{"id":700,"date":"2016-06-01T20:49:13","date_gmt":"2016-06-01T20:49:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=700"},"modified":"2019-02-13T09:37:48","modified_gmt":"2019-02-13T09:37:48","slug":"outcome-graphing-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/outcome-graphing-linear-equations\/","title":{"raw":"6.4 - Linear Inequalities and Systems of Linear Inequalities in Two Variables","rendered":"6.4 &#8211; Linear Inequalities and Systems of Linear Inequalities in Two Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(6.4.1) - Define solutions to a linear inequality in two variables<\/li>\r\n \t<li>(6.4.2) - Graphing the solution set to a linear inequality in two variables<\/li>\r\n \t<li>(6.4.3) - Graph a system of linear inequalities and define the solutions region<\/li>\r\n \t<li>(6.4.4) - Determine whether a point is a solution to a system of inequalities<\/li>\r\n \t<li style=\"list-style-type: none\"><\/li>\r\n \t<li>(6.4.5) - Identify when a system of inequalities has no solution<\/li>\r\n \t<li>(6.4.6) - Applications of systems of linear inequalities<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>(6.4.1) - Define solutions to a linear inequality in two variables<\/h1>\r\n<p id=\"fs-id1167832031143\">Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables. In particular we will look at\u00a0<strong>linear inequalities<\/strong>\u00a0in two variables which are very similar to linear equations in two variables.<\/p>\r\n<p id=\"fs-id1167835236709\">Linear inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs\u2014so that your business made a profit.<\/p>\r\n\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Linear Inequality in Two Variables<\/h3>\r\nA\u00a0<strong>linear inequality <\/strong>in two variables\u00a0is an inequality that can be written in one of the following forms:\r\n<p style=\"text-align: center\">[latex]Ax+By &lt; 0[\/latex]\u00a0 \u00a0 \u00a0 \u00a0[latex]Ax+By&gt;0[\/latex]\u00a0 \u00a0 \u00a0 [latex]Ax+By \\leq 0[\/latex]\u00a0 \u00a0 \u00a0[latex]Ax+By \\geq 0[\/latex]<\/p>\r\nwhere [latex]A[\/latex] and [latex]B[\/latex] are not both zero.\r\n\r\n<\/div>\r\nRecall that an inequality with one variable had many solutions. For example, the solution to the inequality [latex]x &gt; 3[\/latex]\u00a0is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3.\r\n\r\n<img class=\"alignnone size-medium wp-image-5408\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2016\/06\/19213517\/Math101_6_4_1im1-300x27.jpg\" alt=\"\" width=\"300\" height=\"27\" \/>\r\n\r\nSimilarly, linear inequalities in two variables have many solutions. Any ordered pair [latex](x,y)[\/latex]\u00a0that makes an inequality true when we substitute in the values is a\u00a0solution to a linear inequality.\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Solution to a linear inequality in two variables<\/h3>\r\nAn ordered pair\u00a0[latex](x,y)[\/latex] is a\u00a0<strong>solution to a linear inequality<\/strong>\u00a0if the inequality is true when we substitute the values of [latex]x[\/latex] and [latex]y[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nDetermine whether each ordered pair is a solution to the inequality [latex]y&gt;x+4[\/latex].\r\n\r\na)[latex](0,0)[\/latex]\r\n\r\nb)[latex](1,6)[\/latex]\r\n\r\n[reveal-answer q=\"837554\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"837554\"]\r\n\r\na)Substitute 0 for [latex]x[\/latex] and 0 for [latex]y[\/latex]:\r\n<p style=\"text-align: center\">[latex]0 \\overset{?}{&gt;} 0 + 4[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] 0 \\not\\gt 4[\/latex]<\/p>\r\nSo, [latex](0,0)[\/latex] is not a solution to\u00a0[latex]y&gt;x+4[\/latex].\r\n\r\nb)Substitute 1 for [latex]x[\/latex] and 6 for [latex]y[\/latex]:\r\n<p style=\"text-align: center\">[latex]6 \\overset{?}{&gt;} 1 + 4[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex] 6 &gt; 5[\/latex]<\/p>\r\nSo, [latex](1,6)[\/latex] is a solution to\u00a0[latex]y&gt;x+4[\/latex].\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h4>Solution Sets of Inequalities<\/h4>\r\nThe graph below shows the region of values that makes the inequality [latex]3x+2y\\leq6[\/latex] true (shaded red), the boundary line [latex]3x+2y=6[\/latex], as well as a handful of ordered pairs. The boundary line is solid because points on the boundary line [latex]3x+2y=6[\/latex]\u00a0will make the inequality [latex]3x+2y\\leq6[\/latex]\u00a0true.\r\n\r\n<img class=\"aligncenter wp-image-2873 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19172056\/Screen-Shot-2016-04-19-at-10.20.21-AM.png\" alt=\"A solid downward-sloping line running. The region below the line is shaded and is labeled 3x+2y is less than or equal to 6. The region above the line is unshaded and is labeled 3x+2y=6. The points (-5,5) and (-2,-2) are in the shaded region. The points (2,3) and (4,-1) are in the unshaded region. The point (2,0) is on the line.\" width=\"464\" height=\"472\" \/>\r\n\r\nYou can substitute the [latex]x[\/latex]- and [latex]y[\/latex]<i>-<\/i>values in each of the [latex](x,y)[\/latex] ordered pairs into the inequality to find solutions. Sometimes making a table of values makes sense for more complicated inequalities.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Ordered Pair<\/th>\r\n<th>Makes the inequality\r\n\r\n[latex]3x+2y\\leq6[\/latex]\r\n\r\na true statement<\/th>\r\n<th>Makes the inequality\r\n\r\n[latex]3x+2y\\leq6[\/latex]\r\n\r\na false statement<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex](\u22125, 5)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{r}3\\left(\u22125\\right)+2\\left(5\\right)\\leq6\\\\\u221215+10\\leq6\\\\\u22125\\leq6\\end{array}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex](\u22122,\u22122)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{r}3\\left(\u22122\\right)+2\\left(\u20132\\right)\\leq6\\\\\u22126+\\left(\u22124\\right)\\leq6\\\\\u201310\\leq6\\end{array}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex](2,3)[\/latex]<\/td>\r\n<td><\/td>\r\n<td>[latex]\\begin{array}{r}3\\left(2\\right)+2\\left(3\\right)\\leq6\\\\6+6\\leq6\\\\12\\leq6\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex](2,0)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{r}3\\left(2\\right)+2\\left(0\\right)\\leq6\\\\6+0\\leq6\\\\6\\leq6\\end{array}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex](4,\u22121)[\/latex]<\/td>\r\n<td><\/td>\r\n<td>[latex]\\begin{array}{r}3\\left(4\\right)+2\\left(\u22121\\right)\\leq6\\\\12+\\left(\u22122\\right)\\leq6\\\\10\\leq6\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIf substituting [latex](x,y)[\/latex] into the inequality yields a true statement, then the ordered pair is a solution to the inequality, and the point will be plotted within the shaded region or the point will be part of a solid boundary line. A false statement means that the ordered pair is not a solution, and the point will graph outside the shaded region, or the point will be part of a dotted boundary line.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the graph to determine which ordered pairs plotted below are solutions of the inequality\u00a0[latex]x\u2013y&lt;3[\/latex].\r\n\r\n<img class=\"aligncenter wp-image-2876 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19172536\/Screen-Shot-2016-04-19-at-10.25.12-AM.png\" alt=\"Upward-sloping dotted line. The region above the line is shaded and labeled x-y&lt;3. The points (4,0) and (3,-2) are in the unshaded region. The point (1,-2) is on the dotted line. The points (-1,1) and (-2,-2) are in the shaded region.\" width=\"410\" height=\"415\" \/>\r\n\r\n[reveal-answer q=\"840389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"840389\"]\r\n\r\nSolutions will be located in the shaded region. Since this is a \u201cless than\u201d problem, ordered pairs on the boundary line are not included in the solution set.\r\n\r\nThese values are located in the shaded region, so are solutions. (When substituted into the inequality\u00a0[latex]x\u2013y&lt;3[\/latex], they produce true statements.)\r\n<p style=\"text-align: center\">[latex](\u22121,1)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](\u22122,\u22122)[\/latex]<\/p>\r\nThese values are not located in the shaded region, so are not solutions. (When substituted into the inequality [latex]x-y&lt;3[\/latex], they produce false statements.)\r\n<p style=\"text-align: center\">[latex](1,\u22122)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](3,\u22122)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](4,0)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex](\u22121,1)\\,\\,\\,(\u22122,\u22122)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video show an example of determining whether an ordered pair is a solution to an inequality.\r\n\r\nhttps:\/\/youtu.be\/GQVdDRVq5_o\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs [latex](2,\u22123)[\/latex] a solution of the inequality [latex]y&lt;\u22123x+1[\/latex]?\r\n\r\n[reveal-answer q=\"746731\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746731\"]\r\n\r\nIf [latex](2,\u22123)[\/latex] is a solution, then it will yield a true statement when substituted into the inequality\u00a0[latex]y&lt;\u22123x+1[\/latex].\r\n<p style=\"text-align: center\">[latex]y&lt;\u22123x+1[\/latex]<\/p>\r\nSubstitute\u00a0[latex]x=2[\/latex] and [latex]y=\u22123[\/latex]\u00a0into inequality.\r\n<p style=\"text-align: center\">[latex]\u22123&lt;\u22123\\left(2\\right)+1[\/latex]<\/p>\r\nEvaluate.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\u22123&lt;\u22126+1\\\\\u22123&lt;\u22125\\end{array}[\/latex]<\/p>\r\nThis statement is <b>not <\/b>true, so the ordered pair [latex](2,\u22123)[\/latex] is <b>not <\/b>a solution.\r\n<h4>Answer<\/h4>\r\n[latex](2,\u22123)[\/latex] is not a solution.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows another example of determining whether an ordered pair is a solution to an inequality.\r\n\r\nhttps:\/\/youtu.be\/-x-zt_yM0RM\r\n<h1>(6.4.2) - Graphing the solution set to a linear inequality in two variables<\/h1>\r\nSo how do you get from the algebraic form of an inequality, like [latex]y&gt;3x+1[\/latex], to a graph of that inequality? Plotting inequalities is fairly straightforward if you follow a couple steps.\r\n<div class=\"textbox shaded\">\r\n<h3>Graphing Inequalities<\/h3>\r\nTo graph an inequality:\r\n<ul>\r\n \t<li>Graph the related boundary line. Replace the &lt;, &gt;, \u2264 or \u2265 sign in the inequality with = to find the equation of the boundary line.