, graph the second inequality y less than x plus 1. Graph the boundary line y equal to x plus 1. It is a dashed line because the inequality sign is less than. Shade in the side of that boundary line where the inequality is true. Again, we use 0, 0 as a test point. It is a solution so we shade below the line y equals x plus 1.", caption: none) #figure(figph[Step 3. The solution is the region where the shading overlaps. The point where the boundary lines intersect is not a solution because it is not a solution to y less than x plus 1. The solution is all points in the area bounded by the lines on the bottom left.], alt: "Step 3. The solution is the region where the shading overlaps. The point where the boundary lines intersect is not a solution because it is not a solution to y less than x plus 1. The solution is all points in the area bounded by the lines on the bottom left.", caption: none) #figure(figph[Step 4. Check by choosing a test point. We use minus 1, minus 1. Substituting in the inequality y less than 2x minus 1, we get minus 1 less than minus 3 which is true. Hence, it is a solution. Similarly, it is also true for the other inequality. The region containing minus 1, minus 1 is the solution to this system.], alt: "Step 4. Check by choosing a test point. We use minus 1, minus 1. Substituting in the inequality y less than 2x minus 1, we get minus 1 less than minus 3 which is true. Hence, it is a solution. Similarly, it is also true for the other inequality. The region containing minus 1, minus 1 is the solution to this system.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ y less than 3 x plus 2; y greater than − x minus 1 .")[$\{ y < 3 x + 2 \ y > "−" x − 1 .$] #solutionbox[ #figure(figph[The figure shows a graph plotted for the inequalities y less than three times x plus two and y greater than minus x minus one. Two lines intersect each other on the graph. An area to the right of both the lines is colored in grey. It is the solution.], alt: "The figure shows a graph plotted for the inequalities y less than three times x plus two and y greater than minus x minus one. Two lines intersect each other on the graph. An area to the right of both the lines is colored in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ y less than minus the fraction 1 over 2 x plus 3; y less than 3 x minus 4 .")[$\{ y < − frac(1, 2) x + 3 \ y < 3 x − 4 .$] #solutionbox[ #figure(figph[The figure shows the graph plotted for the inequalities y less than minus half of x plus three and y less than three times x minus four. Two intersecting lines are shown on the graph. The area bound by the two lines to the bottom right is shown in grey. It is the solution.], alt: "The figure shows the graph plotted for the inequalities y less than minus half of x plus three and y less than three times x minus four. Two intersecting lines are shown on the graph. The area bound by the two lines to the bottom right is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve a system of linear inequalities by graphing.] + Graph the first inequality. - Graph the boundary line. - Shade in the side of the boundary line where the inequality is true. + On the same grid, graph the second inequality. - Graph the boundary line. - Shade in the side of that boundary line where the inequality is true. + The solution is the region where the shading overlaps. + Check by choosing a test point. ] #examplebox("Example 3")[][ Solve the system by graphing: #math.equation(block: false, alt: "{ x minus y greater than 3; y less than minus the fraction 1 over 5 x plus 4 .")[$\{ x − y > 3 \ y < − frac(1, 5) x + 4 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "{ x minus y greater than 3; y less than minus the fraction 1 over 5 x plus 4")[$\{ x − y > 3 \ y < − frac(1, 5) x + 4$]]), [Graph #emph[x] − #emph[y] \> 3, by graphing #emph[x] − #emph[y] = 3 #linebreak() and testing a point. #linebreak() #linebreak() The intercepts are #emph[x] = 3 and #emph[y] = −3 and the #linebreak() boundary line will be dashed. #linebreak() #linebreak() Test (0, 0) which makes the inequality false so shade #linebreak() (red) the side that does not contain (0, 0).], [#figure(figph[A coordinate plane displays the graph of the linear inequality y \< (1/2)x - 3. A dashed line represents the boundary, and the region below it is shaded in orange.], alt: "A coordinate plane displays the graph of the linear inequality y < (1/2)x - 3. A dashed line represents the boundary, and the region below it is shaded in orange.", caption: none)], )) #figure(table( columns: 2, align: left, inset: 6pt, [Graph #math.equation(block: false, alt: "y less than minus the fraction 1 over 5 x plus 4")[$y < − frac(1, 5) x + 4$] by graphing #math.equation(block: false, alt: "y equals minus the fraction 1 over 5 x plus 4")[$y = − frac(1, 5) x + 4$] #linebreak() using the slope #math.equation(block: false, alt: "m equals minus the fraction 1 over 5")[$m = − frac(1, 5)$] and #emph[y]-intercept #emph[b] = 4. #linebreak() The boundary line will be dashed #linebreak() #linebreak() Test (0, 0) which makes the inequality true, so #linebreak() shade (blue) the side that contains (0, 0). #linebreak() #linebreak() Choose a test point in the solution and verify that it is a solution to both inequalties.], [#figure(figph[A Cartesian coordinate graph displays two linear inequalities. The area above the dashed blue line is shaded light blue, representing one solution set. The area below the dashed red line is shaded dark grey, representing another solution set.], alt: "A Cartesian coordinate graph displays two linear inequalities. The area above the dashed blue line is shaded light blue, representing one solution set. The area below the dashed red line is shaded dark grey, representing another solution set.", caption: none)], )) The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ x plus y less than or equal to 2; y greater than or equal to the fraction 2 over 3 x minus 1 .")[$\{ x + y ≤ 2 \ y ≥ frac(2, 3) x − 1 .$] #solutionbox[ #figure(figph[The figure shows the graph for the inequalities x plus y less than or equal to two and y greater than or equal to two by three of x minus one. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.], alt: "The figure shows the graph for the inequalities x plus y less than or equal to two and y greater than or equal to two by three of x minus one. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ 3 x minus 2 y less than or equal to 6; y greater than − the fraction 1 over 4 x plus 5 .")[$\{ 3 x − 2 y ≤ 6 \ y > "−" frac(1, 4) x + 5 .$] #solutionbox[ #figure(figph[The figure shows graph for the inequalities three times x minus two times y less than or equal to six and y greater than or equal to minus one by four of x plus five. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.], alt: "The figure shows graph for the inequalities three times x minus two times y less than or equal to six and y greater than or equal to minus one by four of x plus five. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] #examplebox("Example 4")[][ Solve the system by graphing: #math.equation(block: false, alt: "{ x minus 2 y less than 5; y greater than −4 .")[$\{ x − 2 y < 5 \ y > −4 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "{ x minus 2 y less than 5; y greater than −4")[$#h(2em) \{ x − 2 y < 5 \ y > −4$]]), [Graph #math.equation(block: false, alt: "x minus 2 y less than 5 ,")[$x − 2 y < 5 ,$] by graphing #math.equation(block: false, alt: "x minus 2 y equals 5")[$x − 2 y = 5$] #linebreak() and testing a point. The intercepts are #linebreak() #emph[x] = 5 and #emph[y] = −2.5 and the #linebreak() boundary line will be dashed. #linebreak() #linebreak() Test (0, 0) which makes the inequality true, so shade #linebreak() (red) the side that contains (0, 0).], [#figure(figph[A graph displays the solution set for the inequality y \> (1/2)x - 2. A dashed red line represents y = (1/2)x - 2, and the region above the line is shaded orange.], alt: "A graph displays the solution set for the inequality y > (1/2)x - 2. A dashed red line represents y = (1/2)x - 2, and the region above the line is shaded orange.", caption: none)], )) #figure(table( columns: 2, align: left, inset: 6pt, [Graph #math.equation(block: false, alt: "y greater than −4 ,")[$y > −4 ,$] by graphing #math.equation(block: false, alt: "y equals −4")[$y = −4$] and #linebreak() recognizing that it is a horizontal line #linebreak() through #math.equation(block: false, alt: "y equals −4 .")[$y = −4 .$] The boundary line will #linebreak() be dashed. #linebreak() #linebreak() Test (0, 0) which makes the inequality #linebreak() true so shade (blue) the side that contains (0, 0).], [#figure(figph[Graph of a system of linear inequalities with two dashed lines, y=1/2x-3 and y=-4, showing three distinct shaded regions: gray, light blue, and light orange.], alt: "Graph of a system of linear inequalities with two dashed lines, y=1/2x-3 and y=-4, showing three distinct shaded regions: gray, light blue, and light orange.", caption: none)], )) The point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ y greater than or equal to 3 x minus 2; y less than minus 1 .")[$\{ y ≥ 3 x − 2 \ y < − 1 .$] #solutionbox[ #figure(figph[The figure shows graph for the inequalities y greater than or equal to three times x minus two and y less than minus one. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution], alt: "The figure shows graph for the inequalities y greater than or equal to three times x minus two and y less than minus one. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution", caption: none) #linebreak() The solution is the grey region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ x greater than −4; x minus 2 y greater than or equal to −4 .")[$\{ x > −4 \ x − 2 y ≥ −4 .$] #solutionbox[ #figure(figph[The figure shows graph for the inequalities x greater than or equal to minus four and x minus two times y greater than minus four. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.], alt: "The figure shows graph for the inequalities x greater than or equal to minus four and x minus two times y greater than minus four. Two intersecting lines are shown and the region bound by both the lines is the marked in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in the next example. #examplebox("Example 5")[][ Solve the system by graphing: #math.equation(block: false, alt: "{ 4 x plus 3 y greater than or equal to 12; y less than minus the fraction 4 over 3 x plus 1 .")