{"id":13625,"date":"2014-02-13T23:26:11","date_gmt":"2014-02-13T23:26:11","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=13625"},"modified":"2026-01-14T10:15:10","modified_gmt":"2026-01-14T16:15:10","slug":"graphing-solutions-to-linear-inequalities","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/graphing-solutions-to-linear-inequalities\/","title":{"rendered":"Graphing Solutions to Linear Inequalities"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_0SDHQOQfbOc\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_0SDHQOQfbOc\" data-source-videoID=\"0SDHQOQfbOc\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Graphing Solutions to Linear Inequalities Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Graphing Solutions to Linear Inequalities\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_0SDHQOQfbOc:hover {cursor:pointer;} img#videoThumbnailImage_0SDHQOQfbOc {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1859-thumb-final-4.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_0SDHQOQfbOc\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_0SDHQOQfbOc\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Graphing Solutions to Linear Inequalities\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_0SDHQOQfbOc\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_0SDHQOQfbOc\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_0SDHQOQfbOc\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction cu8_Function() {\n  var x = document.getElementById(\"cu8\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"cu8_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"cu8\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Reviewing_the_Basics\" class=\"smooth-scroll\">Reviewing the Basics<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Graphing_Inequalities\" class=\"smooth-scroll\">Graphing Inequalities<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Compound_Inequalities\" class=\"smooth-scroll\">Compound Inequalities<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Inequalities_with_Absolute_Value\" class=\"smooth-scroll\">Inequalities with Absolute Value<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Inequalities_with_Two_Variables\" class=\"smooth-scroll\">Inequalities with Two Variables<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hello, and welcome to this video about <strong>graphing solutions to linear inequalities<\/strong>! <\/p>\n<p>Today we\u2019ll explore what a linear inequality is and learn how to graph simple and compound inequalities on a number line. We\u2019ll also talk about how to graph inequalities involving absolute value and inequalities with more than one variable. <\/p>\n<h2><span id=\"Reviewing_the_Basics\" class=\"m-toc-anchor\"><\/span>Reviewing the Basics<\/h2>\n<p>\nBefore we get started, let\u2019s review a few things. First, an inequality compares two values and shows whether they are less than, greater than, less than or equal to, or greater than or equal to each other.<\/p>\n<ul>\n<li>\\(a>b\\) means that \\(a\\) is greater than \\(b\\).<\/li>\n<li>\\(a\\lt b\\) means that \\(a\\) is less than \\(b\\).<\/li>\n<li>\\(a\\geq b\\) means that \\(a\\) is greater than or equal to \\(b\\).<\/li>\n<li>\\(a\\leq b\\) means that \\(a\\) is less than or equal to \\(b\\).<\/li>\n<\/ul>\n<p>For example, \\(x\\lt 15\\) means that the value of \\(x\\) is less than 15. Likewise, \\(x \\geq 6\\) means that the value of \\(x\\) is greater than <em>or<\/em> equal to 6. <\/p>\n<p><strong>Solutions<\/strong> are numbers we can replace the variables for in inequalities that make them true. For instance, one possible solution to \\(x\\lt 15\\) is 5 because 5 is less than 15. <\/p>\n<h2><span id=\"Graphing_Inequalities\" class=\"m-toc-anchor\"><\/span>Graphing Inequalities<\/h2>\n<p>\nSolutions to inequalities are graphed on <strong>number lines<\/strong> and <a class=\"ylist\" href=https:\/\/www.mometrix.com\/academy\/cartesian-coordinate-plane-and-graphing\/>coordinate planes<\/a>, depending on how many variables are in the inequality. These graphs help illustrate all possible solutions, or the <strong>solution set<\/strong>, for the inequality. <\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nLet\u2019s take a look at an example: \\(x \\lt 5\\). If \\(x\\) is less than 5, then all possible solutions for \\(x\\) have to be numbers that are smaller than 5. We can illustrate this on a number line. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solution-to-Inequalities-1.