{"id":98,"date":"2023-11-08T16:09:51","date_gmt":"2023-11-08T16:09:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-graphing-linear-equations-in-different-forms-2\/"},"modified":"2026-02-05T10:14:02","modified_gmt":"2026-02-05T10:14:02","slug":"2-3-solutions-to-systems-of-linear-inequalities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-3-solutions-to-systems-of-linear-inequalities\/","title":{"raw":"2.3 Solutions to Systems of Linear Inequalities","rendered":"2.3 Solutions to Systems of Linear Inequalities"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning OutcomeS<\/h3>\r\n<ul>\r\n \t<li>Determine whether an ordered pair is a solution of a two-variable linear inequality.<\/li>\r\n \t<li>Graph the solution set of a single two-variable linear inequality.<\/li>\r\n \t<li>Graph the solution set of a system of two-variable linear inequalities with two inequalities.<\/li>\r\n \t<li>Graph the solution set of a system of two-variable linear inequalities with more than two inequalities.<\/li>\r\n<\/ul>\r\n<\/div>\r\nA <strong>linear inequality<\/strong> is the same as a linear equation [latex]Ax+By=C,[\/latex] but with the equal sign replaced with an inequality sign. An example is [latex]2x-3y\\leq4.[\/latex] Often a linear inequality is written in \"slope-intercept form,\" for example\u00a0[latex]y\\leq x-2.[\/latex]\r\n\r\nAn ordered pair is a <strong>solution<\/strong> if it makes the inequality true when we substitute [latex]x[\/latex] and\u00a0[latex]y.[\/latex] Here is a quick example.\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<ol>\r\n \t<li>[latex](0,0)[\/latex]<\/li>\r\n \t<li>[latex](1,6)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"837554\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"837554\"]\r\n<ol>\r\n \t<li>Substitute [latex]0[\/latex] for [latex]x[\/latex] and [latex]0[\/latex] 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\nThe ordered pair [latex](0,0)[\/latex] is not a solution to\u00a0[latex]y&gt;x+4.[\/latex]<\/li>\r\n \t<li>Substitute [latex]1[\/latex] for [latex]x[\/latex] and [latex]6[\/latex] 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\nThe ordered pair [latex](1,6)[\/latex] is a solution to\u00a0[latex]y&gt;x+4.[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIt is not a useful strategy to find all solutions to a linear inequality by testing single ordered pairs. We will see that linear inequalities have infinitely many solutions, so a graph in the coordinate plane is a good way to represent the solution set.\r\n\r\nLet's consider a first example, to graph the inequality\r\n<p style=\"text-align: center;\">[latex]x+4y\\leq4[\/latex].<\/p>\r\nAny point on the line [latex]x+4y=4[\/latex] will satisfy the inequality since this inequality symbol allows equality, so we begin by graphing the line. You can use the [latex]x[\/latex]<i>\u00a0<\/i>and [latex]y[\/latex]-intercepts for this equation to graph the line. For example, find the\u00a0[latex]y[\/latex]-intercept by setting [latex]x=0[\/latex] and solving for [latex]y.[\/latex]\r\n<table style=\"width: 80px; height: 42px;\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 150.01px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 162.656px;\">[latex]y[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 150.01px;\">[latex]0[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 162.656px;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 150.01px;\">[latex]4[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 162.656px;\">[latex]0[\/latex]<\/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. This line is called the <strong>boundary line<\/strong>.\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\nNext, let's 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)\\overset{?}{\\leq}4\\\\\u22121+12\\overset{?}{\\leq}4\\\\11\\not\\leq4\\end{array}[\/latex]<\/p>\r\nThis is a false statement so\u00a0[latex](\u22121,3)[\/latex] is not a solution.\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)\\overset{?}{\\leq}4\\\\2+0\\overset{?}{\\leq}4\\\\2\\leq4\\end{array}[\/latex]<\/p>\r\nThis is true! As it turns out, <strong>the boundary line is the dividing line between solutions and non-solutions<\/strong>. We call each side of the line a <strong>half-plane<\/strong>. The half-plane that includes [latex](2,0)[\/latex] should be shaded to indicate this is the region of solutions for the inequality. The half-plane that includes\u00a0[latex](-1,3)[\/latex] has no 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, the graph of the set of solutions for [latex]x+4y\\leq4[\/latex]. Here is our process in summarized form.