<\/li>\r\n \t<li>Identify at least one ordered pair on either side of the boundary line and substitute those [latex](x,y)[\/latex] values into the inequality. Shade the region that contains the ordered pairs that make the inequality a true statement.<b>\u00a0<\/b><\/li>\r\n \t<li>If points on the boundary line are solutions, then use a solid line for drawing the boundary line. This will happen for \u2264 or \u2265 inequalities.<\/li>\r\n \t<li>If points on the boundary line aren\u2019t solutions, then use a dotted line for the boundary line. This will happen for &lt; or &gt; inequalities.<\/li>\r\n<\/ul>\r\n<\/div>\r\nLet\u2019s graph the inequality [latex]x+4y\\leq4[\/latex].\r\n\r\nTo graph the boundary line, find at least two values that lie on the line [latex]x+4y=4[\/latex]. You can use the <i>x<\/i>- and <i>y<\/i>-intercepts for this equation by substituting 0 in for <i>x<\/i> first and finding the value of <i>y<\/i>; then substitute 0 in for <i>y<\/i> and find <i>x<\/i>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><b><i>x<\/i><\/b><\/td>\r\n<td><b><i>y<\/i><\/b><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points [latex](0,1)[\/latex] and [latex](4,0)[\/latex], and draw a line through these two points for the boundary line. The line is solid because \u2264 means \u201cless than or equal to,\u201d so all ordered pairs along the line are included in the solution set.\r\n\r\n<img class=\"aligncenter size-full wp-image-2936\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19230042\/Screen-Shot-2016-04-19-at-4.00.26-PM.png\" alt=\"Solid downward-sloping line that crosses the points (0,1) and (4,0). The point (-1,3) and the point (2,0) are also plotted.\" width=\"417\" height=\"419\" \/>\r\n\r\nThe next step is to find the region that contains the solutions. Is it above or below the boundary line? To identify the region where the inequality holds true, you can test a couple of ordered pairs, one on each side of the boundary line.\r\n\r\nIf you substitute [latex](\u22121,3)[\/latex] into\u00a0[latex]x+4y\\leq4[\/latex]:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\u22121+4\\left(3\\right)\\leq4\\\\\u22121+12\\leq4\\\\11\\leq4\\end{array}[\/latex]<\/p>\r\nThis is a false statement, since 11 is not less than or equal to 4.\r\n\r\nOn the other hand, if you substitute [latex](2,0)[\/latex] into\u00a0[latex]x+4y\\leq4[\/latex]:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2+4\\left(0\\right)\\leq4\\\\2+0\\leq4\\\\2\\leq4\\end{array}[\/latex]<\/p>\r\nThis is true! The region that includes [latex](2,0)[\/latex] should be shaded, as this is the region of solutions.\r\n\r\n<img class=\"aligncenter size-full wp-image-2934\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19225534\/Screen-Shot-2016-04-19-at-3.54.55-PM.png\" alt=\"Solid downward-sloping line marked x+4y=4. The region below the line is shaded and is labeled x+4y is less than or equal to 4.\" width=\"413\" height=\"419\" \/>\r\n\r\nAnd there you have it\u2014the graph of the set of solutions for [latex]x+4y\\leq4[\/latex].\r\n\r\nhttps:\/\/youtu.be\/2VgFg2ztspI\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nGraph the inequality [latex]2y&gt;4x\u20136[\/latex].\r\n\r\n[reveal-answer q=\"138506\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"138506\"]\r\n\r\nSolve for [latex]y[\/latex].\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}2y&gt;4x-6\\\\\\\\\\frac{2y}{2}&gt;\\frac{4x}{2}-\\frac{6}{2}\\\\\\\\y&gt;2x-3\\\\\\end{array}[\/latex]<\/p>\r\nCreate a table of values to find two points on the line [latex] \\displaystyle y=2x-3[\/latex], or graph it based on the slope-intercept method, the [latex]b[\/latex] value of the [latex]y[\/latex]-intercept is [latex]-3[\/latex] and the slope is 2.\r\n\r\nPlot the points, and graph the line. The line is dotted because the sign in the inequality is &gt;, not \u2265 and therefore points on the line are not solutions to the inequality.\r\n\r\n<img class=\"aligncenter size-full wp-image-2937\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19230258\/Screen-Shot-2016-04-19-at-4.02.07-PM.png\" alt=\"Dotted upward-sloping line that crosses the points (2,1) and (0,-3). The points (-3,1) and (4,1) are also plotted.\" width=\"423\" height=\"422\" \/>\r\n<p style=\"text-align: center\">[latex] \\displaystyle y=2x-3[\/latex]<\/p>\r\n\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>x<\/th>\r\n<th>y<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>[latex]\u22123[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind an ordered pair on either side of the boundary line. Insert the [latex]x[\/latex]- and [latex]y[\/latex]-values into the inequality\r\n[latex]2y&gt;4x\u20136[\/latex] and see which ordered pair results in a true statement. Since [latex](\u22123,1)[\/latex] results in a true statement, the region that includes [latex](\u22123,1)[\/latex] should be shaded.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2y&gt;4x\u20136\\\\\\\\\\text{Test }1:\\left(\u22123,1\\right)\\\\2\\left(1\\right)&gt;4\\left(\u22123\\right)\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2&gt;\u201312\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2&gt;\u221218\\\\\\text{TRUE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\2(1)&gt;4\\left(4\\right)\u2013 6\\\\\\,\\,\\,\\,\\,\\,2&gt;16\u20136\\\\\\,\\,\\,\\,\\,\\,2&gt;10\\\\\\text{FALSE}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe graph of the inequality [latex]2y&gt;4x\u20136[\/latex] is:\r\n\r\n<img class=\"aligncenter wp-image-2935 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19225738\/Screen-Shot-2016-04-19-at-3.56.57-PM.png\" alt=\"The dotted upward-sloping line of 2y=4x-6, with the region above the line shaded.\" width=\"387\" height=\"391\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nA quick note about the problem above\u2014notice that you can use the points [latex](0,\u22123)[\/latex] and [latex](2,1)[\/latex] to graph the boundary line, but that these points are not included in the region of solutions, since the region does not include the boundary line!\r\n\r\nhttps:\/\/youtu.be\/Hzxc4HASygU\r\n<h1>(6.4.3) - Graph a system of linear inequalities and define the solutions region<\/h1>\r\nConsider the graph of the inequality [latex]y&lt;2x+5[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064400\/image012.gif\" alt=\"An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1).\" width=\"346\" height=\"343\" \/>\r\n\r\nThe dashed line is [latex]y=2x+5[\/latex]. Every ordered pair in the shaded\u00a0area below the line is a solution to [latex]y&lt;2x+5[\/latex], as all of the points below the line will make the inequality true. If you doubt that, try substituting the [latex]x[\/latex] and [latex]y[\/latex] coordinates of Points A and B into the inequality\u2014you\u2019ll see that they work. So, the shaded area shows all of the solutions for this inequality.\r\n\r\nThe boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don\u2019t satisfy the inequality. If the inequality had been [latex]y\\leq2x+5[\/latex], then the boundary line would have been solid.\r\n\r\nLet\u2019s graph another inequality: [latex]y&gt;\u2212x[\/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064401\/image013.gif\" alt=\"Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1).\" width=\"327\" height=\"324\" \/>\r\n\r\nTo create a system of inequalities, you need to graph two or more inequalities together. Let\u2019s use\u00a0[latex]y&lt;2x+5[\/latex] and [latex]y&gt;\u2212x[\/latex] since we have already graphed each of them.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014.gif\" alt=\"The two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region.\" width=\"318\" height=\"315\" \/>\r\n\r\nThe purple area shows where the solutions of the two inequalities overlap. This area is the solution to the <i>system of inequalities<\/i>. Any point within this purple region will be true for both [latex]y&gt;\u2212x[\/latex] and [latex]y&lt;2x+5[\/latex].\r\n\r\nIn the following video examples, we show how to graph a system of linear inequalities, and define the solution region.\r\n\r\nhttps:\/\/youtu.be\/ACTxJv1h2_c\r\n\r\nhttps:\/\/youtu.be\/cclH2h1NurM\r\n\r\nIn the next section, we will see that points can be solutions to systems of equations and inequalities. \u00a0We will verify algebraically whether a point is a solution to a linear equation or inequality.\r\n<h1>(6.4.4) - Determine whether a point is a solution to a system of inequalities<\/h1>\r\n<img class=\"size-medium wp-image-398 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014-300x297.gif\" alt=\"image014\" width=\"300\" height=\"297\" \/>\r\n\r\nOn the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.\r\n\r\nIn contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y&gt;\u2212x[\/latex] and point A is a solution for the inequality [latex]y&lt;2x+5[\/latex], neither point is a solution for the <i>system<\/i>. The following example shows how to test a point to see whether it is a solution to a system of inequalities.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs the point [latex](2,1)[\/latex] a solution of the system [latex]x+y&gt;1[\/latex] and [latex]2x+y&lt;8[\/latex]?\r\n\r\n[reveal-answer q=\"84880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"84880\"]Check the point with each of the inequalities. Substitute 2 for [latex]x[\/latex] and 1 for [latex]y[\/latex]. Is the point a solution of both inequalities?\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+y&gt;1\\\\2+1&gt;1\\\\3&gt;1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is a solution for [latex]x+y&gt;1[\/latex].<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2x+y&lt;8\\\\2\\left(2\\right)+1&lt;8\\\\4+1&lt;8\\\\5&lt;8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is a solution for [latex]2x+y&lt;8.[\/latex]<\/p>\r\nSince [latex](2,1)[\/latex] is a solution of each inequality, it is also a solution of the system.\r\n<h4>Answer<\/h4>\r\nThe point [latex](2,1)[\/latex] is a solution of the system [latex]x+y&gt;1[\/latex] and [latex]2x+y&lt;8[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is a graph of the system in the example above. Notice that [latex](2,1)[\/latex] lies in the purple area, which is the overlapping area for the two inequalities.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064404\/image015.gif\" alt=\"Two dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.