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"361\" class=\"aligncenter size-full wp-image-119968\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solution-to-Inequalities-1.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solution-to-Inequalities-1-300x166.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>First, find 5 on the number line and draw an open circle above 5. We leave this circle open to indicate that 5 is not part of the solution set for this inequality. The value for \\(x\\) is less than 5, not equal to it. <\/p>\n<p>Next, draw a line on the number line to indicate the possible solutions for \\(x\\). Since \\(x\\) is less than 5, we need to draw a line to the left of 5. <\/p>\n<p>Notice that the line extends to the left of 5 with an arrow at the end. The arrow indicates that even though the number line ends at -10, there are an infinite number of solutions (numbers smaller than -10) that could make this inequality true. <\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s try another one. Graph the solution set for the inequality \\(x\\lt -3\\). If \\(x\\) is greater than -3, then all possible solutions for \\(x\\) have to be numbers that are greater than -3. <\/p>\n<p>Pause the video here, draw a number line, and see if you can graph this one yourself. I know you can do it! <\/p>\n<p>Let\u2019s take a look at the number line together. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-2.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"367\" class=\"aligncenter size-full wp-image-119971\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-2.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-2-300x169.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>Notice that there is an open circle directly above -3 on the number line. Remember, this means that -3 is not part of the solution set because \\(x\\neq -3\\). From -3, there is a line extending to the right with an arrow at the end. This line shows that the possible solutions for \\(x\\) are all numbers greater than -3. Great job!<\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nHere\u2019s another inequality: \\(x\\leq 1\\). Let\u2019s graph the inequality on a number line.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-3.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"365\" class=\"aligncenter size-full wp-image-119974\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-3.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-3-300x168.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>First, find 1 on the number line and draw a shaded circle above it on the number line. The circle is closed to indicate that 1 is part of the solution set for this inequality. Next, starting from 1, draw a line to indicate all other possible solutions for \\(x\\). Since \\(x\\) is less than or equal to 1, we need to draw the line to the left of 1. <\/p>\n<h3><span id=\"Example_4\" class=\"m-toc-anchor\"><\/span>Example #4<\/h3>\n<p>\nNow it\u2019s your turn. Graph the solution set for the inequality \\(x\\geq -4\\).<\/p>\n<p>Pause the video here, draw a number line, and try this one yourself. When you finish, resume the video, and we\u2019ll go over the graph together. <\/p>\n<p>Let\u2019s take a look at the number line for \\(x\\geq -4\\):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-4.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"363\" class=\"aligncenter size-full wp-image-119977\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-4.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-4-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>As you can see, the circle above -4 is closed to indicate that -4 is part of the solution set. Since \\(x\\) is greater than or equal to -4, the line extends to the right, showing all possible solutions for \\(x\\). Nice work!<\/p>\n<h2><span id=\"Compound_Inequalities\" class=\"m-toc-anchor\"><\/span>Compound Inequalities<\/h2>\n<p>\nA compound inequality is two or more inequalities that are joined together with either \u201cand\u201d or \u201cor\u201d. They are sometimes written with the symbols for \u201cand\u201d and  for \u201cor\u201d. <\/p>\n<p>For example, \\(x\\leq -1\\cup\\text{  }x> 5\\) is a compound inequality that means the same thing as \\(x\\leq -1\\text{ or}\\text{ }x> 5\\). The possible solutions are all numbers that are either less than or equal to -1 or greater than 5. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-5.