\r\n<div class=\"textbox shaded\">\r\n<h3><strong>Graphing a linear inequality<\/strong><\/h3>\r\n<ol>\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.\r\n<div class=\"textbox\">\r\n<ul>\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 are not solutions, then use a dashed line for the boundary line. This will happen for &lt; or &gt; inequalities.<\/li>\r\n<\/ul>\r\n<\/div><\/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 half-plane that contains the ordered pairs that make the inequality a true statement.<b>\u00a0<\/b><\/li>\r\n<\/ol>\r\n<\/div>\r\nBelow is a video about how to graph inequalities with two variables.\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\nFirst we need to graph the boundary line. In this example we can do this by writing it in slope intercept form. Remember to switch the inequality sign if dividing by a negative number during this step.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{rl}2y&amp;&gt;\\; 4x-6\\\\\\\\\\dfrac{2y}{2}&amp;&gt;\\; \\dfrac{4x}{2}-\\dfrac{6}{2}\\\\\\\\y&amp;&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]\r\n<table style=\"width: 80px; height: 42px;\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 150.01px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 162.656px;\">[latex]y[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 150.01px;\">[latex]0[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 162.656px;\">[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 150.01px;\">[latex]2[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 162.656px;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points and graph the line. The line is dashed because the sign in the inequality is [latex]&gt;[\/latex], not [latex]\\geq[\/latex] 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: left;\">An ordered pair on each side of the boundary line is chosen above. Test each point in the original inequality.<\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: center;\">\r\n[latex]\\begin{array}{l}\\\\\\text{Test }1:\\left(\u22123,1\\right)\\\\2\\left(1\\right)\\overset{?}{&gt;}4\\left(\u22123\\right)\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2\\overset{?}{&gt;}\u201312\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2&gt;\u221218\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\text{TRUE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\2(1)\\overset{?}{&gt;}4\\left(4\\right)\u2013 6\\\\\\,\\,\\,\\,\\,\\,2\\overset{?}{&gt;}16\u20136\\\\\\,\\,\\,\\,\\,\\,2\\not &gt;10\\\\\\,\\,\\,\\,\\,\\text{FALSE}\\end{array}[\/latex]<\/span><\/p>\r\n<p style=\"text-align: left;\">Since [latex](\u22123,1)[\/latex] results in a true statement, the half-plane that includes [latex](\u22123,1)[\/latex] should be shaded.<\/p>\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\nNote that you can use the points [latex](0,\u22123)[\/latex] and [latex](2,1)[\/latex] to graph the boundary line, but these points are not included in the region of solutions since the region does not include the boundary line!\r\n\r\nAlso note that we only really needed to use one test point instead of two, since we know only one side of the line can be the solution region. If the first test point results in a false statement, shade the other side.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIf you rewrite your inequality in slope-intercept form as in the previous example, it provides an alternative approach to deciding the direction of shading. A\u00a0[latex]&gt;[\/latex] or [latex]\\geq[\/latex] symbol as we had in that example always means shade above the line. A\u00a0[latex]&lt;[\/latex] or [latex]\\leq[\/latex] symbol means shade below. Note this approach does not work unless the inequality is put into slope-intercept form!\r\n<p id=\"video2\">This video gives an example of using this alternative approach to do the shading.<\/p>\r\nhttps:\/\/youtu.be\/Hzxc4HASygU\r\n<h2 id=\"title2\">Systems of Two or More Inequalities<\/h2>\r\nThe graph of a single linear inequality in two variables splits the <b>coordinate plane<\/b> into two regions.\u00a0On one side lie all the solutions to the inequality. Now consider the system of inequalities,\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{l}y&lt;2x+5\\\\y&gt;-x \\end{array} \\right.[\/latex]<\/p>\r\nHere are the graphs of each inequality independently.