\" width=\"321\" height=\"317\" \/>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs the point [latex](2,1)[\/latex] a solution of the system [latex]x+y&gt;1[\/latex] and [latex]3x+y&lt;4[\/latex]?\r\n\r\n[reveal-answer q=\"833522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"833522\"]\r\n\r\nCheck the point with each of the inequalities. Substitute 2 for [latex]x[\/latex] and 1 for [latex]y[\/latex]. Is the point a solution of both inequalities?\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+y&gt;1\\\\2+1&gt;1\\\\3&gt;1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is a solution for [latex]x+y&gt;1[\/latex].<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}3x+y&lt;4\\\\3\\left(2\\right)+1&lt;4\\\\6+1&lt;4\\\\7&lt;4\\\\\\text{FALSE}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is <i>not <\/i>a solution for [latex]3x+y&lt;4[\/latex].<\/p>\r\nSince [latex](2,1)[\/latex] is <i>not <\/i>a solution of one of the inequalities, it is not a solution of the system.\r\n<h4>Answer<\/h4>\r\n<span lang=\"X-NONE\">The point [latex](2,1)[\/latex] is not a solution of the system [latex]x+y&gt;1[\/latex]<\/span>\u00a0and [latex]3x+y&lt;4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is a graph of this system. Notice that [latex](2, 1)[\/latex] is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064405\/image016.gif\" alt=\"A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region.\" width=\"346\" height=\"342\" \/>\r\n\r\nIn the following video we show another example of determining whether a point is in the solution of a system of linear inequalities.\r\n\r\nhttps:\/\/youtu.be\/o9hTFJEBcXs\r\n\r\nAs shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities. \u00a0The general steps are outlined below:\r\n<ul>\r\n \t<li>Graph each inequality as a line and determine whether it will be solid or dashed<\/li>\r\n \t<li>Determine which side of each boundary line represents solutions to the inequality by testing a point on each side<\/li>\r\n \t<li>Shade the region\u00a0that represents solutions for both inequalities<\/li>\r\n<\/ul>\r\n<h1>(6.4.5) - Identify when a system of inequalities has no solution<\/h1>\r\nIn the next\u00a0example, we will show the\u00a0solution to\u00a0a system of two inequalities whose boundary lines are parallel to each other. \u00a0When the graphs of a system of two linear equations are parallel to each other, we found that there was no solution to the system. \u00a0We will get a similar result for the following system of linear inequalities.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nGraph the system\u00a0[latex]\\begin{array}{c}y\\ge2x+1\\\\y\\lt2x-3\\end{array}[\/latex]\r\n[reveal-answer q=\"780322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"780322\"]\r\n\r\nThe boundary lines for this system\u00a0are parallel to each other, note how they have the same slopes.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}y=2x+1\\\\y=2x-3\\end{array}[\/latex]<\/p>\r\nPlotting the boundary lines will give the graph below. Note\u00a0that the inequality [latex]y\\lt2x-3[\/latex] requires that we draw a dashed line, while the inequality [latex]y\\ge2x+1[\/latex] will require a solid line.\r\n\r\n<img class=\"wp-image-4148 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183205\/Screen-Shot-2016-05-13-at-1.56.45-PM-300x300.png\" alt=\"y=2x+1\" width=\"410\" height=\"410\" \/>\r\n\r\nNow we need to add the regions that represent the inequalities. \u00a0For the inequality [latex]y\\ge2x+1[\/latex] we can test a point on either side of the line to see which region to shade. Let's test [latex]\\left(0,0\\right)[\/latex] to make it easy.\r\n\r\nSubstitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\ge2x+1[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}y\\ge2x+1\\\\0\\ge2\\left(0\\right)+1\\\\0\\ge{1}\\end{array}[\/latex]<\/p>\r\nThis is not true, so we know that we need to shade the other side of the boundary line for the inequality\u00a0\u00a0[latex]y\\ge2x+1[\/latex]. The graph will now look like this:\r\n\r\n<img class=\"wp-image-4149 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183206\/Screen-Shot-2016-05-13-at-2.02.49-PM-300x300.png\" alt=\"y=2x+1\" width=\"355\" height=\"355\" \/>\r\n\r\nNow let's shade the region that shows the solutions to the inequality [latex]y\\lt2x-3[\/latex]. \u00a0Again, we can pick\u00a0[latex]\\left(0,0\\right)[\/latex] to test because it makes easy algebra.\r\n\r\nSubstitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\lt2x-3[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}y\\lt2x-3\\\\0\\lt2\\left(0,\\right)x-3\\\\0\\lt{-3}\\end{array}[\/latex]<\/p>\r\nThis is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\\lt2x-3[\/latex]. The graph will now look like this:\r\n\r\n<img class=\"wp-image-4150 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183208\/Screen-Shot-2016-05-13-at-2.07.01-PM-297x300.png\" alt=\"y=2x+1\" width=\"394\" height=\"398\" \/>\r\n\r\nThis system of inequalities shares no points in common.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following examples, we will continue to practice graphing the solution region for systems of linear inequalities. \u00a0We will also\u00a0graph the solutions to a system that includes a compound inequality.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nShade the region of the graph that represents solutions for both inequalities.\u00a0\u00a0[latex]x+y\\geq 1[\/latex] and [latex]y -x\\geq 5[\/latex].\r\n\r\n[reveal-answer q=\"873537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"873537\"]Graph one inequality. First graph the boundary line, using a table of values, intercepts, or any other method you prefer. The boundary line for [latex]x+y\\geq1[\/latex] is [latex]x+y=1[\/latex], or [latex]y=\u2212x+1[\/latex]. Since the equal sign is included with the greater than sign, the boundary line is solid.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064406\/image017-2.jpg\" alt=\"A downward-sloping solid line labeled x+y is greater than 1.\" width=\"370\" height=\"370\" \/>\r\n\r\nFind an ordered pair on either side of the boundary line. Insert the <i>x<\/i>- and <i>y<\/i>-values into the inequality [latex]x+y\\geq1[\/latex] and see which ordered pair results in a true statement.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\text{Test }1:\\left(\u22123,0\\right)\\\\x+y\\geq1\\\\\u22123+0\\geq1\\\\\u22123\\geq1\\\\\\text{FALSE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\x+y\\geq1\\\\4+1\\geq1\\\\5\\geq1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nSince [latex](4, 1)[\/latex] results in a true statement, the region that includes [latex](4, 1)[\/latex] should be shaded.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064409\/image018.gif\" alt=\"A solid downward-sloping line with the region above it shaded and labeled x+y is greater than or equal to 1. The point (4,1) is in the shaded region. The point (-3,0) is not.\" width=\"345\" height=\"342\" \/>\r\n\r\nDo the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y\u2013x=5\\left(\\text{or }y=x+5\\right)[\/latex] and is solid. Test point (\u22123, 0) is not a solution of [latex]y\u2013x\\geq5[\/latex], and test point (0, 6) is a solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064410\/image019.gif\" alt=\"A solid blue line with the region above it shaded and labeled y-x is greater than or equal to 5. A solid red line with the region above it shaded and labeled x+y is greater than 1. The point (-3,0) is not in any shaded region. The point (0,6) is in the overlapping shaded region.\" width=\"337\" height=\"334\" \/>\r\n<h4>Answer<\/h4>\r\nThe purple region in this graph shows the set of all solutions of the system.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064412\/image020-2.jpg\" alt=\"The previous graph, with the purple overlapping shaded region labeled x+y is greater than or equal to 1 and y-x is greater than or equal to 5.\" width=\"329\" height=\"325\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe videos that follow show more\u00a0examples of graphing the solution set of a system of linear inequalities.\r\n\r\nhttps:\/\/youtu.be\/ACTxJv1h2_c\r\n\r\nhttps:\/\/youtu.be\/cclH2h1NurM\r\n\r\nThe system in our last example includes a compound inequality. \u00a0We will see that you can treat a compound inequality like two lines when you are graphing them.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the solution to the system [latex]3x + 2y &lt; 12[\/latex] and [latex]\u22121\\leq y \\leq 5[\/latex].\r\n[reveal-answer q=\"163187\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"163187\"]\r\n\r\nGraph one inequality. First graph the boundary line, then test points.\r\n\r\nRemember, because the inequality [latex]3x + 2y&lt;12[\/latex] does not include the equal sign, draw a dashed border line.\r\n\r\nTesting a point like [latex](0, 0)[\/latex] will show that the area below the line is the solution to this inequality.\r\n\r\n<img class=\"alignnone size-medium wp-image-2427\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223640\/image021-300x297.gif\" alt=\"image021\" width=\"300\" height=\"297\" \/>\r\n\r\nThe inequality [latex]\u22121\\leq y \\leq 5[\/latex] is actually two inequalities: [latex]\u22121\\leq y[\/latex], and [latex]y \\leq 5[\/latex]. Another way to think of this is y must be between \u22121 and 5. The border lines for both are horizontal. The region between those two lines contains the solutions of [latex]\u22121\\leq y\\leq 5[\/latex]. We make the lines solid because we also want to include [latex]y = \u22121[\/latex] and [latex]y = 5[\/latex].\r\n\r\nGraph this region on the same axes as the other inequality.\r\n\r\n<img class=\"alignnone size-medium wp-image-2428\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223642\/image022-300x298.gif\" alt=\"image022\" width=\"300\" height=\"298\" \/>\r\n\r\nThe purple region in this graph shows the set of all solutions of the system.\r\n<h4>Answer<\/h4>\r\n<img class=\"alignnone size-medium wp-image-2429\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223644\/image023-300x297.jpg\" alt=\"image023\" width=\"300\" height=\"297\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we show how to solve another system of inequalities.\r\n\r\nhttps:\/\/youtu.be\/ACTxJv1h2_c\r\n\r\n\r\n<h1>(6.4.6) - Applications of systems of linear inequalities<\/h1>\r\nIn our first\u00a0example we will show how to write and graph a system of linear inequalities that models the amount of sales needed to obtain a specific amount of money.