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"365\" class=\"aligncenter size-full wp-image-119980\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-5.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-5-300x168.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>When graphing compound inequalities on the number line, the same rules apply as with simple inequalities. Both inequality statements are graphed on the same number line. Let\u2019s graph the solutions for \\(x \\leq -1\\text{ }\\cup x\\gt 5\\), starting with the first inequality, \\(x \\leq -1\\). Since \\(x \\leq -1\\), there is a closed circle at \\(-1\\) with a line extending to its left. <\/p>\n<p>Next, graph the solution set for the second inequality, \\(x > 5\\), on the same number line. <\/p>\n<p>Since \\(x \\gt 5\\), there is an open circle at 5 with a line extending to its right. The number line now shows all solutions if \\(x\\) is less than or equal to -1 or greater than 5. Here\u2019s how the solution set can be expressed \\(\\left ( -\\infty ,-1] \\cup (5,\\infty  \\right )\\) with representing infinity, parentheses indicating open circles not part of the solution set, and brackets for closed circles that are part of the solution set. <\/p>\n<p>Consider the compound inequality \\(x \\gt -2\\cap x\\leq 6\\), which can also be written as \\(-2 \\lt x \\leq 6\\). Since these two inequalities are connected by \u201cand,\u201d the graph needs to reflect all solutions for \\(x\\) if it is greater than \\(-2\\) and less than or equal to 6. Following the same steps before, let\u2019s graph both solutions on the same number line, starting with \\(x \\gt -2\\). <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-6.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"362\" class=\"aligncenter size-full wp-image-119983\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-6.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-6-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>Since \\(x > -2\\), there is an open circle at -2 with a line extending to its right. Next, graph the solutions for \\(x \\leq 6\\) on the same number line. <\/p>\n<p>Since \\(x\\leq 6\\), there is a closed circle at 6 with a line extending to its left. However, this graph is not accurate. Recall that these inequalities are connected by \u201cand,\u201d meaning that the number line needs to show where solutions to both inequalities overlap. <\/p>\n<p>The number line now shows all solutions for \\(-2\\lt x \\leq 6\\). <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-7.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"365\" class=\"aligncenter size-full wp-image-119986\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-7.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-7-300x168.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>This solution set can be expressed like this \\((-2,6]\\), with a parenthesis indicating that -2 is not part of the solution set and a bracket meaning that 6 is part of the solution set. <\/p>\n<h2><span id=\"Inequalities_with_Absolute_Value\" class=\"m-toc-anchor\"><\/span>Inequalities with Absolute Value<\/h2>\n<p>\nSolutions to inequalities with absolute value can also be graphed on a number line. Recall that <a class=\"ylist\" href=https:\/\/www.mometrix.com\/academy\/absolute-value\/>absolute value<\/a> is a number\u2019s distance from zero. In other words, it\u2019s the non-negative value of a number. Absolute value is denoted like this: \\(|x|\\). <\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nConsider the inequality \\(\\left | x \\right |> 7\\). Solutions for \\(x\\) include all numbers that are more than 7 spaces away from 0 on the number line. Because possible solutions can be positive or negative, we can rewrite this inequality as two inequalities without absolute value bars. \\(\\left | x \\right |> 7\\) becomes \\(x> 7\\) and \\(x\\lt -7\\). Notice that the sign flipped in the inequality \\(x\\lt -7\\). This is because when a negative is multiplied or divided in an inequality, its sign flips.<\/p>\n<p>Now we\u2019re ready to graph \\(x> 7\\) and \\(x\\lt -7\\). Following the same steps we\u2019ve been practicing, graph both inequalities on the same number line.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solution-to-Inequalities-8.jpg\" alt=\"graph of linear inequality\" width=\"652\" height=\"362\" class=\"aligncenter size-full wp-image-119989\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solution-to-Inequalities-8.