[$\{ 4 x + 3 y ≥ 12 \ y < − frac(4, 3) x + 1 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "{ 4 x plus 3 y greater than or equal to 12; y less than minus the fraction 4 over 3 x plus 1")[$\{ 4 x + 3 y ≥ 12 \ y < − frac(4, 3) x + 1$]]), [Graph #math.equation(block: false, alt: "4 x plus 3 y greater than or equal to 12 ,")[$4 x + 3 y ≥ 12 ,$] by graphing #math.equation(block: false, alt: "4 x plus 3 y equals 12")[$4 x + 3 y = 12$] #linebreak() and testing a point. The intercepts are #emph[x] = 3 #linebreak() and #emph[y] = 4 and the boundary line will be solid. #linebreak() #linebreak() Test (0, 0) which makes the inequality false, so #linebreak() shade (red) the side that does not contain (0, 0).], [#figure(figph[A coordinate plane displaying a linear inequality: the area below the dashed line y = -x + 4 is shaded, indicating the solution set.], alt: "A coordinate plane displaying a linear inequality: the area below the dashed line y = -x + 4 is shaded, indicating the solution set.", caption: none)], [Graph #math.equation(block: false, alt: "y less than minus the fraction 4 over 3 x plus 1")[$y < − frac(4, 3) x + 1$] by graphing #math.equation(block: false, alt: "y equals minus the fraction 4 over 3 x plus 1")[$y = − frac(4, 3) x + 1$] #linebreak() using the slope #math.equation(block: false, alt: "m equals minus the fraction 4 over 3")[$m = − frac(4, 3)$] and #emph[y]-intercept #linebreak() #emph[b] = 1. The boundary line will be dashed. #linebreak() #linebreak() Test (0, 0) which makes the inequality true, so #linebreak() shade (blue) the side that contains (0, 0).], [#figure(figph[A graph showing two linear inequalities. A solid red line represents y = -x + 4, with the region y \> -x + 4 shaded orange. A dashed blue line represents y = -x, with the region y \< -x shaded blue.], alt: "A graph showing two linear inequalities. A solid red line represents y = -x + 4, with the region y > -x + 4 shaded orange. A dashed blue line represents y = -x, with the region y < -x shaded blue.", caption: none)], )) There is no point in both shaded regions, so the system has no solution. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ 3 x minus 2 y greater than or equal to 12; y greater than or equal to the fraction 3 over 2 x plus 1 .")[$\{ 3 x − 2 y ≥ 12 \ y ≥ frac(3, 2) x + 1 .$] #solutionbox[ #figure(figph[The graph of three times x minus two times y greater than or equal to twelve and y greater than or equal to three by two of x plus one is shown. Two intersecting lines are shown. The inequalities do not have a solution.], alt: "The graph of three times x minus two times y greater than or equal to twelve and y greater than or equal to three by two of x plus one is shown. Two intersecting lines are shown. The inequalities do not have a solution.", caption: none) #linebreak() No solution. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ x plus 3 y greater than 8; y less than minus the fraction 1 over 3 x minus 2 .")[$\{ x + 3 y > 8 \ y < − frac(1, 3) x − 2 .$] #solutionbox[ #figure(figph[The graph of x plus three times y greater than eight and y less than minus one by three of x minus two is shown. Two intersecting lines are shown. The inequalities do not have a solution.], alt: "The graph of x plus three times y greater than eight and y less than minus one by three of x minus two is shown. Two intersecting lines are shown. The inequalities do not have a solution.", caption: none) #linebreak() No solution. ] ] Some systems of linear inequalities where the boundary lines are parallel will have a solution. We’ll see this in the next example. #examplebox("Example 6")[][ Solve the system by graphing: #math.equation(block: false, alt: "{ y greater than the fraction 1 over 2 x minus 4; x minus 2 y less than −4 .")[$\{ y > frac(1, 2) x − 4 \ x − 2 y < −4 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "{ y greater than the fraction 1 over 2 x minus 4; x minus 2 y less than −4")[$#h(2em) \{ y > frac(1, 2) x − 4 \ x − 2 y < −4$]]), [Graph #math.equation(block: false, alt: "y greater than the fraction 1 over 2 x minus 4")[$y > frac(1, 2) x − 4$] by graphing #math.equation(block: false, alt: "y equals the fraction 1 over 2 x minus 4")[$y = frac(1, 2) x − 4$] #linebreak() using the slope #math.equation(block: false, alt: "m equals the fraction 1 over 2")[$m = frac(1, 2)$] and the intercept #linebreak() #emph[b] = −4. The boundary line will be dashed. #linebreak() #linebreak() Test (0, 0) which makes the inequality true, so #linebreak() shade (red) the side that contains (0, 0).], [#figure(figph[A graph on a coordinate plane shows a dashed red line passing through (0,-4) and (8,0). The region above the line is shaded in orange, representing the solution set of a linear inequality.], alt: "A graph on a coordinate plane shows a dashed red line passing through (0,-4) and (8,0). The region above the line is shaded in orange, representing the solution set of a linear inequality.", caption: none)], [Graph #math.equation(block: false, alt: "x minus 2 y less than −4")[$x − 2 y < −4$] by graphing #math.equation(block: false, alt: "x minus 2 y equals −4")[$x − 2 y = −4$] #linebreak() and testing a point. The intercepts are #linebreak() #emph[x] = −4 and #emph[y] = 2 and the boundary line will be dashed. #linebreak() #linebreak() Choose a test point in the solution and verify #linebreak() that it is a solution to both inequalties. #linebreak() #linebreak() Test (0, 0) which makes the inequality false, so #linebreak() shade (blue) the side that does not contain (0, 0).], [#figure(figph[A coordinate plane shows two parallel dashed lines, y = 0.5x + 2 and y = 0.5x - 4. The region above the top line is dark gray, and the region between the lines is light orange.], alt: "A coordinate plane shows two parallel dashed lines, y = 0.5x + 2 and y = 0.5x - 4. The region above the top line is dark gray, and the region between the lines is light orange.", caption: none)], )) No point on the boundary lines is included in the solution as both lines are dashed. The solution is the region that is shaded twice which is also the solution to #math.equation(block: false, alt: "x minus 2 y less than −4 .")[$x − 2 y < −4 .$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ y greater than or equal to 3 x plus 1; −3 x plus y greater than or equal to −4 .")[$\{ y ≥ 3 x + 1 \ −3 x + y ≥ −4 .$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y greater than or equal to three times x plus one and minus three times x plus y greater than or equal to minus four. Two parallel lines are shown and the region to the left of both is colored in grey. It is the solution.], alt: "The figure shows the graph of the inequalities y greater than or equal to three times x plus one and minus three times x plus y greater than or equal to minus four. Two parallel lines are shown and the region to the left of both is colored in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by graphing: #math.equation(block: false, alt: "{ y less than or equal to minus the fraction 1 over 4 x plus 2; x plus 4 y less than or equal to 4 .")[$\{ y ≤ − frac(1, 4) x + 2 \ x + 4 y ≤ 4 .$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y less than or equal to minus one fourth of x plus 2 and x plus four times y less than or equal to four. Two parallel lines are shown and the region to the bottom of both is colored in grey. It is the solution.], alt: "The figure shows the graph of the inequalities y less than or equal to minus one fourth of x plus 2 and x plus four times y less than or equal to four. Two parallel lines are shown and the region to the bottom of both is colored in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] ] === Solve Applications of Systems of Inequalities The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so we add inequalities to the system as additional requirements. #examplebox("Example 7")[][ Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least 25 photos to display at her booth. Each small photo she displays costs her \$4 and each large photo costs her \$10. She doesn’t want to spend more than \$200 on photos to display. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could she display 10 small and 20 large photos? #linebreak() ⓓ Could she display 20 small and 10 large photos? #solutionbox[ ⓐ #figure(table( columns: 1, align: left, inset: 6pt, [Let #math.equation(block: false, alt: "x equals the number of small photos .")[$x = "the number of small photos" .$]], [#math.equation(block: false, alt: "y equals the number of large photos")[$#h(1.5em) y = "the number of large photos"$]], [To find the system of equations translate the information.], [#math.equation(block: false, alt: "She wants to have at least 25 photos.")[$#h(3em) "She wants to have at least 25 photos."$]], [#math.equation(block: false, alt: "The number of small plus the number of large should be at least 25.")[$#h(3em) "The number of small plus the number of large should be at least 25."$]], [#math.equation(block: false, alt: "x plus y greater than or equal to 25")[$x + y ≥ 25$]], [#math.equation(block: false, alt: "$4 for each small and $10 for each large must be no more than $200")[$#h(3em) "$4 for each small and $10 for each large must be no more than $200"$]], [#math.equation(block: false, alt: "4 x plus 10 y less than or equal to 200")[$4 x + 10 y ≤ 200$]], [#math.equation(block: false, alt: "The number of small photos must be greater than or equal to 0.")[$#h(3em) "The number of small photos must be greater than or equal to 0."$]], [#math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$]], [#math.equation(block: false, alt: "The number of large photos must be greater than or equal to 0.")[$#h(3em) "The number of large photos must be greater than or equal to 0."