\r\n\r\n<img class=\"aligncenter\" style=\"float: left; margin-right: 50px;\" 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=\"257\" height=\"255\" \/><img class=\"aligncenter\" style=\"float: left;\" 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=\"255\" height=\"253\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nYou can verify by substitution that points [latex]A[\/latex] and [latex]B[\/latex] satisfy the first inequality, while\u00a0points [latex]M[\/latex] and [latex]N[\/latex] satisfy the second inequality.\r\n\r\nA <strong>solution<\/strong> to a system of inequalities must satisfy both inequalities at the same time. The <strong>solution set<\/strong> consists of all points which satisfy both inequalities. The easiest way to find the solution set is to graph both inequalities on the same graph and find the region of overlap.\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 system of inequalities. Any point within this purple region will be true for both [latex]y&lt;2x+5[\/latex] and [latex]y&gt;\u2212x.[\/latex] Note in particular that the blue and red regions do NOT solve the system because they only satisfy one of the two inequalities. In the future, we will usually only shade the overlap region to avoid confusion.\r\n\r\nWe can verify by substitution that the point [latex]M=(-2,3)[\/latex] in the blue region is not a solution to the system since it does not satisfy the first inequality:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y&amp;&lt;2x+5\\\\\r\n3&amp;&lt;2(-2)+5\\\\\r\n3&amp;&lt;-4+5\\\\\r\n3&amp;&lt;1&amp;&amp;\\color{red}{\\textsf{False}}\\end{align}[\/latex]<\/p>\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\nThe general steps are outlined below.\r\n<div class=\"textbox shaded\">\r\n<h3><strong>Graphing a SYSTEm of linear inequalities<\/strong><\/h3>\r\n<ol>\r\n \t<li>Graph all inequalities on the same axes. Determine whether each line is solid or dashed and which side of each line to shade on.<\/li>\r\n \t<li>Shade the region that represents the overlap of all solution regions for inequalities in the system.<\/li>\r\n<\/ol>\r\n<\/div>\r\nThe system in our next example includes a compound inequality. \u00a0We will see that you can treat a compound inequality like two inequalities when you are graphing them, so this example is basically a system of three inequalities.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph the system\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}3x+2y&amp;&lt;12\\\\-1 \\leq y &amp;\\leq 5 \\end{array} \\right.[\/latex]<\/p>\r\n[reveal-answer q=\"163187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"163187\"]\r\n\r\nBegin by graphing the first inequality. The boundary line has intercepts at [latex](0,6)[\/latex] and\u00a0[latex](4,0)[\/latex] and the line should be dashed. Testing the point [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 wp-image-2427 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223640\/image021-300x297.gif\" alt=\"Dashed line with intercepts at (0,6) and (4,0) with shading below the line. Line labeled 3 x + 2 y &lt; 12.\" width=\"300\" height=\"297\" \/>\r\n\r\nThe inequality [latex] -1 \u2264 y \u2264 5[\/latex] is actually two inequalities:\u00a0[latex]\u22121 \u2264 y,[\/latex] and\u00a0[latex]y \u2264 5.[\/latex] Another way to think of this is [latex]y[\/latex] must be between\u00a0[latex]\u22121[\/latex] and\u00a0[latex]5[\/latex]. The border lines for both are horizontal. The region between those two lines contains the solutions of [latex] -1 \u2264 y \u2264 5[\/latex]. We make the lines solid because we also want to include\u00a0[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 wp-image-2428 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223642\/image022-300x298.gif\" alt=\"Dashed line with intercepts at (0,6) and (4,0) with shading red below the line. Horizontal lines at y = 5 and y = negative 1 and shaded blue between the two horizontal lines with negative 1 &lt;= y &lt;= 5 written.\" width=\"300\" height=\"298\" \/>\r\n\r\nThe purple region shows the set of all solutions of the system.\r\n\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\nThe next example is explicitly a system of three inequalities to graph.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nGraph the system\r\n\r\n[latex]\\left\\{ \\begin{array}{rl}x+y&lt;3\\\\y\\geq\\dfrac{2}{3}x+1\\\\y&lt;4x+6 \\end{array} \\right.[\/latex]\r\n\r\n[reveal-answer q=\"142281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"142281\"]\r\n\r\nBegin by graphing all three boundary lines on the same axes. This splits the plane into seven regions. Only one of them can be the solution region.