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCathy is selling ice cream cones at a school fundraiser. She is selling two sizes: small (which has 1 scoop) and large (which has 2 scoops). She knows that she can get a maximum of 70 scoops of ice cream out of her supply. She charges $3 for a small cone and $5 for a large cone.\r\n\r\nCathy wants to earn at least $120 to give back to the school. Write and graph a system of inequalities that models this situation.\r\n[reveal-answer q=\"737192\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"737192\"]\r\n\r\nFirst, identify the variables. There are two variables: the number of small cones and the number of large cones.\r\n<p style=\"text-align: center\">[latex]s[\/latex]\u00a0= small cone<\/p>\r\n<p style=\"text-align: center\">[latex]l[\/latex]\u00a0= large cone<\/p>\r\n<p style=\"text-align: left\">Write the first equation: the maximum number of scoops she can give out. The scoops she has available (70) must be greater than or equal to the number of scoops for the small cones [latex](<i>s<\/i>)[\/latex] and the large cones [latex](2<i>l<\/i>)[\/latex] she sells.<\/p>\r\n<p style=\"text-align: center\">[latex]s+2l\\le70[\/latex]<\/p>\r\n<p style=\"text-align: left\">Write the second equation: the amount of money she raises. She wants the total amount of money earned from small cones [latex](3<i>s<\/i>)[\/latex] and large cones [latex](5<i>l<\/i>)[\/latex] to be at least $120.<\/p>\r\n<p style=\"text-align: center\">[latex]3s+5l\\ge120[\/latex]<\/p>\r\n<p style=\"text-align: left\">Write the system.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{cases}s+2l\\le70\\\\3s+5l\\ge120\\end{cases}[\/latex]<\/p>\r\nNow graph the system. The variables [latex]x[\/latex] and [latex]y[\/latex] have been replaced by [latex]s[\/latex] and [latex]l[\/latex]; graph s along the [latex]x[\/latex]-axis, and [\/latex]l[\/latex]\u00a0along the [\/latex]y[\/latex]-axis.\r\n\r\nFirst graph the region [latex]<i>s<\/i> + 2<i>l<\/i>\u00a0\\leq 70[\/latex]. Graph the boundary line and then test individual points to see which region to shade. The graph is shown below.<img class=\"alignnone size-medium wp-image-2433\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223652\/image027-300x203.jpg\" alt=\"image027\" width=\"300\" height=\"203\" \/>\r\n\r\nNow graph the region [latex]3s+5l\\ge120[\/latex]\u00a0Graph the boundary line and then test individual points to see which region to shade. The graph is shown below.\r\n\r\n<img class=\"alignnone size-medium wp-image-2434\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223654\/image028-300x203.jpg\" alt=\"image028\" width=\"300\" height=\"203\" \/>\r\n\r\nGraphing the regions together, you find the following:\r\n\r\n<img class=\"alignnone size-medium wp-image-2435\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223656\/image029-300x203.jpg\" alt=\"image029\" width=\"300\" height=\"203\" \/>\r\n\r\nAnd represented just as the overlapping region, you have:<img class=\"alignnone size-medium wp-image-2436\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223658\/image030-300x203.jpg\" alt=\"image030\" width=\"300\" height=\"203\" \/>\r\n<h4>Answer<\/h4>\r\nThe region in purple is the solution. As long as the combination of small cones and large cones that Cathy sells can be mapped in the purple region, she will have earned at least $120 and not used more than 70 scoops of ice cream.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn a previous example for finding a solution to a system of linear equations, we introduced a manufacturer\u2019s\u00a0cost and revenue equations:\r\n\r\nCost: [latex]y=0.85x+35,000[\/latex]\r\n\r\nRevenue: [latex]y=1.55x[\/latex]\r\n\r\nThe cost equation is shown in blue in the graph below, and the revenue equation is graphed in orange.The point at which the two lines intersect is called the break-even point, we learned that this is the solution to the system of linear equations that in this case comprise the cost and revenue equations.\r\n\r\nThe shaded region to the right of the break-even point represents quantities for which the company makes a profit. The region to the left represents quantities for which the company suffers a loss.\r\n\r\nIn the next example, you will see how the information you learned about systems of linear inequalities can be applied to answering questions about cost and revenue.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183256\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"487\" height=\"390\" \/>\r\n\r\nNote how the blue shaded region between the Cost and Revenue equations is labeled Profit. This is the \"sweet spot\" that the company wants to achieve\u00a0where they\u00a0produce enough bike frames at a minimal enough cost to\u00a0make money. They don't want more money going out than coming in!\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDefine the profit region for the skateboard\u00a0manufacturing business using inequalities, given the system of linear equations:\r\n\r\nCost: [latex]y=0.85x+35,000[\/latex]\r\n\r\nRevenue: [latex]y=1.55x[\/latex]\r\n[reveal-answer q=\"563864\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"563864\"]\r\n\r\nWe know that graphically, \u00a0solutions to\u00a0linear inequalities are entire regions, and we learned how to graph systems of linear inequalities earlier in this module. Based on the graph below and the equations that define cost and revenue, we can use inequalities to define the region for which the skateboard\u00a0manufacturer will make a profit.\r\n\r\n<img class=\"wp-image-4210 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183258\/Screen-Shot-2016-05-18-at-3.33.11-PM-300x280.png\" alt=\"Cost\/ Revenue with Profit\" width=\"358\" height=\"334\" \/>\r\n\r\nLet's start with the revenue equation. \u00a0We know that the break even point is at (50,000, 77,500) and the profit region is\u00a0the blue area. \u00a0If we choose a point in the region and test it like we did for finding solution regions to inequalities, we will know which kind of inequality sign to use.\r\n\r\nLet's test the point [latex]\\left(65,00,100,000\\right)[\/latex] in both equations to determine which inequality sign to use.\r\n\r\nCost:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y=0.85x+{35,000}\\\\{100,000}\\text{ ? }0.85\\left(65,000\\right)+35,000\\\\100,000\\text{ ? }90,250\\end{array}[\/latex]<\/p>\r\nWe need to use &gt; because 100,000 is greater than 90,250\r\n\r\nThe cost inequality that will ensure the company makes profit - not just break even - is\u00a0[latex]y&gt;0.85x+35,000[\/latex]\r\n\r\nNow test the point in the revenue equation:\r\n\r\nRevenue:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y=1.55x\\\\100,000\\text{ ? }1.55\\left(65,000\\right)\\\\100,000\\text{ ? }100,750\\end{array}[\/latex]<\/p>\r\nWe need to use &lt; because 100,000 is less than 100,750\r\n\r\nThe revenue inequality that will ensure the company makes profit - not just break even - is\u00a0[latex]y&lt;1.55x[\/latex]\r\n\r\nThe systems of inequalities that defines the profit region for the bike manufacturer:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y&gt;0.85x+35,000\\\\y&lt;1.55x\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. The system of linear inequalities that represents the number of units that the company must produce in order to earn a profit is:\r\n\r\n[latex]\\begin{array}{l}y&gt;0.85x+35,000\\\\y&lt;1.55x\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video you will see an example of how to find the break even point for a small sno-cone business.\r\n\r\nhttps:\/\/youtu.be\/qey3FmE8saQ\r\n\r\nAnd here is one more video example of solving an application\u00a0using a sustem of linear inequalities.\r\n\r\nhttps:\/\/youtu.be\/gbHl6K-dJ8o\r\n\r\nWe have seen that systems of linear equations and inequalities can help to define market behaviors that are very helpful to businesses. \u00a0The intersection of cost and revenue equations gives the break even point, and also helps define the region for which a company will make a profit.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(6.4.1) &#8211; Define solutions to a linear inequality in two variables<\/li>\n<li>(6.4.2) &#8211; Graphing the solution set to a linear inequality in two variables<\/li>\n<li>(6.4.3) &#8211; Graph a system of linear inequalities and define the solutions region<\/li>\n<li>(6.4.4) &#8211; Determine whether a point is a solution to a system of inequalities<\/li>\n<li style=\"list-style-type: none\"><\/li>\n<li>(6.4.5) &#8211; Identify when a system of inequalities has no solution<\/li>\n<li>(6.4.6) &#8211; Applications of systems of linear inequalities<\/li>\n<\/ul>\n<\/div>\n<h1>(6.4.1) &#8211; Define solutions to a linear inequality in two variables<\/h1>\n<p id=\"fs-id1167832031143\">Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables. In particular we will look at\u00a0<strong>linear inequalities<\/strong>\u00a0in two variables which are very similar to linear equations in two variables.<\/p>\n<p id=\"fs-id1167835236709\">Linear inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs\u2014so that your business made a profit.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Linear Inequality in Two Variables<\/h3>\n<p>A\u00a0<strong>linear inequality <\/strong>in two variables\u00a0is an inequality that can be written in one of the following forms:<\/p>\n<p style=\"text-align: center\">[latex]Ax+By < 0[\/latex]\u00a0 \u00a0 \u00a0 \u00a0[latex]Ax+By>0[\/latex]\u00a0 \u00a0 \u00a0 [latex]Ax+By \\leq 0[\/latex]\u00a0 \u00a0 \u00a0[latex]Ax+By \\geq 0[\/latex]<\/p>\n<p>where [latex]A[\/latex] and [latex]B[\/latex] are not both zero.<\/p>\n<\/div>\n<p>Recall that an inequality with one variable had many solutions. For example, the solution to the inequality [latex]x > 3[\/latex]\u00a0is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-5408\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2016\/06\/19213517\/Math101_6_4_1im1-300x27.jpg\" alt=\"\" width=\"300\" height=\"27\" \/><\/p>\n<p>Similarly, linear inequalities in two variables have many solutions. Any ordered pair [latex](x,y)[\/latex]\u00a0that makes an inequality true when we substitute in the values is a\u00a0solution to a linear inequality.