$]], [#math.equation(block: false, alt: "y greater than or equal to 0")[$y ≥ 0$]], [We have our system of equations.], [#math.equation(block: false, alt: "{ x plus y greater than or equal to 25; 4 x plus 10 y less than or equal to 200; x greater than or equal to 0; y greater than or equal to 0")[$#h(1em) \{ x + y ≥ 25 \ 4 x + 10 y ≤ 200 \ x ≥ 0 \ y ≥ 0 #h(3em)$]], )) ⓑ #linebreak() Since #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")[$y ≥ 0$] (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, [To graph #math.equation(block: false, alt: "x plus y greater than or equal to 25 ,")[$x + y ≥ 25 ,$] graph #math.equation(block: false, alt: "x plus y equals 25")[$x + y = 25$] as a solid line. #linebreak() Choose (0, 0) as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point (0, 0). #linebreak() #linebreak() To graph #math.equation(block: false, alt: "4 x plus 10 y less than or equal to 200 ,")[$4 x + 10 y ≤ 200 ,$] graph #math.equation(block: false, alt: "4 x plus 10 y equals 200")[$4 x + 10 y = 200$] as a solid line. #linebreak() Choose (0, 0) as a test point. Since it does make the inequality true, shade (blue) the side that include the point (0, 0).], [#figure(figph[A graph on a coordinate plane shows two downward-sloping lines representing linear inequalities. Individual shaded regions highlight their solutions, with a darker overlapping area indicating the feasible region.], alt: "A graph on a coordinate plane shows two downward-sloping lines representing linear inequalities. Individual shaded regions highlight their solutions, with a darker overlapping area indicating the feasible region.", caption: none)], )) The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly-shaded section are included in the solution as are the points on the #emph[x]-axis from (25, 0) to (55, 0). ⓒ To determine if 10 small and 20 large photos would work, we look at the graph to see if the point (10, 20) is in the solution region. We could also test the point to see if it is a solution of both equations. It is not, Christy would not display 10 small and 20 large photos. ⓓ To determine if 20 small and 10 large photos would work, we look at the graph to see if the point (20, 10) is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so Christy could choose to display 20 small and 10 large photos. Notice that we could also test the possible solutions by substituting the values into each inequality. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A trailer can carry a maximum weight of 160 pounds and a maximum volume of 15 cubic feet. A microwave oven weighs 30 pounds and has 2 cubic feet of volume, while a printer weighs 20 pounds and has 3 cubic feet of space. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could 4 microwaves and 2 printers be carried on this trailer? #linebreak() ⓓ Could 7 microwaves and 3 printers be carried on this trailer? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ 30 m plus 20 p less than or equal to 160; 2 m plus 3 p less than or equal to 15")[$\{ 30 m + 20 p ≤ 160 \ 2 m + 3 p ≤ 15$] #linebreak() ⓑ #linebreak() #figure(figph[The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.], alt: "The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.", caption: none) #linebreak() ⓒ yes #linebreak() ⓓ no ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Mary needs to purchase supplies of answer sheets and pencils for a standardized test to be given to the juniors at her high school. The number of the answer sheets needed is at least 5 more than the number of pencils. The pencils cost \$2 and the answer sheets cost \$1. Mary’s budget for these supplies allows for a maximum cost of \$400. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could Mary purchase 100 pencils and 100 answer sheets? #linebreak() ⓓ Could Mary purchase 150 pencils and 150 answer sheets? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ a greater than or equal to p plus 5; a plus 2 p less than or equal to 400")[$\{ a ≥ p + 5 \ a + 2 p ≤ 400$] #linebreak() ⓑ #linebreak() #figure(figph[The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.], alt: "The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.", caption: none) #linebreak() ⓒ no #linebreak() ⓓ no ] ] When we use variables other than #emph[x] and #emph[y] to define an unknown quantity, we must change the names of the axes of the graph as well. #examplebox("Example 8")[][ Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than \$5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs \$1.40. Each cookie has 160 calories and costs \$0.50. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could he eat 3 hamburgers and 1 cookie? #linebreak() ⓓ Could he eat 2 hamburgers and 4 cookies? #solutionbox[ ⓐ #figure(table( columns: 1, align: left, inset: 6pt, [Let #math.equation(block: false, alt: "h equals the number of hamburgers .")[$h = "the number of hamburgers" .$]], [#math.equation(block: false, alt: "c equals the number of cookies")[$#h(1.5em) c = "the number of cookies"$]], [To find the system of equations translate the information.], [The calories from hamburgers at 240 calories each, plus the calories from cookies at 160 calories each], [must be more that 800.], [#math.equation(block: false, alt: "240 h plus 160 c greater than or equal to 800")[$240 h + 160 c ≥ 800$]], [#math.equation(block: false, alt: "The amount spent on hamburgers at $1.40 each, plus the amount spent on cookies at $0.50 each")[$#h(3em) "The amount spent on hamburgers at $1.40 each, plus the amount spent on cookies at $0.50 each"$]], [#math.equation(block: false, alt: "must be no more than $5.00.")[$#h(3em) "must be no more than $5.00."$]], [#math.equation(block: false, alt: "1.40 h plus 0.50 c less than or equal to 5")[$1.40 h + 0.50 c ≤ 5$]], [#math.equation(block: false, alt: "The number of hamburgers must be greater than or equal to 0.")[$#h(3em) "The number of hamburgers must be greater than or equal to 0."$]], [#math.equation(block: false, alt: "h greater than or equal to 0")[$h ≥ 0$]], [#math.equation(block: false, alt: "The number of cookies must be greater than or equal to 0.")[$#h(3em) "The number of cookies must be greater than or equal to 0."$]], [#math.equation(block: false, alt: "c greater than or equal to 0")[$c ≥ 0$]], [We have our system of equations. #math.equation(block: false, alt: "{ 240 h plus 160 c greater than or equal to 800; 1.40 h plus 0.50 c less than or equal to 5; h greater than or equal to 0; c greater than or equal to 0")[$#h(2em) \{ 240 h + 160 c ≥ 800 \ 1.40 h + 0.50 c ≤ 5 \ h ≥ 0 \ c ≥ 0$]], )) ⓑ #linebreak() Since #math.equation(block: false, alt: "h greater than equals 0")[$h > = 0$] and #math.equation(block: false, alt: "c greater than equals 0")[$c > = 0$] (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, [To graph #math.equation(block: false, alt: "240 h plus 160 c greater than or equal to 800 ,")[$240 h + 160 c ≥ 800 ,$] graph #math.equation(block: false, alt: "240 h plus 160 c equals 800")[$240 h + 160 c = 800$] as a solid line. #linebreak() #linebreak() #linebreak() Choose (0, 0) as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point (0, 0).], [#figure(figph[A graph on an h-c coordinate plane shows two intersecting lines and three distinct shaded regions, representing solutions to a system of inequalities.], alt: "A graph on an h-c coordinate plane shows two intersecting lines and three distinct shaded regions, representing solutions to a system of inequalities.", caption: none)], )) Graph #math.equation(block: false, alt: "1.40 h plus 0.50 c less than or equal to 5 .")[$1.40 h + 0.50 c ≤ 5 .$] The boundary line is #math.equation(block: false, alt: "1.40 h plus 0.50 c equals 5 .")[$1.40 h + 0.50 c = 5 .$] We test (0, 0) and it makes the inequality true. We shade the side of the line that includes (0, 0). The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the #emph[x]-axis from (5, 0) to (10, 0). ⓒ To determine if 3 hamburgers and 1 cookie would meet Omar’s criteria, we see if the point (3, 2) is in the solution region. It is, so Omar might choose to eat 3 hamburgers and 1 cookie. ⓓ To determine if 2 hamburgers and 4 cookies would meet Omar’s criteria, we see if the point (2, 4) is in the solution region. It is, Omar might choose to eat 2 hamburgers and 4 cookies. We could also test the possible solutions by substituting the values into each inequality. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Tenison needs to eat at least an extra 1,000 calories a day to prepare for running a marathon. He has only \$25 to spend on the extra food he needs and will spend it on \$0.75 donuts which have 360 calories each and \$2 energy drinks which have 110 calories. ⓐ Write a system of inequalities that models this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can he buy 8 donuts and 4 energy drinks and satisfy his caloric needs? #linebreak() ⓓ Can he buy 1 donut and 3 energy drinks and satisfy his caloric needs? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ 0.75 d plus 2 e less than or equal to 25; 360 d plus 110 e greater than or equal to 1000")[$\{ 0.75 d + 2 e ≤ 25 \ 360 d + 110 e ≥ 1000$] #linebreak() ⓑ #linebreak() #figure(figph[The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.], alt: "The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.", caption: none) #linebreak() ⓒ yes #linebreak() ⓓ no ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Philip’s doctor tells him he should add at least 1,000 more calories per day to his usual diet. Philip wants to buy protein bars that cost \$1.80 each and have 140 calories and juice that costs \$1.25 per bottle and have 125 calories. He doesn’t want to spend more than \$12. ⓐ Write a system of inequalities that models this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can he buy 3 protein bars and 5 bottles of juice? #linebreak() ⓓ Can he buy 5 protein bars and 3 bottles of juice? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ 140 p plus 125 j greater than or equal to 1000; 1.80 p plus 1.25 j less than or equal to 12")[$\{ 140 p + 125 j ≥ 1000 \ 1.80 p + 1.25 j ≤ 12$] #linebreak() ⓑ #linebreak() #figure(figph[The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.], alt: "The graph of two intersecting lines, one red and one blue, is shown. The area bound by the two lines is shown in grey.", caption: none) #linebreak() ⓒ yes #linebreak() ⓓ no ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with solving systems of linear inequalities by graphing. - #link("https://openstax.org/l/37sysineqgraph")[Solving Systems of Linear Inequalities by Graphing] - #link("https://openstax.org/l/37sysineqgraph2")[Systems of Linear Inequalities] ] === Key Concepts - #strong[Solutions of a System of Linear Inequalities:] Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the #emph[x, y] coordinate system that includes all the points whose ordered pairs make the inequalities true. - #strong[How to solve a system of linear inequalities by graphing.] + Graph the first inequality. #linebreak() Graph the boundary line. #linebreak() Shade in the side of the boundary line where the inequality is true. + On the same grid, graph the second inequality. #linebreak() Graph the boundary line. #linebreak() Shade in the side of that boundary line where the inequality is true. + The solution is the region where the shading overlaps. + Check by choosing a test point. === Section Exercises ==== Practice Makes Perfect #strong[Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities] In the following exercises, determine whether each ordered pair is a solution to the system. #math.equation(block: true, alt: "{ 3 x plus y greater than 5; 2 x minus y less than or equal to 10")[$\{ 3 x + y > 5 \ 2 x − y ≤ 10$] ⓐ #math.equation(block: false, alt: "open parenthesis 3 , −3 close parenthesis")[$( 3 , −3 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 7 , 1 close parenthesis")[$( 7 , 1 )$] #math.equation(block: true, alt: "{ 4 x minus y less than 10; −2 x plus 2 y greater than −8")[$\{ 4 x − y < 10 \ −2 x + 2 y > −8$] ⓐ #math.equation(block: false, alt: "open parenthesis 5 , −2 close parenthesis")[$( 5 , −2 )$] ⓑ #math.equation(block: false, alt: "open parenthesis −1 , 3 close parenthesis")[$( −1 , 3 )$] #solutionbox[ ⓐ false ⓑ true ] #math.equation(block: true, alt: "{ y greater than the fraction 2 over 3 x minus 5; x plus the fraction 1 over 2 y less than or equal to 4")[$\{ y > frac(2, 3) x − 5 \ x + frac(1, 2) y ≤ 4$] ⓐ #math.equation(block: false, alt: "(6, −4)")[$"(6, −4)"$] ⓑ #math.equation(block: false, alt: "(3, 0)")[$"(3, 0)"$] #math.equation(block: true, alt: "{ y less than the fraction 3 over 2 x plus 3; the fraction 3 over 4 x minus 2 y less than 5")[$\{ y < frac(3, 2) x + 3 \ frac(3, 4) x − 2 y < 5$] ⓐ #math.equation(block: false, alt: "open parenthesis −4 , −1 close parenthesis")[$( −4 , −1 )$] ⓑ #math.equation(block: false, alt: "(8, 3)")[$"(8, 3)"$] #solutionbox[ ⓐ false ⓑ true ] #math.equation(block: true, alt: "{ 7 x plus 2 y greater than 14; 5 x minus y less than or equal to 8")[$\{ 7 x + 2 y > 14 \ 5 x − y ≤ 8$] ⓐ #math.equation(block: false, alt: "(2, 3)")[$"(2, 3)"$] ⓑ #math.equation(block: false, alt: "(7, −1)")[$"(7, −1)"$] #math.equation(block: true, alt: "{ 6 x minus 5 y less than 20; −2 x plus 7 y greater than −8")[$\{ 6 x − 5 y < 20 \ −2 x + 7 y > −8$] ⓐ #math.equation(block: false, alt: "(1, −3)")[$"(1, −3)"$] ⓑ #math.equation(block: false, alt: "(−4, 4)")[$"(−4, 4)"$] #solutionbox[ ⓐ false ⓑ true ] #strong[Solve a System of Linear Inequalities by Graphing] In the following exercises, solve each system by graphing. #math.equation(block: true, alt: "{ y less than or equal to 3 x plus 2; y greater than x minus 1")[$\{ y ≤ 3 x + 2 \ y > x − 1$] #math.equation(block: true, alt: "{ y less than minus 2 x plus 2; y greater than or equal to minus x minus 1")[$\{ y < − 2 x + 2 \ y ≥ − x − 1$] #solutionbox[ #figure(figph[The figure shows the graph of inequalities y less than minus two times x plus two and y greater than or equal to minus x minus one. Two intersecting lines are shown, one in red and the other in blue. The area bound by the two lines is shown in grey.], alt: "The figure shows the graph of inequalities y less than minus two times x plus two and y greater than or equal to minus x minus one. Two intersecting lines are shown, one in red and the other in blue. The area bound by the two lines is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ y less than 2 x minus 1; y less than or equal to minus the fraction 1 over 2 x plus 4")[$\{ y < 2 x − 1 \ y ≤ − frac(1, 2) x + 4$] #math.equation(block: true, alt: "{ y greater than or equal to minus the fraction 2 over 3 x plus 2; y greater than 2 x minus 3")[$\{ y ≥ − frac(2, 3) x + 2 \ y > 2 x − 3$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y greater than or equal to minus two by three x plus two and y greater than two times x minus three. Two intersecting lines, one in red and the other in blue, are shown. The region bound by them is shown in grey.], alt: "The figure shows the graph of the inequalities y greater than or equal to minus two by three x plus two and y greater than two times x minus three. Two intersecting lines, one in red and the other in blue, are shown. The region bound by them is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "x minus y greater than 1; y less than minus the fraction 1 over 4 x plus 3")[$x − y > 1 \ y < − frac(1, 4) x + 3$] #math.equation(block: true, alt: "{ x plus 2 y less than 4; y less than x minus 2")[$\{ x + 2 y < 4 \ y < x − 2$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities x minus two times y less than four and y less than x minus two. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey.], alt: "The figure shows the graph of the inequalities x minus two times y less than four and y less than x minus two. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ 3 x minus y greater than or equal to 6; y greater than or equal to minus the fraction 1 over 2 x")[$\{ 3 x − y ≥ 6 \ y ≥ − frac(1, 2) x$] #math.equation(block: true, alt: "{ 2 x plus 4 y greater than or equal to 8; y less than or equal to the fraction 3 over 4 x")[$\{ 2 x + 4 y ≥ 8 \ y ≤ frac(3, 4) x$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities two times x plus four times y greater than or equal to eight and y less than or equal to minus three fourth of x. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities two times x plus four times y greater than or equal to eight and y less than or equal to minus three fourth of x. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ 2 x minus 5 y less than 10; 3 x plus 4 y greater than or equal to 12")[$\{ 2 x − 5 y < 10 \ 3 x + 4 y ≥ 12$] #math.equation(block: true, alt: "{ 3 x minus 2 y less than or equal to 6; −4 x minus 2 y greater than 8")[$\{ 3 x − 2 y ≤ 6 \ −4 x − 2 y > 8$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities three times x minus two times y less than or equal to six and minus four times x minus two times y greater than eight. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities three times x minus two times y less than or equal to six and minus four times x minus two times y greater than eight. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ 2 x plus 2 y greater than −4; − x plus 3 y greater than or equal to 9")[$\{ 2 x + 2 y > −4 \ "−" x + 3 y ≥ 9$] #math.equation(block: true, alt: "{ 2 x plus y greater than −6; − x plus 2 y greater than or equal to −4")[$\{ 2 x + y > −6 \ "−" x + 2 y ≥ −4$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities two times x plus y greater than minus six and minus x plus two times y greater than or equal to minus four. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities two times x plus y greater than minus six and minus x plus two times y greater than or equal to minus four. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ x minus 2 y less than 3; y less than or equal to 1")[$\{ x − 2 y < 3 \ y ≤ 1$] #math.equation(block: true, alt: "{ x minus 3 y greater than 4; y less than or equal to minus 1")[$\{ x − 3 y > 4 \ y ≤ − 1$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities x minus three times y greater than four and y less than or equal to minus one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities x minus three times y greater than four and y less than or equal to minus one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ y greater than or equal to minus the fraction 1 over 2 x minus 3; x less than or equal to 2")[$\{ y ≥ − frac(1, 2) x − 3 \ x ≤ 2$] #math.equation(block: true, alt: "{ y less than or equal to minus the fraction 2 over 3 x plus 5; x greater than or equal to 3")[$\{ y ≤ − frac(2, 3) x + 5 \ x ≥ 3$] #solutionbox[ #figure(figph[The figure shows the graph of the inequality y less than or equal to minus two by three times x plus five and x greater than or equal to three. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequality y less than or equal to minus two by three times x plus five and x greater than or equal to three. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ y greater than or equal to the fraction 3 over 4 x minus 2; y less than 2")[$\{ y ≥ frac(3, 4) x − 2 \ y < 2$] #math.equation(block: true, alt: "{ y less than or equal to minus the fraction 1 over 2 x plus 3; y less than 1")[$\{ y ≤ − frac(1, 2) x + 3 \ y < 1$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y less than or equal to minus half x plus three and y less than one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities y less than or equal to minus half x plus three and y less than one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ 3 x minus 4 y less than 8; x less than 1")[$\{ 3 x − 4 y < 8 \ x < 1$] #math.equation(block: true, alt: "{ −3 x plus 5 y greater than 10; x greater than −1")[$\{ −3 x + 5 y > 10 \ x > −1$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities minus three times x plus five times y greater than ten and x greater than minus one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities minus three times x plus five times y greater than ten and x greater than minus one. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ x greater than or equal to 3; y less than or equal to 2")[$\{ x ≥ 3 \ y ≤ 2$] #math.equation(block: true, alt: "{ x less than or equal to −1; y greater than or equal to 3")[$\{ x ≤ −1 \ y ≥ 3$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities x less than or equal to minus one and y greater than or equal to three. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.], alt: "The figure shows the graph of the inequalities x less than or equal to minus one and y greater than or equal to three. Two intersecting lines, one in blue and the other in red, are shown. The area bound by the lines is shown in grey. It is the solution.