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solution-to-Inequalities-8-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>The first inequality states that \\(x> 7\\), so the graph shows an open circle at 7 with a line extending to the right. The second inequality states that \\(x> -7\\), so the graph shows an open circle at -7 with a line extending to the left. <\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s try one more: \\(\\left | 1+b \\right |\\leq 8\\). The absolute value bars indicate that possible solutions can be positive or negative. Therefore, we need to rewrite the inequality as two inequalities without the absolute value: \\(1+b\\leq 8\\) and \\(1+b\\geq -8\\). Notice that the sign flipped in the second inequality since a negative value was introduced. <\/p>\n<p>From here, isolate the variable \\(b\\) in both inequalities.<\/p>\n<p>Subtract 1 from both sides, and we\u2019ll get \\(b\\leq 7\\).<\/p>\n<div class=\"examplesentence\">\\(1+b\\leq 8\\)<br \/>\n\\(1-1+b\\leq 8-1\\)<br \/>\n\\(b\\leq7\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we\u2019ll do the same thing over here. Subtract 1 from both sides, and we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\(1+b\\geq-8\\)<br \/>\n\\(1-1+b\\geq-8-1\\)<br \/>\n\\(b\\geq-9\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe solution set for this inequality includes all values that are less than or equal to <em>7 and greater<\/em> than or equal to -9.<\/p>\n<p>Note that these inequalities are connected by the term \u201cand.\u201d We are looking for solutions that are true for both inequalities, so the number line needs to show where solutions to both inequalities overlap. <\/p>\n<p>The number line now shows all solutions for \\(-9\\leq b\\leq 7\\). The solution set can be expressed like this \\([-9,7]\\), with brackets indicating that -9 and 7 are part of the solution set. <\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nNow it\u2019s your turn. Following the same steps as we\u2019ve practiced, graph the solution set for \\(\\left | 2x-3 \\right |> 5\\). <\/p>\n<p>Pause the video here, draw a number line, and try this one yourself. When you finish, resume the video, and we\u2019ll go over it together. <\/p>\n<p>Let\u2019s take a look. First, rewrite the inequality as two inequalities without absolute value bars. After that, isolate the variable \\(x\\).<\/p>\n<p>So we have \\(2x-3> 5\\) and \\(2x-3\\lt -5\\).<\/p>\n<div class=\"examplesentence\">\\(2x-3> 5\\)<\/div>\n<p>\n&nbsp;<br \/>\nFrom here, we\u2019re going to add 3 to both sides. So we have:<\/p>\n<div class=\"examplesentence\">\\(2x-3+3> 5+3\\)<br \/>\n\\(2x> 8\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd divide by 2 on both sides.<\/p>\n<div class=\"examplesentence\">\\(2x\\div 2> 8\\div 2\\)<br \/>\n\\(x> 4\\)<\/div>\n<p>\n&nbsp;<br \/>\nOver here we\u2019re going to follow the same steps. We\u2019ll add 3 to both sides.<\/p>\n<div class=\"examplesentence\">\\(2x-3\\lt -5\\)<br \/>\n\\(2x-3+3\\lt -5+3\\)<br \/>\n\\(2x\\lt -2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd divide by 2 on both sides.<\/p>\n<div class=\"examplesentence\">\\(2x\\div 2\\lt -2\\div 2\\)<br \/>\n\\(x\\lt -1\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can graph the solution set for \\(x>4\\) or \\(x\\lt-1\\).<\/p>\n<p>This solution set can be expressed like this \\((-\\infty , -1)\\cup (4,\\infty )\\) with parentheses indicating open circles not part of the solution set. Nice work! <\/p>\n<h2><span id=\"Inequalities_with_Two_Variables\" class=\"m-toc-anchor\"><\/span>Inequalities with Two Variables<\/h2>\n<p>\nSo far, we\u2019ve looked at inequalities that contain one variable. What if an inequality has two variables?<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nConsider \\(y> 3x+1\\). Since there are two variables, we need to illustrate the solution sets on a coordinate plane. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-9.jpg\" alt=\"graph of y greater than 3x plus 1\" width=\"652\" height=\"361\" class=\"aligncenter size-full wp-image-119992\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-9.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-9-300x166.