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Solution to a linear inequality in two variables<\/h3>\n<p>An ordered pair\u00a0[latex](x,y)[\/latex] is a\u00a0<strong>solution to a linear inequality<\/strong>\u00a0if the inequality is true when we substitute the values of [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Determine whether each ordered pair is a solution to the inequality [latex]y>x+4[\/latex].<\/p>\n<p>a)[latex](0,0)[\/latex]<\/p>\n<p>b)[latex](1,6)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q837554\">Show Answer<\/span><\/p>\n<div id=\"q837554\" class=\"hidden-answer\" style=\"display: none\">\n<p>a)Substitute 0 for [latex]x[\/latex] and 0 for [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]0 \\overset{?}{>} 0 + 4[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]0 \\not\\gt 4[\/latex]<\/p>\n<p>So, [latex](0,0)[\/latex] is not a solution to\u00a0[latex]y>x+4[\/latex].<\/p>\n<p>b)Substitute 1 for [latex]x[\/latex] and 6 for [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]6 \\overset{?}{>} 1 + 4[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]6 > 5[\/latex]<\/p>\n<p>So, [latex](1,6)[\/latex] is a solution to\u00a0[latex]y>x+4[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h4>Solution Sets of Inequalities<\/h4>\n<p>The graph below shows the region of values that makes the inequality [latex]3x+2y\\leq6[\/latex] true (shaded red), the boundary line [latex]3x+2y=6[\/latex], as well as a handful of ordered pairs. The boundary line is solid because points on the boundary line [latex]3x+2y=6[\/latex]\u00a0will make the inequality [latex]3x+2y\\leq6[\/latex]\u00a0true.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2873 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19172056\/Screen-Shot-2016-04-19-at-10.20.21-AM.png\" alt=\"A solid downward-sloping line running. The region below the line is shaded and is labeled 3x+2y is less than or equal to 6. The region above the line is unshaded and is labeled 3x+2y=6. The points (-5,5) and (-2,-2) are in the shaded region. The points (2,3) and (4,-1) are in the unshaded region. The point (2,0) is on the line.\" width=\"464\" height=\"472\" \/><\/p>\n<p>You can substitute the [latex]x[\/latex]&#8211; and [latex]y[\/latex]<i>&#8211;<\/i>values in each of the [latex](x,y)[\/latex] ordered pairs into the inequality to find solutions. Sometimes making a table of values makes sense for more complicated inequalities.<\/p>\n<table>\n<thead>\n<tr>\n<th>Ordered Pair<\/th>\n<th>Makes the inequality<\/p>\n<p>[latex]3x+2y\\leq6[\/latex]<\/p>\n<p>a true statement<\/th>\n<th>Makes the inequality<\/p>\n<p>[latex]3x+2y\\leq6[\/latex]<\/p>\n<p>a false statement<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex](\u22125, 5)[\/latex]<\/td>\n<td>[latex]\\begin{array}{r}3\\left(\u22125\\right)+2\\left(5\\right)\\leq6\\\\\u221215+10\\leq6\\\\\u22125\\leq6\\end{array}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex](\u22122,\u22122)[\/latex]<\/td>\n<td>[latex]\\begin{array}{r}3\\left(\u22122\\right)+2\\left(\u20132\\right)\\leq6\\\\\u22126+\\left(\u22124\\right)\\leq6\\\\\u201310\\leq6\\end{array}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex](2,3)[\/latex]<\/td>\n<td><\/td>\n<td>[latex]\\begin{array}{r}3\\left(2\\right)+2\\left(3\\right)\\leq6\\\\6+6\\leq6\\\\12\\leq6\\end{array}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex](2,0)[\/latex]<\/td>\n<td>[latex]\\begin{array}{r}3\\left(2\\right)+2\\left(0\\right)\\leq6\\\\6+0\\leq6\\\\6\\leq6\\end{array}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex](4,\u22121)[\/latex]<\/td>\n<td><\/td>\n<td>[latex]\\begin{array}{r}3\\left(4\\right)+2\\left(\u22121\\right)\\leq6\\\\12+\\left(\u22122\\right)\\leq6\\\\10\\leq6\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>If substituting [latex](x,y)[\/latex] into the inequality yields a true statement, then the ordered pair is a solution to the inequality, and the point will be plotted within the shaded region or the point will be part of a solid boundary line. A false statement means that the ordered pair is not a solution, and the point will graph outside the shaded region, or the point will be part of a dotted boundary line.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the graph to determine which ordered pairs plotted below are solutions of the inequality\u00a0[latex]x\u2013y<3[\/latex].\n\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2876 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19172536\/Screen-Shot-2016-04-19-at-10.25.12-AM.png\" alt=\"Upward-sloping dotted line. The region above the line is shaded and labeled x-y&lt;3. The points (4,0) and (3,-2) are in the unshaded region. The point (1,-2) is on the dotted line. The points (-1,1) and (-2,-2) are in the shaded region.\" width=\"410\" height=\"415\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q840389\">Show Solution<\/span><\/p>\n<div id=\"q840389\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solutions will be located in the shaded region. Since this is a \u201cless than\u201d problem, ordered pairs on the boundary line are not included in the solution set.<\/p>\n<p>These values are located in the shaded region, so are solutions. (When substituted into the inequality\u00a0[latex]x\u2013y<3[\/latex], they produce true statements.)\n\n\n<p style=\"text-align: center\">[latex](\u22121,1)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](\u22122,\u22122)[\/latex]<\/p>\n<p>These values are not located in the shaded region, so are not solutions. (When substituted into the inequality [latex]x-y<3[\/latex], they produce false statements.)\n\n\n<p style=\"text-align: center\">[latex](1,\u22122)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](3,\u22122)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](4,0)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex](\u22121,1)\\,\\,\\,(\u22122,\u22122)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>The following video show an example of determining whether an ordered pair is a solution to an inequality.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Use a Graph Determine Ordered Pair Solutions of a Linear Inequality in Two Variable\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GQVdDRVq5_o?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is [latex](2,\u22123)[\/latex] a solution of the inequality [latex]y<\u22123x+1[\/latex]?\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q746731\">Show Solution<\/span><\/p>\n<div id=\"q746731\" class=\"hidden-answer\" style=\"display: none\">\n<p>If [latex](2,\u22123)[\/latex] is a solution, then it will yield a true statement when substituted into the inequality\u00a0[latex]y<\u22123x+1[\/latex].\n\n\n<p style=\"text-align: center\">[latex]y<\u22123x+1[\/latex]<\/p>\n<p>Substitute\u00a0[latex]x=2[\/latex] and [latex]y=\u22123[\/latex]\u00a0into inequality.<\/p>\n<p style=\"text-align: center\">[latex]\u22123<\u22123\\left(2\\right)+1[\/latex]<\/p>\n<p>Evaluate.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\u22123<\u22126+1\\\\\u22123<\u22125\\end{array}[\/latex]<\/p>\n<p>This statement is <b>not <\/b>true, so the ordered pair [latex](2,\u22123)[\/latex] is <b>not <\/b>a solution.<\/p>\n<h4>Answer<\/h4>\n<p>[latex](2,\u22123)[\/latex] is not a solution.<\/p><\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of determining whether an ordered pair is a solution to an inequality.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Determine if Ordered Pairs Satisfy a Linear Inequality\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-x-zt_yM0RM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(6.4.2) &#8211; Graphing the solution set to a linear inequality in two variables<\/h1>\n<p>So how do you get from the algebraic form of an inequality, like [latex]y>3x+1[\/latex], to a graph of that inequality? Plotting inequalities is fairly straightforward if you follow a couple steps.<\/p>\n<div class=\"textbox shaded\">\n<h3>Graphing Inequalities<\/h3>\n<p>To graph an inequality:<\/p>\n<ul>\n<li>Graph the related boundary line. Replace the &lt;, &gt;, \u2264 or \u2265 sign in the inequality with = to find the equation of the boundary line.<\/li>\n<li>Identify at least one ordered pair on either side of the boundary line and substitute those [latex](x,y)[\/latex] values into the inequality. Shade the region that contains the ordered pairs that make the inequality a true statement.<b>\u00a0<\/b><\/li>\n<li>If points on the boundary line are solutions, then use a solid line for drawing the boundary line. This will happen for \u2264 or \u2265 inequalities.<\/li>\n<li>If points on the boundary line aren\u2019t solutions, then use a dotted line for the boundary line. This will happen for &lt; or &gt; inequalities.<\/li>\n<\/ul>\n<\/div>\n<p>Let\u2019s graph the inequality [latex]x+4y\\leq4[\/latex].<\/p>\n<p>To graph the boundary line, find at least two values that lie on the line [latex]x+4y=4[\/latex]. You can use the <i>x<\/i>&#8211; and <i>y<\/i>-intercepts for this equation by substituting 0 in for <i>x<\/i> first and finding the value of <i>y<\/i>; then substitute 0 in for <i>y<\/i> and find <i>x<\/i>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><b><i>x<\/i><\/b><\/td>\n<td><b><i>y<\/i><\/b><\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points [latex](0,1)[\/latex] and [latex](4,0)[\/latex], and draw a line through these two points for the boundary line. The line is solid because \u2264 means \u201cless than or equal to,\u201d so all ordered pairs along the line are included in the solution set.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2936\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19230042\/Screen-Shot-2016-04-19-at-4.00.26-PM.png\" alt=\"Solid downward-sloping line that crosses the points (0,1) and (4,0). The point (-1,3) and the point (2,0) are also plotted.\" width=\"417\" height=\"419\" \/><\/p>\n<p>The next step is to find the region that contains the solutions. Is it above or below the boundary line? To identify the region where the inequality holds true, you can test a couple of ordered pairs, one on each side of the boundary line.<\/p>\n<p>If you substitute [latex](\u22121,3)[\/latex] into\u00a0[latex]x+4y\\leq4[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\u22121+4\\left(3\\right)\\leq4\\\\\u22121+12\\leq4\\\\11\\leq4\\end{array}[\/latex]<\/p>\n<p>This is a false statement, since 11 is not less than or equal to 4.<\/p>\n<p>On the other hand, if you substitute [latex](2,0)[\/latex] into\u00a0[latex]x+4y\\leq4[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2+4\\left(0\\right)\\leq4\\\\2+0\\leq4\\\\2\\leq4\\end{array}[\/latex]<\/p>\n<p>This is true! The region that includes [latex](2,0)[\/latex] should be shaded, as this is the region of solutions.