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ 2 x plus 4 y greater than 4; y less than or equal to minus the fraction 1 over 2 x minus 2")[$\{ 2 x + 4 y > 4 \ y ≤ − frac(1, 2) x − 2$] #math.equation(block: true, alt: "{ x minus 3 y greater than or equal to 6; y greater than the fraction 1 over 3 x plus 1")[$\{ x − 3 y ≥ 6 \ y > frac(1, 3) x + 1$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities x minus three times y greater than or equal to six and y greater than one third of x plus one. Two non intersecting lines, one in blue and the other in red, are shown.], alt: "The figure shows the graph of the inequalities x minus three times y greater than or equal to six and y greater than one third of x plus one. Two non intersecting lines, one in blue and the other in red, are shown.", caption: none) #linebreak() No solution. ] #math.equation(block: true, alt: "{ −2 x plus 6 y less than 0; 6 y greater than 2 x plus 4")[$\{ −2 x + 6 y < 0 \ 6 y > 2 x + 4$] #math.equation(block: true, alt: "{ −3 x plus 6 y greater than 12; 4 y less than or equal to 2 x minus 4")[$\{ −3 x + 6 y > 12 \ 4 y ≤ 2 x − 4$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities minus three times x plus six times y greater than twelve and four times y less than or equal to two times x minus four. Two non intersecting lines, one in blue and the other in red, are shown.], alt: "The figure shows the graph of the inequalities minus three times x plus six times y greater than twelve and four times y less than or equal to two times x minus four. Two non intersecting lines, one in blue and the other in red, are shown.", caption: none) #linebreak() No solution. ] #math.equation(block: true, alt: "{ y greater than or equal to −3 x plus 2; 3 x plus y greater than 5")[$\{ y ≥ −3 x + 2 \ 3 x + y > 5$] #math.equation(block: true, alt: "{ y greater than or equal to the fraction 1 over 2 x minus 1; −2 x plus 4 y greater than or equal to 4")[$\{ y ≥ frac(1, 2) x − 1 \ −2 x + 4 y ≥ 4$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y greater than or equal to minus half x minus one and minus two times x plus four times y greater than or equal to four. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.], alt: "The figure shows the graph of the inequalities y greater than or equal to minus half x minus one and minus two times x plus four times y greater than or equal to four. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ y less than or equal to minus the fraction 1 over 4 x minus 2; x plus 4 y less than 6")[$\{ y ≤ − frac(1, 4) x − 2 \ x + 4 y < 6$] #math.equation(block: true, alt: "{ y greater than or equal to 3 x minus 1; −3 x plus y greater than −4")[$\{ y ≥ 3 x − 1 \ −3 x + y > −4$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y greater than or equal to three times x minus one and minus three times x plus y greater than minus four. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.], alt: "The figure shows the graph of the inequalities y greater than or equal to three times x minus one and minus three times x plus y greater than minus four. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ 3 y greater than x plus 2; −2 x plus 6 y greater than 8")[$\{ 3 y > x + 2 \ −2 x + 6 y > 8$] #math.equation(block: true, alt: "{ y less than the fraction 3 over 4 x minus 2; −3 x plus 4 y less than 7")[$\{ y < frac(3, 4) x − 2 \ −3 x + 4 y < 7$] #solutionbox[ #figure(figph[The figure shows the graph of the inequalities y less than three by fourth x minus two and minus three x plus four y less than seven. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.], alt: "The figure shows the graph of the inequalities y less than three by fourth x minus two and minus three x plus four y less than seven. Two non intersecting lines, one in blue and the other in red, are shown. The solution area is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #strong[Solve Applications of Systems of Inequalities] In the following exercises, translate to a system of inequalities and solve. Caitlyn sells her drawings at the county fair. She wants to sell at least 60 drawings and has portraits and landscapes. She sells the portraits for \$15 and the landscapes for \$10. She needs to sell at least \$800 worth of drawings in order to earn a profit. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Will she make a profit if she sells 20 portraits and 35 landscapes? #linebreak() ⓓ Will she make a profit if she sells 50 portraits and 20 landscapes? Jake does not want to spend more than \$50 on bags of fertilizer and peat moss for his garden. Fertilizer costs \$2 a bag and peat moss costs \$5 a bag. Jake’s van can hold at most 20 bags. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can he buy 15 bags of fertilizer and 4 bags of peat moss? #linebreak() ⓓ Can he buy 10 bags of fertilizer and 10 bags of peat moss? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ f greater than or equal to 0; p greater than or equal to 0; f plus p less than or equal to 20; 2 f plus 5 p less than or equal to 50")[$\{ f ≥ 0 \ p ≥ 0 \ f + p ≤ 20 \ 2 f + 5 p ≤ 50$] #linebreak() ⓑ #linebreak() #figure(figph[The figure shows the graph of the inequalities f plus p less than or equal to twenty and two f and five p less than or equal to fifty. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.], alt: "The figure shows the graph of the inequalities f plus p less than or equal to twenty and two f and five p less than or equal to fifty. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.", caption: none) #linebreak() ⓒ yes #linebreak() ⓓ no ] Reiko needs to mail her Christmas cards and packages and wants to keep her mailing costs to no more than \$500. The number of cards is at least 4 more than twice the number of packages. The cost of mailing a card (with pictures enclosed) is \$3 and for a package the cost is \$7. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can she mail 60 cards and 26 packages? #linebreak() ⓓ Can she mail 90 cards and 40 packages? Juan is studying for his final exams in chemistry and algebra. he knows he only has 24 hours to study, and it will take him at least three times as long to study for algebra than chemistry. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can he spend 4 hours on chemistry and 20 hours on algebra? #linebreak() ⓓ Can he spend 6 hours on chemistry and 18 hours on algebra? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ c greater than or equal to 0; a greater than or equal to 0; c plus a less than or equal to 24; a greater than or equal to 3 c")[$\{ c ≥ 0 \ a ≥ 0 \ c + a ≤ 24 \ a ≥ 3 c$] #linebreak() ⓑ #linebreak() #figure(figph[The figure shows the graph of the inequalities c plus a less than or equal to twenty four and a greater than or equal to three times c. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.], alt: "The figure shows the graph of the inequalities c plus a less than or equal to twenty four and a greater than or equal to three times c. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.", caption: none) #linebreak() ⓒ yes #linebreak() ⓓ no ] Jocelyn is pregnant and so she needs to eat at least 500 more calories a day than usual. When buying groceries one day with a budget of \$15 for the extra food, she buys bananas that have 90 calories each and chocolate granola bars that have 150 calories each. The bananas cost \$0.35 each and the granola bars cost \$2.50 each. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could she buy 5 bananas and 6 granola bars? #linebreak() ⓓ Could she buy 3 bananas and 4 granola bars? Mark is attempting to build muscle mass and so he needs to eat at least an additional 80 grams of protein a day. A bottle of protein water costs \$3.20 and a protein bar costs \$1.75. The protein water supplies 27 grams of protein and the bar supplies 16 gram. If he has \$10 dollars to spend ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could he buy 3 bottles of protein water and 1 protein bar? #linebreak() ⓓ Could he buy no bottles of protein water and 5 protein bars? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ w greater than or equal to 0; b greater than or equal to 0; 27 w plus 16 b greater than 80; 3.20 w plus 1.75 b less than or equal to 10")[$\{ w ≥ 0 \ b ≥ 0 \ 27 w + 16 b > 80 \ 3.20 w + 1.75 b ≤ 10$] #linebreak() ⓑ #linebreak() #figure(figph[The figure shows the graph of the inequalities twenty seven times w plus sixteen times b greater than eighty and three point two times w plus one point seven five b less than or equal to ten. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.], alt: "The figure shows the graph of the inequalities twenty seven times w plus sixteen times b greater than eighty and three point two times w plus one point seven five b less than or equal to ten. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.", caption: none) #linebreak() ⓒ no #linebreak() ⓓ yes ] Jocelyn desires to increase both her protein consumption and caloric intake. She desires to have at least 35 more grams of protein each day and no more than an additional 200 calories daily. An ounce of cheddar cheese has 7 grams of protein and 110 calories. An ounce of parmesan cheese has 11 grams of protein and 22 calories. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could she eat 1 ounce of cheddar cheese and 3 ounces of parmesan cheese? #linebreak() ⓓ Could she eat 2 ounces of cheddar cheese and 1 ounce of parmesan cheese? Mark is increasing his exercise routine by running and walking at least 4 miles each day. His goal is to burn a minimum of 1500 calories from this exercise. Walking burns 270 calories/mile and running burns 650 calories. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Could he meet his goal by walking 3 miles and running 1 mile? #linebreak() ⓓ Could he meet his goal by walking 2 miles and running 2 miles? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ w greater than or equal to 0; r greater than or equal to 0; w plus r greater than or equal to 4; 270 w plus 650 r greater than or equal to 1500")[$\{ w ≥ 0 \ r ≥ 0 \ w + r ≥ 4 \ 270 w + 650 r ≥ 1500$] #linebreak() ⓑ #linebreak() #figure(figph[The figure shows the graph of the inequalities w plus r greater than or equals to four and two seventy w plus six fifty r greater than or equal to fifteen hundred. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.], alt: "The figure shows the graph of the inequalities w plus r greater than or equals to four and two seventy w plus six fifty r greater than or equal to fifteen hundred. Two intersecting lines, one in blue and the other in red, are shown. An area is shown in grey.", caption: none) #linebreak() ⓒ no #linebreak() ⓓ yes ] ==== Writing Exercises Graph the inequality #math.equation(block: false, alt: "x minus y greater than or equal to 3 .")[$x − y ≥ 3 .