\r\n\r\n<img class=\"aligncenter wp-image-1838 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-300x300.png\" alt=\"Three lines intersect on a coordinate system forming a triangle shape: a solid blue line with positive slope passing through (negative 1.5,0) and (0,1), an orange dashed line with negative slope passing through (3,0) and (0,3), and a green dashed line with positive slope passing through (negative 1.5,0) and approximately (negative 1,3).\" width=\"300\" height=\"300\" \/>\r\n\r\nThe point [latex](0,0)[\/latex] is not on any of the lines, so use it as a test point.\r\n\r\nInequality 1 (red):\u00a0 [latex]0+0\\leq 3[\/latex] TRUE\r\n\r\nInequality 2 (blue):\u00a0 [latex]0\\geq \\dfrac{2}{3}\\cdot 0 + 1, \\textsf{ or } 0 \\geq 1[\/latex] FALSE\r\n\r\nInequality 3 (green):\u00a0\u00a0[latex]0&lt;4\\cdot 0 + 6, \\textsf{ or } 0&lt;6[\/latex] TRUE\r\n\r\nFor lines 1 (red) and 3 (green) we shade toward the point [latex](0, 0),[\/latex] which means down. For line 2 (blue) we shade away from the point [latex](0,0),[\/latex] which means up.\r\n\r\n<img class=\"aligncenter wp-image-1768 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-300x300.png\" alt=\"Three lines on a coordinate plane: a blue increasing line y is greater than or equal to quantity two-thirds x plus 1 and shaded above, a green increasing line y is less than 4 x plus 6 shaded above\/right, and an orange dashed decreasing line x plus y is less than 3 shaded below. Overlapping triangular area is shaded with all three colors.\" width=\"300\" height=\"300\" \/>\r\n\r\nThe solution region is the one shaded all three times, which is the center triangle. If we want to be extra sure, we can test another point which is inside that triangle and verify that it satisfies all three inequalities. Test point [latex](0,2):[\/latex]\r\n\r\nInequality 1:\u00a0 [latex]0+2\\leq 3[\/latex] TRUE\r\n\r\nInequality 2:\u00a0 [latex]2\\geq \\dfrac{2}{3}\\cdot 0 + 1, \\textsf{ or } 2 \\geq 1[\/latex] TRUE\r\n\r\nInequality 3:\u00a0\u00a0[latex]2&lt;4\\cdot 0 + 6, \\textsf{ or } 2&lt;6[\/latex] TRUE\r\n\r\nThe solution region is the center triangle. Two edges of the triangle are dashed (not solutions), and the bottom edge which is part of the blue line is in the solution set.\r\n\r\n<img class=\"aligncenter wp-image-1769 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-300x300.png\" alt=\"Three lines and their equations are shown: a blue increasing line representing y is greater than or equal to quantity two-thirds x plus 1, a green increasing line representing y is less than 4 x plus 6, and an orange dashed decreasing line representing x plus y is less than 3. Only the triangle between lines is shaded purple.\" width=\"300\" height=\"300\" \/>\r\n\r\nNote that by modifying the original problem and switching the direction of some of the inequality symbols, we could have made any of the seven regions be the solution region. For example, switching only Inequality 2 to a [latex]\\leq[\/latex] sign would have caused our first test point [latex](0,0)[\/latex] to satisfy all three inequalities, so the solution region would have been the bottom, unbounded region containing\u00a0[latex](0,0).[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nWhen linear inequalities are graphed on a coordinate plane, the solution takes the form of all points on one side of the boundary line, and possibly also the line itself. You can tell which region to shade by testing some points in the inequality, or observing the direction of the sign if the inequality is in slope-intercept form. When graphing a system of linear inequalities, graph all inequalities on the same axes. The solution set is the overlap of all the individual solution sets.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning OutcomeS<\/h3>\n<ul>\n<li>Determine whether an ordered pair is a solution of a two-variable linear inequality.<\/li>\n<li>Graph the solution set of a single two-variable linear inequality.<\/li>\n<li>Graph the solution set of a system of two-variable linear inequalities with two inequalities.<\/li>\n<li>Graph the solution set of a system of two-variable linear inequalities with more than two inequalities.<\/li>\n<\/ul>\n<\/div>\n<p>A <strong>linear inequality<\/strong> is the same as a linear equation [latex]Ax+By=C,[\/latex] but with the equal sign replaced with an inequality sign. An example is [latex]2x-3y\\leq4.[\/latex] Often a linear inequality is written in &#8220;slope-intercept form,&#8221; for example\u00a0[latex]y\\leq x-2.[\/latex]<\/p>\n<p>An ordered pair is a <strong>solution<\/strong> if it makes the inequality true when we substitute [latex]x[\/latex] and\u00a0[latex]y.