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2934\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19225534\/Screen-Shot-2016-04-19-at-3.54.55-PM.png\" alt=\"Solid downward-sloping line marked x+4y=4. The region below the line is shaded and is labeled x+4y is less than or equal to 4.\" width=\"413\" height=\"419\" \/><\/p>\n<p>And there you have it\u2014the graph of the set of solutions for [latex]x+4y\\leq4[\/latex].<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 2:  Graphing Linear Inequalities in Two Variables (Standard Form)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/2VgFg2ztspI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Graph the inequality [latex]2y>4x\u20136[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q138506\">Show Solution<\/span><\/p>\n<div id=\"q138506\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}2y>4x-6\\\\\\\\\\frac{2y}{2}>\\frac{4x}{2}-\\frac{6}{2}\\\\\\\\y>2x-3\\\\\\end{array}[\/latex]<\/p>\n<p>Create a table of values to find two points on the line [latex]\\displaystyle y=2x-3[\/latex], or graph it based on the slope-intercept method, the [latex]b[\/latex] value of the [latex]y[\/latex]-intercept is [latex]-3[\/latex] and the slope is 2.<\/p>\n<p>Plot the points, and graph the line. The line is dotted because the sign in the inequality is &gt;, not \u2265 and therefore points on the line are not solutions to the inequality.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2937\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19230258\/Screen-Shot-2016-04-19-at-4.02.07-PM.png\" alt=\"Dotted upward-sloping line that crosses the points (2,1) and (0,-3). The points (-3,1) and (4,1) are also plotted.\" width=\"423\" height=\"422\" \/><\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle y=2x-3[\/latex]<\/p>\n<table>\n<thead>\n<tr>\n<th>x<\/th>\n<th>y<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>[latex]\u22123[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Find an ordered pair on either side of the boundary line. Insert the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-values into the inequality<br \/>\n[latex]2y>4x\u20136[\/latex] and see which ordered pair results in a true statement. Since [latex](\u22123,1)[\/latex] results in a true statement, the region that includes [latex](\u22123,1)[\/latex] should be shaded.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2y>4x\u20136\\\\\\\\\\text{Test }1:\\left(\u22123,1\\right)\\\\2\\left(1\\right)>4\\left(\u22123\\right)\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2>\u201312\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2>\u221218\\\\\\text{TRUE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\2(1)>4\\left(4\\right)\u2013 6\\\\\\,\\,\\,\\,\\,\\,2>16\u20136\\\\\\,\\,\\,\\,\\,\\,2>10\\\\\\text{FALSE}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The graph of the inequality [latex]2y>4x\u20136[\/latex] is:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2935 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/19225738\/Screen-Shot-2016-04-19-at-3.56.57-PM.png\" alt=\"The dotted upward-sloping line of 2y=4x-6, with the region above the line shaded.\" width=\"387\" height=\"391\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>A quick note about the problem above\u2014notice that you can use the points [latex](0,\u22123)[\/latex] and [latex](2,1)[\/latex] to graph the boundary line, but that these points are not included in the region of solutions, since the region does not include the boundary line!<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 1:  Graphing Linear Inequalities in Two Variables (Slope Intercept Form)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Hzxc4HASygU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(6.4.3) &#8211; Graph a system of linear inequalities and define the solutions region<\/h1>\n<p>Consider the graph of the inequality [latex]y<2x+5[\/latex].\n\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064400\/image012.gif\" alt=\"An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1).\" width=\"346\" height=\"343\" \/><\/p>\n<p>The dashed line is [latex]y=2x+5[\/latex]. Every ordered pair in the shaded\u00a0area below the line is a solution to [latex]y<2x+5[\/latex], as all of the points below the line will make the inequality true. If you doubt that, try substituting the [latex]x[\/latex] and [latex]y[\/latex] coordinates of Points A and B into the inequality\u2014you\u2019ll see that they work. So, the shaded area shows all of the solutions for this inequality.\n\nThe boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don\u2019t satisfy the inequality. If the inequality had been [latex]y\\leq2x+5[\/latex], then the boundary line would have been solid.\n\nLet\u2019s graph another inequality: [latex]y>\u2212x[\/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064401\/image013.gif\" alt=\"Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1).\" width=\"327\" height=\"324\" \/><\/p>\n<p>To create a system of inequalities, you need to graph two or more inequalities together. Let\u2019s use\u00a0[latex]y<2x+5[\/latex] and [latex]y>\u2212x[\/latex] since we have already graphed each of them.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014.gif\" alt=\"The two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region.\" width=\"318\" height=\"315\" \/><\/p>\n<p>The purple area shows where the solutions of the two inequalities overlap. This area is the solution to the <i>system of inequalities<\/i>. Any point within this purple region will be true for both [latex]y>\u2212x[\/latex] and [latex]y<2x+5[\/latex].\n\nIn the following video examples, we show how to graph a system of linear inequalities, and define the solution region.\n\n<iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex 1:  Graph a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ACTxJv1h2_c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex 2:  Graph a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/cclH2h1NurM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next section, we will see that points can be solutions to systems of equations and inequalities. \u00a0We will verify algebraically whether a point is a solution to a linear equation or inequality.<\/p>\n<h1>(6.4.4) &#8211; Determine whether a point is a solution to a system of inequalities<\/h1>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-398 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014-300x297.gif\" alt=\"image014\" width=\"300\" height=\"297\" \/><\/p>\n<p>On the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.<\/p>\n<p>In contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y>\u2212x[\/latex] and point A is a solution for the inequality [latex]y<2x+5[\/latex], neither point is a solution for the <i>system<\/i>. The following example shows how to test a point to see whether it is a solution to a system of inequalities.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is the point [latex](2,1)[\/latex] a solution of the system [latex]x+y>1[\/latex] and [latex]2x+y<8[\/latex]?\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q84880\">Show Solution<\/span><\/p>\n<div id=\"q84880\" class=\"hidden-answer\" style=\"display: none\">Check the point with each of the inequalities. Substitute 2 for [latex]x[\/latex] and 1 for [latex]y[\/latex]. Is the point a solution of both inequalities?<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+y>1\\\\2+1>1\\\\3>1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is a solution for [latex]x+y>1[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2x+y<8\\\\2\\left(2\\right)+1<8\\\\4+1<8\\\\5<8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is a solution for [latex]2x+y<8.[\/latex]<\/p>\n<p>Since [latex](2,1)[\/latex] is a solution of each inequality, it is also a solution of the system.<\/p>\n<h4>Answer<\/h4>\n<p>The point [latex](2,1)[\/latex] is a solution of the system [latex]x+y>1[\/latex] and [latex]2x+y<8[\/latex].\n\n<\/div>\n<\/div>\n<\/div>\n<p>Here is a graph of the system in the example above. Notice that [latex](2,1)[\/latex] lies in the purple area, which is the overlapping area for the two inequalities.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064404\/image015.gif\" alt=\"Two dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.\" width=\"321\" height=\"317\" \/><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is the point [latex](2,1)[\/latex] a solution of the system [latex]x+y>1[\/latex] and [latex]3x+y<4[\/latex]?\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q833522\">Show Solution<\/span><\/p>\n<div id=\"q833522\" class=\"hidden-answer\" style=\"display: none\">\n<p>Check the point with each of the inequalities. Substitute 2 for [latex]x[\/latex] and 1 for [latex]y[\/latex]. Is the point a solution of both inequalities?<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+y>1\\\\2+1>1\\\\3>1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is a solution for [latex]x+y>1[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}3x+y<4\\\\3\\left(2\\right)+1<4\\\\6+1<4\\\\7<4\\\\\\text{FALSE}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex](2,1)[\/latex] is <i>not <\/i>a solution for [latex]3x+y<4[\/latex].<\/p>\n<p>Since [latex](2,1)[\/latex] is <i>not <\/i>a solution of one of the inequalities, it is not a solution of the system.<\/p>\n<h4>Answer<\/h4>\n<p><span lang=\"X-NONE\">The point [latex](2,1)[\/latex] is not a solution of the system [latex]x+y>1[\/latex]<\/span>\u00a0and [latex]3x+y<4[\/latex].\n\n<\/div>\n<\/div>\n<\/div>\n<p>Here is a graph of this system. Notice that [latex](2, 1)[\/latex] is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064405\/image016.gif\" alt=\"A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region.\" width=\"346\" height=\"342\" \/><\/p>\n<p>In the following video we show another example of determining whether a point is in the solution of a system of linear inequalities.