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>Notice that \\(y> 3x+1\\) is in <a class=\"ylist\" href=https:\/\/www.mometrix.com\/academy\/slope-intercept-and-point-slope-forms\/>slope-intercept form<\/a>, \\(y=\\text{ }mx+b\\). When an inequality has two variables and is in slope-intercept form, the first step is to graph the line on a coordinate plane. Since the \\(y\\)-intercept (\\(b\\)) is 1, the line will intersect the \\(y\\)-axis at 1, or \\((0,1)\\). The slope (\\(m\\)) is 3, or \\(\\frac{3}{1}\\). Starting at \\((0,1)\\), use the slope to find another point on the line. Since slope is \\(\\frac{\\text{rise}}{\\text{run}}\\), move up 3 units and right 1 unit to \\((1,4)\\). <\/p>\n<p>Next, draw a line to connect the points. Since the inequality sign in the original expression is less than (and not less than <em>or equal to<\/em>), connect the ordered pairs with a dashed line. The dashed line means that points on this line are not possible solutions. <\/p>\n<p>The solution set for a linear inequality is shown as a shaded region on the coordinate plane. This shaded region covers the area either to the right or to the left of the line graphed. To figure out which region to shade, select a test point to substitute into the inequality and solve. An easy test point to work with is the point of origin \\((0, 0)\\). Substitute 0 for both \\(x\\) and \\(y\\) in the inequality. <\/p>\n<p>So we if we have our inequality \\(y> 3x+1\\) and we substitute in 0 for both \\(x\\) and \\(y\\), we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\((0)> 3(0)+1\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo this becomes:<\/p>\n<div class=\"examplesentence\">\\(0>0+1\\)<\/div>\n<p>\n&nbsp;<br \/>\nOr:<\/p>\n<div class=\"examplesentence\">\\(0 > 1\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich we know is not true.<\/p>\n<p>Since 0 is not greater than 1, then this point \\((0,0)\\) cannot be part of the solution set. That means the values to the right of the graphed line are not solutions to the inequality. Instead, shade the area to the left of the line.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-10.jpg\" alt=\"graph with left side shaded\" width=\"652\" height=\"362\" class=\"aligncenter size-full wp-image-119995\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-10.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-10-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>The finished graph has a dashed line with a shaded region to its left. All possible solutions are points to the left of the line graphed. <\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s try one more.<\/p>\n<p>Consider the inequality \\(y\\leq \\frac{2}{5}x+2\\). First, graph the line on a coordinate plane. Since the \\(y\\)-intercept (\\(b\\)) is 2, the line will intersect the \\(y\\)-axis at 2, or \\((0,2)\\). The slope (\\(m\\)) is 25. Starting at \\((0,2)\\), use the slope to find another point on the line. Since the slope is \\(\\frac{\\text{rise}}{\\text{run}}\\), move up 2 units and right 5 units to point \\((5,4)\\). <\/p>\n<p>Next, draw a line to connect the points. Since the inequality sign is less than or <em>equal to<\/em>, connect the ordered pairs with a solid line. This indicates that points on this line are possible solutions. <\/p>\n<p>Next, substitute the origin point \\((0,0)\\) into the inequality as a test point to determine which portion of the coordinate plane to shade. <\/p>\n<p>So if we plug in \\((0,0)\\) for our \\(x\\) and \\(y\\) here, we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\((0)\\leq \\frac{2}{5}(0)+2\\)<\/span><\/div>\n<p>\n&nbsp;<br \/>\nWhich will give us:<\/p>\n<div class=\"examplesentence\">\\(0\\leq 0+2\\)<\/div>\n<p>\n&nbsp;<br \/>\nOr:<\/p>\n<div class=\"examplesentence\">\\(0\\leq 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis is true!<\/p>\n<p>Since \\(0 \\lt 2\\), point \\((0,0)\\) is part of the solution set. That means the values to the right of the graphed line are solutions to the inequality. The finished graph has a solid line with a shaded region to its right. All possible solutions are points to the right of the line or on the line graphed.  <\/p>\n<h3><span id=\"Example_3_2\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nAre you ready for a challenge? Let\u2019s take a look at a compound inequality containing two variables: <\/p>\n<div class=\"examplesentence\">\\(y\\geq x\\text{ }+1\\text{ }\\cup y\\lt x-8\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember that  connects the two inequalities and means \u201cor.\u201d We can solve this compound inequality by following the same steps we\u2019ve been practicing. The only difference is that now we have two lines to graph on the coordinate plane. Let\u2019s get started. <\/p>\n<p>So we have the line \\(y\\geq x+1\\) and the line \\(y \\lt x-8\\).