[\/latex] Here is a quick example.<\/p>\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<ol>\n<li>[latex](0,0)[\/latex]<\/li>\n<li>[latex](1,6)[\/latex]<\/li>\n<\/ol>\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<ol>\n<li>Substitute [latex]0[\/latex] for [latex]x[\/latex] and [latex]0[\/latex] for [latex]y[\/latex]:\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>The ordered pair [latex](0,0)[\/latex] is not a solution to\u00a0[latex]y>x+4.[\/latex]<\/li>\n<li>Substitute [latex]1[\/latex] for [latex]x[\/latex] and [latex]6[\/latex] for [latex]y[\/latex]:\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>The ordered pair [latex](1,6)[\/latex] is a solution to\u00a0[latex]y>x+4.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>It is not a useful strategy to find all solutions to a linear inequality by testing single ordered pairs. We will see that linear inequalities have infinitely many solutions, so a graph in the coordinate plane is a good way to represent the solution set.<\/p>\n<p>Let&#8217;s consider a first example, to graph the inequality<\/p>\n<p style=\"text-align: center;\">[latex]x+4y\\leq4[\/latex].<\/p>\n<p>Any point on the line [latex]x+4y=4[\/latex] will satisfy the inequality since this inequality symbol allows equality, so we begin by graphing the line. You can use the [latex]x[\/latex]<i>\u00a0<\/i>and [latex]y[\/latex]-intercepts for this equation to graph the line. For example, find the\u00a0[latex]y[\/latex]-intercept by setting [latex]x=0[\/latex] and solving for [latex]y.[\/latex]<\/p>\n<table style=\"width: 80px; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 150.01px;\">[latex]x[\/latex]<\/td>\n<td style=\"height: 14px; width: 162.656px;\">[latex]y[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 150.01px;\">[latex]0[\/latex]<\/td>\n<td style=\"height: 14px; width: 162.656px;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 150.01px;\">[latex]4[\/latex]<\/td>\n<td style=\"height: 14px; width: 162.656px;\">[latex]0[\/latex]<\/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. This line is called the <strong>boundary line<\/strong>.<\/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>Next, let&#8217;s 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)\\overset{?}{\\leq}4\\\\\u22121+12\\overset{?}{\\leq}4\\\\11\\not\\leq4\\end{array}[\/latex]<\/p>\n<p>This is a false statement so\u00a0[latex](\u22121,3)[\/latex] is not a solution.<\/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)\\overset{?}{\\leq}4\\\\2+0\\overset{?}{\\leq}4\\\\2\\leq4\\end{array}[\/latex]<\/p>\n<p>This is true! As it turns out, <strong>the boundary line is the dividing line between solutions and non-solutions<\/strong>. We call each side of the line a <strong>half-plane<\/strong>. The half-plane that includes [latex](2,0)[\/latex] should be shaded to indicate this is the region of solutions for the inequality. The half-plane that includes\u00a0[latex](-1,3)[\/latex] has no 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, the graph of the set of solutions for [latex]x+4y\\leq4[\/latex]. Here is our process in summarized form.<\/p>\n<div class=\"textbox shaded\">\n<h3><strong>Graphing a linear inequality<\/strong><\/h3>\n<ol>\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.\n<div class=\"textbox\">\n<ul>\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 are not solutions, then use a dashed line for the boundary line. This will happen for &lt; or &gt; inequalities.<\/li>\n<\/ul>\n<\/div>\n<\/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 half-plane that contains the ordered pairs that make the inequality a true statement.<b>\u00a0<\/b><\/li>\n<\/ol>\n<\/div>\n<p>Below is a video about how to graph inequalities with two variables.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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>First we need to graph the boundary line. In this example we can do this by writing it in slope intercept form. Remember to switch the inequality sign if dividing by a negative number during this step.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{rl}2y&>\\; 4x-6\\\\\\\\\\dfrac{2y}{2}&>\\; \\dfrac{4x}{2}-\\dfrac{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]<\/p>\n<table style=\"width: 80px; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 150.01px;\">[latex]x[\/latex]<\/td>\n<td style=\"height: 14px; width: 162.656px;\">[latex]y[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 150.01px;\">[latex]0[\/latex]<\/td>\n<td style=\"height: 14px; width: 162.656px;\">[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 150.01px;\">[latex]2[\/latex]<\/td>\n<td style=\"height: 14px; width: 162.656px;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points and graph the line. The line is dashed because the sign in the inequality is [latex]>[\/latex], not [latex]\\geq[\/latex] 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: left;\">An ordered pair on each side of the boundary line is chosen above. Test each point in the original inequality.