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Determine if an Ordered Pair is a Solution to a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/o9hTFJEBcXs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>As shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities. \u00a0The general steps are outlined below:<\/p>\n<ul>\n<li>Graph each inequality as a line and determine whether it will be solid or dashed<\/li>\n<li>Determine which side of each boundary line represents solutions to the inequality by testing a point on each side<\/li>\n<li>Shade the region\u00a0that represents solutions for both inequalities<\/li>\n<\/ul>\n<h1>(6.4.5) &#8211; Identify when a system of inequalities has no solution<\/h1>\n<p>In the next\u00a0example, we will show the\u00a0solution to\u00a0a system of two inequalities whose boundary lines are parallel to each other. \u00a0When the graphs of a system of two linear equations are parallel to each other, we found that there was no solution to the system. \u00a0We will get a similar result for the following system of linear inequalities.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>Graph the system\u00a0[latex]\\begin{array}{c}y\\ge2x+1\\\\y\\lt2x-3\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q780322\">Show Solution<\/span><\/p>\n<div id=\"q780322\" class=\"hidden-answer\" style=\"display: none\">\n<p>The boundary lines for this system\u00a0are parallel to each other, note how they have the same slopes.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}y=2x+1\\\\y=2x-3\\end{array}[\/latex]<\/p>\n<p>Plotting the boundary lines will give the graph below. Note\u00a0that the inequality [latex]y\\lt2x-3[\/latex] requires that we draw a dashed line, while the inequality [latex]y\\ge2x+1[\/latex] will require a solid line.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4148 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183205\/Screen-Shot-2016-05-13-at-1.56.45-PM-300x300.png\" alt=\"y=2x+1\" width=\"410\" height=\"410\" \/><\/p>\n<p>Now we need to add the regions that represent the inequalities. \u00a0For the inequality [latex]y\\ge2x+1[\/latex] we can test a point on either side of the line to see which region to shade. Let&#8217;s test [latex]\\left(0,0\\right)[\/latex] to make it easy.<\/p>\n<p>Substitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\ge2x+1[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}y\\ge2x+1\\\\0\\ge2\\left(0\\right)+1\\\\0\\ge{1}\\end{array}[\/latex]<\/p>\n<p>This is not true, so we know that we need to shade the other side of the boundary line for the inequality\u00a0\u00a0[latex]y\\ge2x+1[\/latex]. The graph will now look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4149 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183206\/Screen-Shot-2016-05-13-at-2.02.49-PM-300x300.png\" alt=\"y=2x+1\" width=\"355\" height=\"355\" \/><\/p>\n<p>Now let&#8217;s shade the region that shows the solutions to the inequality [latex]y\\lt2x-3[\/latex]. \u00a0Again, we can pick\u00a0[latex]\\left(0,0\\right)[\/latex] to test because it makes easy algebra.<\/p>\n<p>Substitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\lt2x-3[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}y\\lt2x-3\\\\0\\lt2\\left(0,\\right)x-3\\\\0\\lt{-3}\\end{array}[\/latex]<\/p>\n<p>This is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\\lt2x-3[\/latex]. The graph will now look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4150 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183208\/Screen-Shot-2016-05-13-at-2.07.01-PM-297x300.png\" alt=\"y=2x+1\" width=\"394\" height=\"398\" \/><\/p>\n<p>This system of inequalities shares no points in common.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following examples, we will continue to practice graphing the solution region for systems of linear inequalities. \u00a0We will also\u00a0graph the solutions to a system that includes a compound inequality.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Shade the region of the graph that represents solutions for both inequalities.\u00a0\u00a0[latex]x+y\\geq 1[\/latex] and [latex]y -x\\geq 5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q873537\">Show Solution<\/span><\/p>\n<div id=\"q873537\" class=\"hidden-answer\" style=\"display: none\">Graph one inequality. First graph the boundary line, using a table of values, intercepts, or any other method you prefer. The boundary line for [latex]x+y\\geq1[\/latex] is [latex]x+y=1[\/latex], or [latex]y=\u2212x+1[\/latex]. Since the equal sign is included with the greater than sign, the boundary line is solid.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064406\/image017-2.jpg\" alt=\"A downward-sloping solid line labeled x+y is greater than 1.\" width=\"370\" height=\"370\" \/><\/p>\n<p>Find an ordered pair on either side of the boundary line. Insert the <i>x<\/i>&#8211; and <i>y<\/i>-values into the inequality [latex]x+y\\geq1[\/latex] and see which ordered pair results in a true statement.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\text{Test }1:\\left(\u22123,0\\right)\\\\x+y\\geq1\\\\\u22123+0\\geq1\\\\\u22123\\geq1\\\\\\text{FALSE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\x+y\\geq1\\\\4+1\\geq1\\\\5\\geq1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>Since [latex](4, 1)[\/latex] results in a true statement, the region that includes [latex](4, 1)[\/latex] should be shaded.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064409\/image018.gif\" alt=\"A solid downward-sloping line with the region above it shaded and labeled x+y is greater than or equal to 1. The point (4,1) is in the shaded region. The point (-3,0) is not.\" width=\"345\" height=\"342\" \/><\/p>\n<p>Do the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y\u2013x=5\\left(\\text{or }y=x+5\\right)[\/latex] and is solid. Test point (\u22123, 0) is not a solution of [latex]y\u2013x\\geq5[\/latex], and test point (0, 6) is a solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064410\/image019.gif\" alt=\"A solid blue line with the region above it shaded and labeled y-x is greater than or equal to 5. A solid red line with the region above it shaded and labeled x+y is greater than 1. The point (-3,0) is not in any shaded region. The point (0,6) is in the overlapping shaded region.\" width=\"337\" height=\"334\" \/><\/p>\n<h4>Answer<\/h4>\n<p>The purple region in this graph shows the set of all solutions of the system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064412\/image020-2.jpg\" alt=\"The previous graph, with the purple overlapping shaded region labeled x+y is greater than or equal to 1 and y-x is greater than or equal to 5.\" width=\"329\" height=\"325\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The videos that follow show more\u00a0examples of graphing the solution set of a system of linear inequalities.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Ex 1:  Graph a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ACTxJv1h2_c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Ex 2:  Graph a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/cclH2h1NurM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The system in our last example includes a compound inequality. \u00a0We will see that you can treat a compound inequality like two lines when you are graphing them.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the solution to the system [latex]3x + 2y < 12[\/latex] and [latex]\u22121\\leq y \\leq 5[\/latex].\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q163187\">Show Answer<\/span><\/p>\n<div id=\"q163187\" class=\"hidden-answer\" style=\"display: none\">\n<p>Graph one inequality. First graph the boundary line, then test points.<\/p>\n<p>Remember, because the inequality [latex]3x + 2y<12[\/latex] does not include the equal sign, draw a dashed border line.\n\nTesting a point like [latex](0, 0)[\/latex] will show that the area below the line is the solution to this inequality.\n\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2427\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223640\/image021-300x297.gif\" alt=\"image021\" width=\"300\" height=\"297\" \/><\/p>\n<p>The inequality [latex]\u22121\\leq y \\leq 5[\/latex] is actually two inequalities: [latex]\u22121\\leq y[\/latex], and [latex]y \\leq 5[\/latex]. Another way to think of this is y must be between \u22121 and 5. The border lines for both are horizontal. The region between those two lines contains the solutions of [latex]\u22121\\leq y\\leq 5[\/latex]. We make the lines solid because we also want to include [latex]y = \u22121[\/latex] and [latex]y = 5[\/latex].<\/p>\n<p>Graph this region on the same axes as the other inequality.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2428\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223642\/image022-300x298.gif\" alt=\"image022\" width=\"300\" height=\"298\" \/><\/p>\n<p>The purple region in this graph shows the set of all solutions of the system.<\/p>\n<h4>Answer<\/h4>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2429\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223644\/image023-300x297.jpg\" alt=\"image023\" width=\"300\" height=\"297\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we show how to solve another system of inequalities.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Ex 1:  Graph a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ACTxJv1h2_c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(6.4.6) &#8211; Applications of systems of linear inequalities<\/h1>\n<p>In our first\u00a0example we will show how to write and graph a system of linear inequalities that models the amount of sales needed to obtain a specific amount of money.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Cathy is selling ice cream cones at a school fundraiser. She is selling two sizes: small (which has 1 scoop) and large (which has 2 scoops). She knows that she can get a maximum of 70 scoops of ice cream out of her supply. She charges $3 for a small cone and $5 for a large cone.<\/p>\n<p>Cathy wants to earn at least $120 to give back to the school. Write and graph a system of inequalities that models this situation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737192\">Show Answer<\/span><\/p>\n<div id=\"q737192\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, identify the variables. There are two variables: the number of small cones and the number of large cones.<\/p>\n<p style=\"text-align: center\">[latex]s[\/latex]\u00a0= small cone<\/p>\n<p style=\"text-align: center\">[latex]l[\/latex]\u00a0= large cone<\/p>\n<p style=\"text-align: left\">Write the first equation: the maximum number of scoops she can give out. The scoops she has available (70) must be greater than or equal to the number of scoops for the small cones [latex](<i>s<\/i>)[\/latex] and the large cones [latex](2<i>l<\/i>)[\/latex] she sells.