$] How do you know which side of the line #math.equation(block: false, alt: "x minus y equals 3")[$x − y = 3$] should be shaded? Graph the system #math.equation(block: false, alt: "{ x plus 2 y less than or equal to 6; y greater than or equal to minus the fraction 1 over 2 x minus 4 .")[$\{ x + 2 y ≤ 6 \ y ≥ − frac(1, 2) x − 4 .$] What does the solution mean? #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[The figure shows a table with four columns and four rows. The first row is the title row. The four titles are I can…, confidently, with some help and No – I don’t get it! In the second row of the first column, the text says ‘determine whether an ordered pair is a solution of a system of linear inequalities.’ In the third row of the first column, the text says ‘solve applications of systems of inequalities.’], alt: "The figure shows a table with four columns and four rows. The first row is the title row. The four titles are I can…, confidently, with some help and No – I don’t get it! In the second row of the first column, the text says ‘determine whether an ordered pair is a solution of a system of linear inequalities.’ In the third row of the first column, the text says ‘solve applications of systems of inequalities.’", caption: none) ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve? === Chapter Review Exercises ==== Solve Systems of Linear Equations with Two Variables #strong[Determine Whether an Ordered Pair is a Solution of a System of Equations]. In the following exercises, determine if the following points are solutions to the given system of equations. #math.equation(block: false, alt: "{ x plus 3 y equals −9; 2 x minus 4 y equals 12")[$\{ x + 3 y = −9 \ 2 x − 4 y = 12$] #linebreak() ⓐ #math.equation(block: false, alt: "open parenthesis −3 , −2 close parenthesis")[$( −3 , −2 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 0 , −3 close parenthesis")[$( 0 , −3 )$] #math.equation(block: false, alt: "{ x plus y equals 8; y equals x minus 4")[$\{ x + y = 8 \ y = x − 4$] #linebreak() ⓐ #math.equation(block: false, alt: "open parenthesis 6 , 2 close parenthesis")[$( 6 , 2 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 9 , −1 close parenthesis")[$( 9 , −1 )$] #solutionbox[ ⓐ yes ⓑ no ] #strong[Solve a System of Linear Equations by Graphing] In the following exercises, solve the following systems of equations by graphing. #math.equation(block: true, alt: "{ 3 x plus y equals 6; x plus 3 y equals −6")[$\{ 3 x + y = 6 \ x + 3 y = −6$] #math.equation(block: true, alt: "{ x plus 4 y equals −1; x equals 3")[$\{ x + 4 y = −1 \ x = 3$] #solutionbox[ #figure(figph[The figure shows the graph of equations x plus four times y equal to minus one and x equal to three. Two intersecting lines are shown.], alt: "The figure shows the graph of equations x plus four times y equal to minus one and x equal to three. Two intersecting lines are shown.", caption: none) #linebreak() #math.equation(block: true, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$] ] #math.equation(block: true, alt: "{ 2 x minus y equals 5; 4 x minus 2 y equals 10")[$\{ 2 x − y = 5 \ 4 x − 2 y = 10$] #math.equation(block: true, alt: "{ − x plus 2 y equals 4; y equals the fraction 1 over 2 x minus 3")[$\{ "−" x + 2 y = 4 \ y = frac(1, 2) x − 3$] #solutionbox[ #figure(figph[The figure shows the graph for the equations minus x plus two times y equal to four and y equal to half x minus three. Two parallel lines are shown.], alt: "The figure shows the graph for the equations minus x plus two times y equal to four and y equal to half x minus three. Two parallel lines are shown.", caption: none) #linebreak() no solution ] In the following exercises, without graphing determine the number of solutions and then classify the system of equations. #math.equation(block: true, alt: "{ y equals the fraction 2 over 5 x plus 2; −2 x plus 5 y equals 10")[$\{ y = frac(2, 5) x + 2 \ −2 x + 5 y = 10$] #math.equation(block: true, alt: "{ 3 x plus 2 y equals 6; y equals −3 x plus 4")[$\{ 3 x + 2 y = 6 \ y = −3 x + 4$] #solutionbox[ one solution, consistent system, independent equations ] #math.equation(block: true, alt: "{ 5 x minus 4 y equals 0; y equals the fraction 5 over 4 x minus 5")[$\{ 5 x − 4 y = 0 \ y = frac(5, 4) x − 5$] #strong[Solve a System of Equations by Substitution] In the following exercises, solve the systems of equations by substitution. #math.equation(block: true, alt: "{ 3 x minus 2 y equals 2; y equals the fraction 1 over 2 x plus 3")[$\{ 3 x − 2 y = 2 \ y = frac(1, 2) x + 3$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 , 5 close parenthesis")[$( 4 , 5 )$] ] #math.equation(block: true, alt: "{ x minus y equals 0; 2 x plus 5 y equals −14")[$\{ x − y = 0 \ 2 x + 5 y = −14$] #math.equation(block: true, alt: "{ y equals −2 x plus 7; y equals the fraction 2 over 3 x minus 1")[$\{ y = −2 x + 7 \ y = frac(2, 3) x − 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] ] #math.equation(block: true, alt: "{ y equals −5 x; 5 x plus y equals 6")[$\{ y = −5 x \ 5 x + y = 6$] #math.equation(block: true, alt: "{ y equals minus the fraction 1 over 3 x plus 2; x plus 3 y equals 6")[$\{ y = − frac(1, 3) x + 2 \ x + 3 y = 6$] #solutionbox[ infinitely many solutions ] #strong[Solve a System of Equations by Elimination] In the following exercises, solve the systems of equations by elimination #math.equation(block: true, alt: "{ x plus y equals 12; x minus y equals −10")[$\{ x + y = 12 \ x − y = −10$] #math.equation(block: true, alt: "{ 3 x minus 8 y equals 20; x plus 3 y equals 1")[$\{ 3 x − 8 y = 20 \ x + 3 y = 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 , −1 close parenthesis")[$( 4 , −1 )$] ] #math.equation(block: true, alt: "{ 9 x plus 4 y equals 2; 5 x plus 3 y equals 5")[$\{ 9 x + 4 y = 2 \ 5 x + 3 y = 5$] #math.equation(block: true, alt: "{ the fraction 1 over 3 x minus the fraction 1 over 2 y equals 1; the fraction 3 over 4 x minus y equals the fraction 5 over 2")[$\{ frac(1, 3) x − frac(1, 2) y = 1 \ frac(3, 4) x − y = frac(5, 2)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 6 , 2 close parenthesis")[$( 6 , 2 )$] ] #math.equation(block: true, alt: "{ − x plus 3 y equals 8; 2 x minus 6 y equals −20")[$\{ "−" x + 3 y = 8 \ 2 x − 6 y = −20$] #strong[Choose the Most Convenient Method to Solve a System of Linear Equations] In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination. #math.equation(block: true, alt: "{ 6 x minus 5 y equals 27; 3 x plus 10 y equals −24")[$\{ 6 x − 5 y = 27 \ 3 x + 10 y = −24$] #solutionbox[ elimination ] #math.equation(block: true, alt: "{ y equals 3 x minus 9; 4 x minus 5 y equals 23")[$\{ y = 3 x − 9 \ 4 x − 5 y = 23$] ==== Solve Applications with Systems of Equations #strong[Solve Direct Translation Applications] In the following exercises, translate to a system of equations and solve. Mollie wants to plant 200 bulbs in her garden, all irises and tulips. She wants to plant three times as many tulips as irises. How many irises and how many tulips should she plant? #solutionbox[ 50 irises and 150 tulips ] Ashanti has been offered positions by two phone companies. The first company pays a salary of \$22,000 plus a commission of \$100 for each contract sold. The second pays a salary of \$28,000 plus a commission of \$25 for each contract sold. How many contract would need to be sold to make the total pay the same? Leroy spent 20 minutes jogging and 40 minutes cycling and burned 600 calories. The next day, Leroy swapped times, doing 40 minutes of jogging and 20 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling? #solutionbox[ 10 calories jogging and 10 calories cycling ] Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for \$116. Lisa bought two notebooks and three thumb drives for \$68. Find the cost of each notebook and each thumb drive. #strong[Solve Geometry Applications] In the following exercises, translate to a system of equations and solve. The difference of two supplementary angles is 58 degrees. Find the measures of the angles. #solutionbox[ 119 degrees and 61 degrees ] Two angles are complementary. The measure of the larger angle is five more than four times the measure of the smaller angle. Find the measures of both angles. The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles. #solutionbox[ 35 degrees and 55 degrees ] Becca is hanging a 28 foot floral garland on the two sides and top of a pergola to prepare for a wedding. The height is four feet less than the width. Find the height and width of the pergola. The perimeter of a city rectangular park is 1428 feet. The length is 78 feet more than twice the width. Find the length and width of the park. #solutionbox[ Length = 502 feet, Width = 212 feet ] #strong[Solve Uniform Motion Applications] In the following exercises, translate to a system of equations and solve. Sheila and Lenore were driving to their grandmother’s house. Lenore left one hour after Sheila. Sheila drove at a rate of 45 mph, and Lenore drove at a rate of 60 mph. How long will it take for Lenore to catch up to Sheila? Bob left home, riding his bike at a rate of 10 miles per hour to go to the lake. Cheryl, his wife, left 45 minutes #math.equation(block: false, alt: "open parenthesis the fraction 3 over 4")[$\( frac(3, 4)$] hour) later, driving her car at a rate of 25 miles per hour. How long will it take Cheryl to catch up to Bob? #solutionbox[ #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] an hour ] Marcus can drive his boat 36 miles down the river in three hours but takes four hours to return upstream. Find the rate of the boat in still water and the rate of the current. A passenger jet can fly 804 miles in 2 hours with a tailwind but only 776 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind. #solutionbox[ the rate of the jet is 395 mph, the rate of the wind is 7 mph ] ==== Solve Mixture Applications with Systems of Equations #strong[Solve Mixture Applications with Systems of Equations] For the following exercises, translate to a system of equations and solve. Lynn paid a total of \$2,780 for 261 tickets to the theater. Student tickets cost \$10 and adult tickets cost \$15. How many student tickets and how many adult tickets did Lynn buy? Priam has dimes and pennies in a cup holder in his car. The total value of the coins is \$4.21. The number of dimes is three less than four times the number of pennies. How many dimes