<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: center;\"><br \/>\n[latex]\\begin{array}{l}\\\\\\text{Test }1:\\left(\u22123,1\\right)\\\\2\\left(1\\right)\\overset{?}{>}4\\left(\u22123\\right)\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2\\overset{?}{>}\u201312\u20136\\\\\\,\\,\\,\\,\\,\\,\\,2>\u221218\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\text{TRUE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\2(1)\\overset{?}{>}4\\left(4\\right)\u2013 6\\\\\\,\\,\\,\\,\\,\\,2\\overset{?}{>}16\u20136\\\\\\,\\,\\,\\,\\,\\,2\\not >10\\\\\\,\\,\\,\\,\\,\\text{FALSE}\\end{array}[\/latex]<\/span><\/p>\n<p style=\"text-align: left;\">Since [latex](\u22123,1)[\/latex] results in a true statement, the half-plane that includes [latex](\u22123,1)[\/latex] should be shaded.<\/p>\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<p>Note that you can use the points [latex](0,\u22123)[\/latex] and [latex](2,1)[\/latex] to graph the boundary line, but these points are not included in the region of solutions since the region does not include the boundary line!<\/p>\n<p>Also note that we only really needed to use one test point instead of two, since we know only one side of the line can be the solution region. If the first test point results in a false statement, shade the other side.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>If you rewrite your inequality in slope-intercept form as in the previous example, it provides an alternative approach to deciding the direction of shading. A\u00a0[latex]>[\/latex] or [latex]\\geq[\/latex] symbol as we had in that example always means shade above the line. A\u00a0[latex]<[\/latex] or [latex]\\leq[\/latex] symbol means shade below. Note this approach does not work unless the inequality is put into slope-intercept form!\n\n\n<p id=\"video2\">This video gives an example of using this alternative approach to do the shading.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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<h2 id=\"title2\">Systems of Two or More Inequalities<\/h2>\n<p>The graph of a single linear inequality in two variables splits the <b>coordinate plane<\/b> into two regions.\u00a0On one side lie all the solutions to the inequality. Now consider the system of inequalities,<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{l}y<2x+5\\\\y>-x \\end{array} \\right.[\/latex]<\/p>\n<p>Here are the graphs of each inequality independently.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" style=\"float: left; margin-right: 50px;\" 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=\"257\" height=\"255\" \/><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" style=\"float: left;\" 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=\"255\" height=\"253\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>You can verify by substitution that points [latex]A[\/latex] and [latex]B[\/latex] satisfy the first inequality, while\u00a0points [latex]M[\/latex] and [latex]N[\/latex] satisfy the second inequality.<\/p>\n<p>A <strong>solution<\/strong> to a system of inequalities must satisfy both inequalities at the same time. The <strong>solution set<\/strong> consists of all points which satisfy both inequalities. The easiest way to find the solution set is to graph both inequalities on the same graph and find the region of overlap.<\/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 system of inequalities. Any point within this purple region will be true for both [latex]y<2x+5[\/latex] and [latex]y>\u2212x.[\/latex] Note in particular that the blue and red regions do NOT solve the system because they only satisfy one of the two inequalities. In the future, we will usually only shade the overlap region to avoid confusion.<\/p>\n<p>We can verify by substitution that the point [latex]M=(-2,3)[\/latex] in the blue region is not a solution to the system since it does not satisfy the first inequality:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y&<2x+5\\\\  3&<2(-2)+5\\\\  3&<-4+5\\\\  3&<1&&\\color{red}{\\textsf{False}}\\end{align}[\/latex]<\/p>\n<p>In the following video examples, we show how to graph a system of linear inequalities and define the solution region.