<\/p>\n<p style=\"text-align: center\">[latex]s+2l\\le70[\/latex]<\/p>\n<p style=\"text-align: left\">Write the second equation: the amount of money she raises. She wants the total amount of money earned from small cones [latex](3<i>s<\/i>)[\/latex] and large cones [latex](5<i>l<\/i>)[\/latex] to be at least $120.<\/p>\n<p style=\"text-align: center\">[latex]3s+5l\\ge120[\/latex]<\/p>\n<p style=\"text-align: left\">Write the system.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{cases}s+2l\\le70\\\\3s+5l\\ge120\\end{cases}[\/latex]<\/p>\n<p>Now graph the system. The variables [latex]x[\/latex] and [latex]y[\/latex] have been replaced by [latex]s[\/latex] and [latex]l[\/latex]; graph s along the [latex]x[\/latex]-axis, and [\/latex]l[\/latex]\u00a0along the [\/latex]y[\/latex]-axis.<\/p>\n<p>First graph the region [latex]<i>s<\/i> + 2<i>l<\/i>\u00a0\\leq 70[\/latex]. Graph the boundary line and then test individual points to see which region to shade. The graph is shown below.<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2433\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223652\/image027-300x203.jpg\" alt=\"image027\" width=\"300\" height=\"203\" \/><\/p>\n<p>Now graph the region [latex]3s+5l\\ge120[\/latex]\u00a0Graph the boundary line and then test individual points to see which region to shade. The graph is shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2434\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223654\/image028-300x203.jpg\" alt=\"image028\" width=\"300\" height=\"203\" \/><\/p>\n<p>Graphing the regions together, you find the following:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2435\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223656\/image029-300x203.jpg\" alt=\"image029\" width=\"300\" height=\"203\" \/><\/p>\n<p>And represented just as the overlapping region, you have:<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2436\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223658\/image030-300x203.jpg\" alt=\"image030\" width=\"300\" height=\"203\" \/><\/p>\n<h4>Answer<\/h4>\n<p>The region in purple is the solution. As long as the combination of small cones and large cones that Cathy sells can be mapped in the purple region, she will have earned at least $120 and not used more than 70 scoops of ice cream.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In a previous example for finding a solution to a system of linear equations, we introduced a manufacturer\u2019s\u00a0cost and revenue equations:<\/p>\n<p>Cost: [latex]y=0.85x+35,000[\/latex]<\/p>\n<p>Revenue: [latex]y=1.55x[\/latex]<\/p>\n<p>The cost equation is shown in blue in the graph below, and the revenue equation is graphed in orange.The point at which the two lines intersect is called the break-even point, we learned that this is the solution to the system of linear equations that in this case comprise the cost and revenue equations.<\/p>\n<p>The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The region to the left represents quantities for which the company suffers a loss.<\/p>\n<p>In the next example, you will see how the information you learned about systems of linear inequalities can be applied to answering questions about cost and revenue.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183256\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"487\" height=\"390\" \/><\/p>\n<p>Note how the blue shaded region between the Cost and Revenue equations is labeled Profit. This is the &#8220;sweet spot&#8221; that the company wants to achieve\u00a0where they\u00a0produce enough bike frames at a minimal enough cost to\u00a0make money. They don&#8217;t want more money going out than coming in!<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Define the profit region for the skateboard\u00a0manufacturing business using inequalities, given the system of linear equations:<\/p>\n<p>Cost: [latex]y=0.85x+35,000[\/latex]<\/p>\n<p>Revenue: [latex]y=1.55x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q563864\">Show Solution<\/span><\/p>\n<div id=\"q563864\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that graphically, \u00a0solutions to\u00a0linear inequalities are entire regions, and we learned how to graph systems of linear inequalities earlier in this module. Based on the graph below and the equations that define cost and revenue, we can use inequalities to define the region for which the skateboard\u00a0manufacturer will make a profit.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4210 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183258\/Screen-Shot-2016-05-18-at-3.33.11-PM-300x280.png\" alt=\"Cost\/ Revenue with Profit\" width=\"358\" height=\"334\" \/><\/p>\n<p>Let&#8217;s start with the revenue equation. \u00a0We know that the break even point is at (50,000, 77,500) and the profit region is\u00a0the blue area. \u00a0If we choose a point in the region and test it like we did for finding solution regions to inequalities, we will know which kind of inequality sign to use.<\/p>\n<p>Let&#8217;s test the point [latex]\\left(65,00,100,000\\right)[\/latex] in both equations to determine which inequality sign to use.<\/p>\n<p>Cost:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y=0.85x+{35,000}\\\\{100,000}\\text{ ? }0.85\\left(65,000\\right)+35,000\\\\100,000\\text{ ? }90,250\\end{array}[\/latex]<\/p>\n<p>We need to use &gt; because 100,000 is greater than 90,250<\/p>\n<p>The cost inequality that will ensure the company makes profit &#8211; not just break even &#8211; is\u00a0[latex]y>0.85x+35,000[\/latex]<\/p>\n<p>Now test the point in the revenue equation:<\/p>\n<p>Revenue:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y=1.55x\\\\100,000\\text{ ? }1.55\\left(65,000\\right)\\\\100,000\\text{ ? }100,750\\end{array}[\/latex]<\/p>\n<p>We need to use &lt; because 100,000 is less than 100,750<\/p>\n<p>The revenue inequality that will ensure the company makes profit &#8211; not just break even &#8211; is\u00a0[latex]y<1.55x[\/latex]\n\nThe systems of inequalities that defines the profit region for the bike manufacturer:\n\n\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y>0.85x+35,000\\\\y<1.55x\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. The system of linear inequalities that represents the number of units that the company must produce in order to earn a profit is:<\/p>\n<p>[latex]\\begin{array}{l}y>0.85x+35,000\\\\y<1.55x\\end{array}[\/latex]\n\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video you will see an example of how to find the break even point for a small sno-cone business.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"System of Equations App:  Break-Even Point\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qey3FmE8saQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>And here is one more video example of solving an application\u00a0using a sustem of linear inequalities.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-12\" title=\"Ex:  Linear Inequality in Two Variables Application Problem  (Phone Cost:  Day and Night Minutes)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/gbHl6K-dJ8o?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We have seen that systems of linear equations and inequalities can help to define market behaviors that are very helpful to businesses. \u00a0The intersection of cost and revenue equations gives the break even point, and also helps define the region for which a company will make a profit.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-700\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Ex 1: Graphing Linear Inequalities in Two Variables (Slope Intercept Form). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Hzxc4HASygU\">https:\/\/youtu.be\/Hzxc4HASygU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>System of Equations App: Break-Even Point.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qey3FmE8saQ.%20\">https:\/\/youtu.be\/qey3FmE8saQ.%20<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 2: Graphing Linear Inequalities in Two Variables (Standard Form). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/2VgFg2ztspI\">https:\/\/youtu.be\/2VgFg2ztspI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 13: Graphing, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Use a Graph Determine Ordered Pair Solutions of a Linear Inequalty in Two Variable. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/GQVdDRVq5_o\">https:\/\/youtu.be\/GQVdDRVq5_o<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Determine if Ordered Pairs Satisfy a Linear Inequality. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-x-zt_yM0RM\">https:\/\/youtu.be\/-x-zt_yM0RM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Graph a System of Linear Inequalities. . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) .. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ACTxJv1h2_c\">https:\/\/youtu.be\/ACTxJv1h2_c<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Graph a System of Linear Inequalities.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/cclH2h1NurM\">https:\/\/youtu.be\/cclH2h1NurM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program.. <strong>Provided by<\/strong>: Monterey Institute of Technology. . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext.\">http:\/\/nrocnetwork.org\/dm-opentext.<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Determine the Solution to a System of Inequalities (Compound).. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/D-Cnt6m8l18.\">https:\/\/youtu.be\/D-Cnt6m8l18.<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>College Algebra. <strong>Authored by<\/strong>: Jay Abrams, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstaxcollege.org\/textbooks\/college-algebra.\">https:\/\/openstaxcollege.org\/textbooks\/college-algebra.<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Intermediate Algebra . <strong>Authored by<\/strong>: Lynn Marecek et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@4.13\">http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@4.13<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@4.13<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":5,"template":"","meta":{"_candela_citation":"{\"0\":{\"type\":\"cc\",\"description\":\"Ex 2: Graphing Linear Inequalities in Two Variables (Standard Form)\",\"author\":\"James Sousa (Mathispower4u.com) 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