and how many pennies are in the cup? #solutionbox[ 41 dimes and 11 pennies ] Yumi wants to make 12 cups of party mix using candies and nuts. Her budget requires the party mix to cost her \$1.29 per cup. The candies are \$2.49 per cup and the nuts are \$0.69 per cup. How many cups of candies and how many cups of nuts should she use? A scientist needs 70 liters of a 40% solution of alcohol. He has a 30% and a 60% solution available. How many liters of the 30% and how many liters of the 60% solutions should he mix to make the 40% solution? #solutionbox[ #math.equation(block: false, alt: "46 the fraction 2 over 3")[$46 frac(2, 3)$] liters of 30% solution, #math.equation(block: false, alt: "23 the fraction 1 over 3")[$23 frac(1, 3)$] liters of 60% solution ] #strong[Solve Interest Applications] For the following exercises, translate to a system of equations and solve. Jack has \$12,000 to invest and wants to earn 7.5% interest per year. He will put some of the money into a savings account that earns 4% per year and the rest into CD account that earns 9% per year. How much money should he put into each account? When she graduates college, Linda will owe \$43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was \$1,585. What is the amount of each loan? #solutionbox[ \$29,000 for the federal loan, \$14,000 for the private loan ] ==== Solve Systems of Equations with Three Variables #strong[Solve Systems of Equations with Three Variables] In the following exercises, determine whether the ordered triple is a solution to the system. #math.equation(block: false, alt: "{ 3 x minus 4 y minus 3 z equals 2; 2 x minus 6 y plus z equals 3; 2 x plus 3 y minus 2 z equals 3")[$\{ 3 x − 4 y − 3 z = 2 \ 2 x − 6 y + z = 3 \ 2 x + 3 y − 2 z = 3$] #linebreak() ⓐ #math.equation(block: false, alt: "open parenthesis 2 , 3 , −1 close parenthesis")[$( 2 , 3 , −1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 3 , 1 , 3 close parenthesis")[$( 3 , 1 , 3 )$] #math.equation(block: false, alt: "{ y equals the fraction 2 over 3 x minus 2; x plus 3 y minus z equals 15; x minus 3 y plus z equals −2")[$\{ y = frac(2, 3) x − 2 \ x + 3 y − z = 15 \ x − 3 y + z = −2$] #linebreak() ⓐ #math.equation(block: false, alt: "open parenthesis −6 , 5 , the fraction 1 over 2 close parenthesis")[$( −6 , 5 , frac(1, 2) )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 5 , the fraction 4 over 3 , −3 close parenthesis")[$( 5 , frac(4, 3) , −3 )$] #solutionbox[ ⓐ no ⓑ no ] #strong[Solve a System of Linear Equations with Three Variables] In the following exercises, solve the system of equations. #math.equation(block: true, alt: "{ 3 x minus 5 y plus 4 z equals 5; 5 x plus 2 y plus z equals 0; 2 x plus 3 y minus 2 z equals 3")[$\{ 3 x − 5 y + 4 z = 5 \ 5 x + 2 y + z = 0 \ 2 x + 3 y − 2 z = 3$] #math.equation(block: true, alt: "{ x plus the fraction 5 over 2 y plus z equals −2; 2 x plus 2 y plus the fraction 1 over 2 z equals −4; the fraction 1 over 3 x minus y minus z equals 1")[$\{ x + frac(5, 2) y + z = −2 \ 2 x + 2 y + frac(1, 2) z = −4 \ frac(1, 3) x − y − z = 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −3 , 2 , −4 close parenthesis")[$( −3 , 2 , −4 )$] ] #math.equation(block: true, alt: "{ 5 x plus 3 y equals −6; 2 y plus 3 z equals −1; 7 x plus z equals 1")[$\{ 5 x + 3 y = −6 \ 2 y + 3 z = −1 \ 7 x + z = 1$] #math.equation(block: true, alt: "{ 2 x plus 3 y plus z equals 12; x plus y plus z equals 9; 3 x plus 4 y plus 2 z equals 20")[$\{ 2 x + 3 y + z = 12 \ x + y + z = 9 \ 3 x + 4 y + 2 z = 20$] #solutionbox[ no solution ] #math.equation(block: true, alt: "{ − x minus 3 y plus 2 z equals 14; − x plus 2 y minus 3 z equals −4; 3 x plus y minus 2 z equals 6")[$\{ "−" x − 3 y + 2 z = 14 \ "−" x + 2 y − 3 z = −4 \ 3 x + y − 2 z = 6$] #strong[Solve Applications using Systems of Linear Equations with Three Variables] After attending a major league baseball game, the patrons often purchase souvenirs. If a family purchases 4 t-shirts, a cap and 1 stuffed animal their total is \$135. A couple buys 2 t-shirts, a cap and 3 stuffed animals for their nieces and spends \$115. Another couple buys 2 t-shirts, a cap and 1 stuffed animal and their total is \$85. What is the cost of each item? #solutionbox[ 25, 20, 15 ] ==== Solve Systems of Equations Using Matrices #strong[Write the Augmented Matrix for a System of Equations.] Write each system of linear equations as an augmented matrix. #math.equation(block: true, alt: "{ 3 x minus y equals −1; −2 x plus 2 y equals 5")[$\{ 3 x − y = −1 \ −2 x + 2 y = 5$] #math.equation(block: true, alt: "{ 4 x plus 3 y equals −2; x minus 2 y minus 3 z equals 7; 2 x minus y plus 2 z equals −6")[$\{ 4 x + 3 y = −2 \ x − 2 y − 3 z = 7 \ 2 x − y + 2 z = −6$] #solutionbox[ #math.equation(block: true, alt: "[ 4, 3, 0, −2; 1, −2, −3, 7; 2, −1, 2, −6 ]")[$[ 4 & & & 3 & & & 0 & & & −2 \ 1 & & & −2 & & & −3 & & & 7 \ 2 & & & −1 & & & 2 & & & −6 ]$] ] Write the system of equations that that corresponds to the augmented matrix. #math.equation(block: true, alt: "[ 2, −4; 3, −3 | −2; −1 ]")[$[ 2 & & & −4 \ 3 & & & −3 #h(1em) | #h(1em) −2 \ −1 ]$] #math.equation(block: true, alt: "[ 1, 0, −3; 1, −2, 0; 0, −1, 2 | −1; −2; 3 ]")[$[ 1 & & & 0 & & & −3 \ 1 & & & −2 & & & 0 \ 0 & & & −1 & & & 2 #h(1em) | #h(1em) −1 \ −2 \ 3 ]$] #solutionbox[ #math.equation(block: true, alt: "{ x minus 3 z equals −1; x minus 2 y equals −2; − y plus 2 z equals 3")[$\{ x − 3 z = −1 \ x − 2 y = −2 \ "−" y + 2 z = 3$] ] In the following exercises, perform the indicated operations on the augmented matrices. #math.equation(block: true, alt: "[ 4, −6; 3, 2 | −3; 1 ]")[$[ 4 & & & −6 \ 3 & & & 2 #h(1em) | #h(1em) −3 \ 1 ]$] ⓐ Interchange rows 2 and 1. #linebreak() ⓑ Multiply row 1 by 4. #linebreak() ⓒ Multiply row 2 by 3 and add to row 1. #math.equation(block: true, alt: "[ 1, −3, −2; 2, 2, −1; 4, −2, −3 | 4; −3; −1 ]")[$[ 1 & & & −3 & & & −2 \ 2 & & & 2 & & & −1 \ 4 & & & −2 & & & −3 #h(1em) | #h(1em) 4 \ −3 \ −1 ]$] ⓐ Interchange rows 2 and 3. #linebreak() ⓑ Multiply row 1 by 2. #linebreak() ⓒ Multiply row 3 by #math.equation(block: false, alt: "−2")[$−2$] and add to row 2. #solutionbox[ ⓐ #math.equation(block: false, alt: "[ 1, −3, −2, 4; 4, −2, −3, −1; 2, 2, −1, −3 ]")[$[ 1 & & & −3 & & & −2 & & & 4 \ 4 & & & −2 & & & −3 & & & −1 \ 2 & & & 2 & & & −1 & & & −3 ]$] #linebreak() ⓑ #math.equation(block: false, alt: "[ 2, −6, −4, 8; 4, −2, −3, −1; 2, 2, −1, −3 ]")[$[ 2 & & & −6 & & & −4 & & & 8 \ 4 & & & −2 & & & −3 & & & −1 \ 2 & & & 2 & & & −1 & & & −3 ]$] #linebreak() ⓒ #math.equation(block: false, alt: "[ 2, −6, −4, 8; 4, −2, −3, −1; 0, −6, −1, 5 ]")[$[ 2 & & & −6 & & & −4 & & & 8 \ 4 & & & −2 & & & −3 & & & −1 \ 0 & & & −6 & & & −1 & & & 5 ]$] ] #strong[Solve Systems of Equations Using Matrices] In the following exercises, solve each system of equations using a matrix. #math.equation(block: true, alt: "{ 4 x plus y equals 6; x minus y equals 4")[$\{ 4 x + y = 6 \ x − y = 4$] #math.equation(block: true, alt: "{ 2 x minus y plus 3 z equals −3; − x plus 2 y minus z equals 10; x plus y plus z equals 5")[$\{ 2 x − y + 3 z = −3 \ "−" x + 2 y − z = 10 \ x + y + z = 5$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −2 , 5 , 2 close parenthesis")[$( −2 , 5 , 2 )$] ] #math.equation(block: true, alt: "{ 2 y plus 3 z equals −1; 5 x plus 3 y equals −6; 7 x plus z equals 1")[$\{ 2 y + 3 z = −1 \ 5 x + 3 y = −6 \ 7 x + z = 1$] #math.equation(block: true, alt: "{ x plus 2 y minus 3 z equals −1; x minus 3 y plus z equals 1; 2 x minus y minus 2 z equals 2")[$\{ x + 2 y − 3 z = −1 \ x − 3 y + z = 1 \ 2 x − y − 2 z = 2$] #solutionbox[ no solution ] #math.equation(block: true, alt: "{ x plus y minus 3 z equals −1; y minus z equals 0; − x plus 2 y equals 1")[$\{ x + y − 3 z = −1 \ y − z = 0 \ "−" x + 2 y = 1$] ==== Solve Systems of Equations Using Determinants #strong[Evaluate the Determinant of a 2 × 2 Matrix] In the following exercise, evaluate the determinate of the square matrix. #math.equation(block: true, alt: "[ 8, −4; 5, −3 ]")[$[ 8 & & & −4 \ 5 & & & −3 ]$] #solutionbox[ #math.equation(block: true, alt: "−4")[$−4$] ] #strong[Evaluate the Determinant of a 3 × 3 Matrix] In the following exercise, find and then evaluate the indicated minors. #math.equation(block: false, alt: "| −1, −3, 2; 4, −2, −1; −2, 0, −3 | ;")[$| −1 & & & −3 & & & 2 \ 4 & & & −2 & & & −1 \ −2 & & & 0 & & & −3 | ;$] Find the minor ⓐ #math.equation(block: false, alt: "a sub 1")[$a_(1)$] ⓑ #math.equation(block: false, alt: "b sub 1")[$b_(1)$] ⓒ #math.equation(block: false, alt: "c sub 2")[$c_(2)$] In the following exercise, evaluate each determinant by expanding by minors along the first row. #math.equation(block: true, alt: "| −2, −3, −4; 5, −6, 7; −1, 2, 0 |")[$| −2 & & & −3 & & & −4 \ 5 & & & −6 & & & 7 \ −1 & & & 2 & & & 0 |$] #solutionbox[ 33 ] In the following exercise, evaluate each determinant by expanding by minors. #math.equation(block: true, alt: "| 3, 5, 4; −1, 3, 0; −2, 6, 1 |")[$| 3 & & & 5 & & & 4 \ −1 & & & 3 & & & 0 \ −2 & & & 6 & & & 1 |$] #strong[Use Cramer’s Rule to Solve Systems of Equations] In the following exercises, solve each system of equations using Cramer’s rule #math.equation(block: true, alt: "{ x minus 3 y equals −9; 2 x plus 5 y equals 4")[$\{ x − 3 y = −9 \ 2 x + 5 y = 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −3 , 2 close parenthesis")[$( −3 , 2 )$] ] #math.equation(block: true, alt: "{ 4 x minus 3 y plus z equals 7; 2 x minus 5 y minus 4 z equals 3; 3 x minus 2 y minus 2 z equals −7")[$\{ 4 x − 3 y + z = 7 \ 2 x − 5 y − 4 z = 3 \ 3 x − 2 y − 2 z = −7$] #math.equation(block: true, alt: "{ 2 x plus 5 y equals 4; 3 y minus z equals 3; 4 x plus 3 z equals −3")[$\{ 2 x + 5 y = 4 \ 3 y − z = 3 \ 4 x + 3 z = −3$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −3 , 2 , 3 close parenthesis")[$( −3 , 2 , 3 )$] ] #math.equation(block: true, alt: "{ x plus y minus 3 z equals −1; y minus z equals 0; − x plus 2 y equals 1")[$\{ x + y − 3 z = −1 \ y − z = 0 \ "−" x + 2 y = 1$] #math.equation(block: true, alt: "{ 3 x plus 4 y minus 3 z equals −2; 2 x plus 3 y minus z equals −12; x plus y minus 2 z equals 6")[$\{ 3 x + 4 y − 3 z = −2 \ 2 x + 3 y − z = −12 \ x + y − 2 z = 6$] #solutionbox[ inconsistent ] #strong[Solve Applications Using Determinants] In the following exercises, determine whether the given points are collinear. #math.equation(block: false, alt: "open parenthesis 