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" 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-4\" 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 general steps are outlined below.<\/p>\n<div class=\"textbox shaded\">\n<h3><strong>Graphing a SYSTEm of linear inequalities<\/strong><\/h3>\n<ol>\n<li>Graph all inequalities on the same axes. Determine whether each line is solid or dashed and which side of each line to shade on.<\/li>\n<li>Shade the region that represents the overlap of all solution regions for inequalities in the system.<\/li>\n<\/ol>\n<\/div>\n<p>The system in our next example includes a compound inequality. \u00a0We will see that you can treat a compound inequality like two inequalities when you are graphing them, so this example is basically a system of three inequalities.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph the system<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}3x+2y&<12\\\\-1 \\leq y &\\leq 5 \\end{array} \\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q163187\">Show Solution<\/span><\/p>\n<div id=\"q163187\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by graphing the first inequality. The boundary line has intercepts at [latex](0,6)[\/latex] and\u00a0[latex](4,0)[\/latex] and the line should be dashed. Testing the point [latex](0,0)[\/latex] will show that the area below the line is the solution to this inequality.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2427 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223640\/image021-300x297.gif\" alt=\"Dashed line with intercepts at (0,6) and (4,0) with shading below the line. Line labeled 3 x + 2 y &lt; 12.\" width=\"300\" height=\"297\" \/><\/p>\n<p>The inequality [latex]-1 \u2264 y \u2264 5[\/latex] is actually two inequalities:\u00a0[latex]\u22121 \u2264 y,[\/latex] and\u00a0[latex]y \u2264 5.[\/latex] Another way to think of this is [latex]y[\/latex] must be between\u00a0[latex]\u22121[\/latex] and\u00a0[latex]5[\/latex]. The border lines for both are horizontal. The region between those two lines contains the solutions of [latex]-1 \u2264 y \u2264 5[\/latex]. We make the lines solid because we also want to include\u00a0[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 wp-image-2428 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/11223642\/image022-300x298.gif\" alt=\"Dashed line with intercepts at (0,6) and (4,0) with shading red below the line. Horizontal lines at y = 5 and y = negative 1 and shaded blue between the two horizontal lines with negative 1 &lt;= y &lt;= 5 written.\" width=\"300\" height=\"298\" \/><\/p>\n<p>The purple region shows the set of all solutions of the system.<\/p>\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>The next example is explicitly a system of three inequalities to graph.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Graph the system<\/p>\n<p>[latex]\\left\\{ \\begin{array}{rl}x+y<3\\\\y\\geq\\dfrac{2}{3}x+1\\\\y<4x+6 \\end{array} \\right.[\/latex]\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q142281\">Show Solution<\/span><\/p>\n<div id=\"q142281\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by graphing all three boundary lines on the same axes. This splits the plane into seven regions. Only one of them can be the solution region.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1838 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-300x300.png\" alt=\"Three lines intersect on a coordinate system forming a triangle shape: a solid blue line with positive slope passing through (negative 1.5,0) and (0,1), an orange dashed line with negative slope passing through (3,0) and (0,3), and a green dashed line with positive slope passing through (negative 1.5,0) and approximately (negative 1,3).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-1024x1024.png 1024w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3LinesNew.png 1414w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>The point [latex](0,0)[\/latex] is not on any of the lines, so use it as a test point.<\/p>\n<p>Inequality 1 (red):\u00a0 [latex]0+0\\leq 3[\/latex] TRUE<\/p>\n<p>Inequality 2 (blue):\u00a0 [latex]0\\geq \\dfrac{2}{3}\\cdot 0 + 1, \\textsf{ or } 0 \\geq 1[\/latex] FALSE<\/p>\n<p>Inequality 3 (green):\u00a0\u00a0[latex]0<4\\cdot 0 + 6, \\textsf{ or } 0<6[\/latex] TRUE\n\nFor lines 1 (red) and 3 (green) we shade toward the point [latex](0, 0),[\/latex] which means down. For line 2 (blue) we shade away from the point [latex](0,0),[\/latex] which means up.\n\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1768 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-300x300.png\" alt=\"Three lines on a coordinate plane: a blue increasing line y is greater than or equal to quantity two-thirds x plus 1 and shaded above, a green increasing line y is less than 4 x plus 6 shaded above\/right, and an orange dashed decreasing line x plus y is less than 3 shaded below. Overlapping triangular area is shaded with all three colors.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-1024x1024.png 1024w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Shading.png 1414w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>The solution region is the one shaded all three times, which is the center triangle. If we want to be extra sure, we can test another point which is inside that triangle and verify that it satisfies all three inequalities. Test point [latex](0,2):[\/latex]<\/p>\n<p>Inequality 1:\u00a0 [latex]0+2\\leq 3[\/latex] TRUE<\/p>\n<p>Inequality 2:\u00a0 [latex]2\\geq \\dfrac{2}{3}\\cdot 0 + 1, \\textsf{ or } 2 \\geq 1[\/latex] TRUE<\/p>\n<p>Inequality 3:\u00a0\u00a0[latex]2<4\\cdot 0 + 6, \\textsf{ or } 2<6[\/latex] TRUE\n\nThe solution region is the center triangle. Two edges of the triangle are dashed (not solutions), and the bottom edge which is part of the blue line is in the solution set.\n\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1769 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-300x300.png\" alt=\"Three lines and their equations are shown: a blue increasing line representing y is greater than or equal to quantity two-thirds x plus 1, a green increasing line representing y is less than 4 x plus 6, and an orange dashed decreasing line representing x plus y is less than 3. Only the triangle between lines is shaded purple.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-1024x1024.png 1024w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/2.3SystemOf3Solution.png 1414w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Note that by modifying the original problem and switching the direction of some of the inequality symbols, we could have made any of the seven regions be the solution region. For example, switching only Inequality 2 to a [latex]\\leq[\/latex] sign would have caused our first test point [latex](0,0)[\/latex] to satisfy all three inequalities, so the solution region would have been the bottom, unbounded region containing\u00a0[latex](0,0).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>When linear inequalities are graphed on a coordinate plane, the solution takes the form of all points on one side of the boundary line, and possibly also the line itself. You can tell which region to shade by testing some points in the inequality, or observing the direction of the sign if the inequality is in slope-intercept form. When graphing a system of linear inequalities, graph all inequalities on the same axes. The solution set is the overlap of all the individual solution sets.<\/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-98\">\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><\/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><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><\/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":395986,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex 2: Graphing Linear Inequalities in Two Variables (Standard Form)\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/2VgFg2ztspI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex 1: Graphing Linear Inequalities in Two Variables (Slope Intercept Form)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Hzxc4HASygU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"cebc62e4-baa3-413f-bb7a-144ed9cf688e","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-98","chapter","type-chapter","status-publish","hentry"],"part":115,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/98","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/98\/revisions"}],"predecessor-version":[{"id":2110,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/98\/revisions\/2110"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/115"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/98\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=98"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=98"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=98"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=98"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}