0 , 2 close parenthesis ,")[$( 0 , 2 ) ,$] #math.equation(block: false, alt: "open parenthesis −1 , −1 close parenthesis ,")[$( −1 , −1 ) ,$] and #math.equation(block: false, alt: "open parenthesis −2 , 4 close parenthesis")[$( −2 , 4 )$] ==== Graphing Systems of Linear Inequalities #strong[Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities] In the following exercises, determine whether each ordered pair is a solution to the system. #math.equation(block: true, alt: "{ 4 x plus y greater than 6; 3 x minus y less than or equal to 12")[$\{ 4 x + y > 6 \ 3 x − y ≤ 12$] ⓐ #math.equation(block: false, alt: "open parenthesis 2 , −1 close parenthesis")[$( 2 , −1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 3 , −2 close parenthesis")[$( 3 , −2 )$] #solutionbox[ ⓐ yes ⓑ yes ] #math.equation(block: true, alt: "{ y greater than the fraction 1 over 3 x plus 2; x minus the fraction 1 over 4 y less than or equal to 10")[$\{ y > frac(1, 3) x + 2 \ x − frac(1, 4) y ≤ 10$] ⓐ #math.equation(block: false, alt: "open parenthesis 6 , 5 close parenthesis")[$( 6 , 5 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 15 , 8 close parenthesis")[$( 15 , 8 )$] #strong[Solve a System of Linear Inequalities by Graphing] In the following exercises, solve each system by graphing. #math.equation(block: true, alt: "{ y less than or equal to 3 x plus 2; y greater than minus x minus 1")[$\{ y ≤ 3 x + 2 \ y > − x − 1$] #solutionbox[ #figure(figph[The figure shows the graph of inequalities y less than three times x plus one and y greater than or equal to minus x minus two. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.], alt: "The figure shows the graph of inequalities y less than three times x plus one and y greater than or equal to minus x minus two. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ x minus y greater than −1; y less than the fraction 1 over 3 x minus 2")[$\{ x − y > −1 \ y < frac(1, 3) x − 2$] #math.equation(block: true, alt: "{ 2 x minus 3 y less than 6; 3 x plus 4 y greater than or equal to 12")[$\{ 2 x − 3 y < 6 \ 3 x + 4 y ≥ 12$] #solutionbox[ #figure(figph[The figure shows the graph of inequalities two times x minus three times y less six and three times x plus four times y greater than or equal to twelve. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.], alt: "The figure shows the graph of inequalities two times x minus three times y less six and three times x plus four times y greater than or equal to twelve. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.", caption: none) #linebreak() The solution is the grey region. ] #math.equation(block: true, alt: "{ y less than or equal to minus the fraction 3 over 4 x plus 1; x greater than or equal to −5")[$\{ y ≤ − frac(3, 4) x + 1 \ x ≥ −5$] #math.equation(block: true, alt: "{ x plus 3 y less than 5; y greater than or equal to minus the fraction 1 over 3 x plus 6")[$\{ x + 3 y < 5 \ y ≥ − frac(1, 3) x + 6$] #solutionbox[ #figure(figph[The figure shows the graph of inequalities x plus three times y less than five and y greater than or equal to minus one third x plus six. Two parallel lines, one in red and the other in blue, are shown. An area is shown in grey.], alt: "The figure shows the graph of inequalities x plus three times y less than five and y greater than or equal to minus one third x plus six. Two parallel lines, one in red and the other in blue, are shown. An area is shown in grey.", caption: none) #linebreak() No solution. ] #math.equation(block: true, alt: "{ y greater than or equal to 2 x minus 5; −6 x plus 3 y greater than −4")[$\{ y ≥ 2 x − 5 \ −6 x + 3 y > −4$] #strong[Solve Applications of Systems of Inequalities] In the following exercises, translate to a system of inequalities and solve. Roxana makes bracelets and necklaces and sells them at the farmers’ market. She sells the bracelets for \$12 each and the necklaces for \$18 each. At the market next weekend she will have room to display no more than 40 pieces, and she needs to sell at least \$500 worth in order to earn a profit. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Should she display 26 bracelets and 14 necklaces? #linebreak() ⓓ Should she display 39 bracelets and 1 necklace? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ b greater than or equal to 0; n greater than or equal to 0; b plus n less than or equal to 40; 12 b plus 18 n greater than or equal to 500")[$\{ b ≥ 0 \ n ≥ 0 \ b + n ≤ 40 \ 12 b + 18 n ≥ 500$] #linebreak() ⓑ #linebreak() #figure(figph[The figure shows the graph of b plus n equal to forty and twelve b plus eighteen n equal to five hundred. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.], alt: "The figure shows the graph of b plus n equal to forty and twelve b plus eighteen n equal to five hundred. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.", caption: none) #linebreak() ⓒ yes #linebreak() ⓓ no ] Annie has a budget of \$600 to purchase paperback books and hardcover books for her classroom. She wants the number of hardcover to be at least 5 more than three times the number of paperback books. Paperback books cost \$4 each and hardcover books cost \$15 each. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can she buy 8 paperback books and 40 hardcover books? #linebreak() ⓓ Can she buy 10 paperback books and 37 hardcover books? === Chapter Practice Test In the following exercises, solve the following systems by graphing. #math.equation(block: true, alt: "{ x minus y equals 5; x plus 2 y equals −4")[$\{ x − y = 5 \ x + 2 y = −4$] #solutionbox[ #figure(figph[The figure shows the graph of inequalities h equal to three p plus five and four times p plus fifteen times h equal to six hundred. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.], alt: "The figure shows the graph of inequalities h equal to three p plus five and four times p plus fifteen times h equal to six hundred. Two intersecting lines, one in red and the other in blue, are shown. An area is shown in grey.", caption: none) #linebreak() #math.equation(block: true, alt: "open parenthesis 2 , −3 close parenthesis")[$( 2 , −3 )$] ] #math.equation(block: true, alt: "{ x minus y greater than −2; y less than or equal to 3 x plus 1")[$\{ x − y > −2 \ y ≤ 3 x + 1$] In the following exercises, solve each system of equations. Use either substitution or elimination. #math.equation(block: true, alt: "{ x plus 4 y equals 6; −2 x plus y equals −3")[$\{ x + 4 y = 6 \ −2 x + y = −3$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$] ] #math.equation(block: true, alt: "{ −3 x plus 4 y equals 25; x minus 5 y equals −23")[$\{ −3 x + 4 y = 25 \ x − 5 y = −23$] #math.equation(block: true, alt: "{ x plus y minus z equals −1; 2 x minus y plus 2 z equals 8; −3 x plus 2 y plus z equals −9")[$\{ x + y − z = −1 \ 2 x − y + 2 z = 8 \ −3 x + 2 y + z = −9$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 , −2 , 1 close parenthesis")[$( 2 , −2 , 1 )$] ] Solve the system of equations using a matrix. #math.equation(block: true, alt: "{ 2 x plus y equals 7; x minus 2 y equals 6")[$\{ 2 x + y = 7 \ x − 2 y = 6$] #math.equation(block: true, alt: "{ −3 x plus y plus z equals −4; − x plus 2 y minus 2 z equals 1; 2 x minus y minus z equals −1")[$\{ −3 x + y + z = −4 \ "−" x + 2 y − 2 z = 1 \ 2 x − y − z = −1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 , 7 , 4 close parenthesis")[$( 5 , 7 , 4 )$] ] Solve using Cramer’s rule. #math.equation(block: true, alt: "{ 3 x plus y equals −3; 2 x plus 3 y equals 6")[$\{ 3 x + y = −3 \ 2 x + 3 y = 6$] Evaluate the determinant by expanding by minors: #linebreak() #math.equation(block: false, alt: "| 3, −2, −2; 2, −1, 4; −1, 0, −3 | .")[$| 3 & & & −2 & & & −2 \ 2 & & & −1 & & & 4 \ −1 & & & 0 & & & −3 | .$] #solutionbox[ #math.equation(block: true, alt: "7")[$7$] ] In the following exercises, translate to a system of equations and solve. Greg is paddling his canoe upstream, against the current, to a fishing spot 10 miles away. If he paddles upstream for 2.5 hours and his return trip takes 1.25 hours, find the speed of the current and his paddling speed in still water. A pharmacist needs 20 liters of a 2% saline solution. He has a 1% and a 5% solution available. How many liters of the 1% and how many liters of the 5% solutions should she mix to make the 2% solution? #solutionbox[ 15 liters of 1% solution, 5 liters of 5% solution ] Arnold invested \$64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received \$4,500 in interest in one year? The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of \$65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of \$140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 can of popcorn for a total sales of \$250. What is the cost of each item? #solutionbox[ The candy cost \$20; the cookies cost \$5; and the popcorn cost \$10. ] The manufacturer of a granola bar spends \$1.20 to make each bar and sells them for \$2. The manufacturer also has fixed costs each month of \$8,000. ⓐ Find the cost function #emph[C] when #emph[x] granola bars are manufactured #linebreak() ⓑ Find the revenue function #emph[R] when #emph[x] granola bars are sold. #linebreak() ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. #linebreak() ⓓ Find the break-even point. Interpret what the break-even point means. Translate to a system of inequalities and solve. Andi wants to spend no more than \$50 on Halloween treats. She wants to buy candy bars that cost \$1 each and lollipops that cost \$0.50 each, and she wants the number of lollipops to be at least three times the number of candy bars. ⓐ Write a system of inequalities to model this situation. #linebreak() ⓑ Graph the system. #linebreak() ⓒ Can she buy 20 candy bars and 40 lollipops? #solutionbox[ ⓐ #math.equation(block: false, alt: "{ C greater than or equal to 0; L greater than or equal to 0; C plus 0.5 L less than or equal to 50; L greater than or equal to 3 C")[$\{ C ≥ 0 \ L ≥ 0 \ C + 0.5 L ≤ 50 \ L ≥ 3 C$] #linebreak() ⓑ #linebreak() #figure(figph[The figure shows the graph of two equations. Two intersecting lines, one in red and the other in blue, are shown. The red line passes through origin. An area is shown in grey.], alt: "The figure shows the graph of two equations. Two intersecting lines, one in red and the other in blue, are shown. The red line passes through origin. An area is shown in grey.", caption: none) #linebreak() ⓒ no #linebreak() ⓓ yes ]