<\/p>\n<p>\\(y\\geq x+1\\) has a \\(y\\)-intercept at 1 \\((0,1)\\) and a slope of 1. It is graphed with a solid line to show that points on the line are included in the solution set. \\(y\\lt x-8\\) has a \\(y\\)-intercept at -8 \\((0,-8)\\) and a slope of 1. It is graphed with a dashed line to indicate that points on the line are not included in the solution set. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-11.jpg\" alt=\"graph of two inequalities\" width=\"500\" height=\"499\" class=\"aligncenter size-full wp-image-119998\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-11.jpg 500w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-11-300x300.jpg 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-11-150x150.jpg 150w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<p>Once the two lines are graphed, substitute the origin \\((0,0)\\) into both inequalities as a test point. Doing so will determine which portion of the coordinate plane to shade. <\/p>\n<p>So, if we plug in \\((0,0)\\) here, we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\(y\\geq x+1\\)<br \/>\n\\((0)\\geq (0)+1\\)<\/div>\n<p>\n&nbsp;<br \/>\nOr:<\/p>\n<div class=\"examplesentence\">\\(0\\geq1\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe know this is not true. Now we\u2019ll come over here and plug in \\((0,0)\\). We\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\(y\\lt x-8\\)<br \/>\n\\((0)\\lt (0)-8\\) <\/span><\/div>\n<p>\n&nbsp;<br \/>\nWhich simplifies to:<\/p>\n<div class=\"examplesentence\">>\\(0\\lt -8\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe know this is also not true.<\/p>\n<p>Since both statements are false when solved, shade in both lines on the sides that do not include the origin. \\(y\\geq x+1\\) has a solid line with a shaded region to its left. \\(y\\lt x-8\\) has a dashed line with a shaded region to its right. Both shaded portions are part of the solution set.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-12.jpg\" alt=\"inequality graph with shaded portions\" width=\"500\" height=\"497\" class=\"aligncenter size-full wp-image-120004\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-12.jpg 500w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-12-300x298.jpg 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-12-150x150.jpg 150w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<h3><span id=\"Example_4_1\" class=\"m-toc-anchor\"><\/span>Example #4<\/h3>\n<p>\nWe have learned a lot about inequalities today, and I have one more problem for you to try. It\u2019s a little more challenging, but I know you can do it. The inequality statement is: <\/p>\n<div class=\"examplesentence\">\\(2x+4\\lt y\\leq 3x+8\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember, compound linear inequalities are graphed on the same coordinate plane. Your first step is to graph the two lines. After graphing, use the test point \\((0,0)\\) to determine where to shade your graph. <\/p>\n<p>Pause the video here and try this one yourself. When you\u2019re done, resume the video, and we\u2019ll go over it together. <\/p>\n<p>Let\u2019s take a look at this problem together. The two inequalities are \\(y>2x+4\\) and \\(y\\leq 3x+8\\). When graphed, \\(y>2x+4\\) has a dashed line, and \\(y\\leq 3x+8\\) has a solid line. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-13.jpg\" alt=\"inequality graph with shaded portions overlapping\" width=\"500\" height=\"496\" class=\"aligncenter size-full wp-image-120007\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-13.jpg 500w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-13-300x298.jpg 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/04\/Graphing-Solutions-to-Inequalities-13-150x150.jpg 150w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<p>Next, substitute the origin \\((0,0)\\) into both inequalities as a test point to determine which portion of the coordinate plane to shade.<\/p>\n<p>So if we plug in \\((0,0)\\) here, we get:<\/p>\n<div class=\"examplesentence\">\\(y> 2x+4\\)<br \/>\n\\((0)> 2(0)+4\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich gives us:<\/p>\n<div class=\"examplesentence\">\\(0>0+4\\)<\/div>\n<p>\n&nbsp;<br \/>\nOr:<\/p>\n<div class=\"examplesentence\">\\(0>4\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe know this is not true. Now let\u2019s try it over here. So we have:<\/p>\n<div class=\"examplesentence\">\\(y\\leq 3x+8\\)<br \/>\n\\((0)\\leq3(0)+8\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich simplifies to:<\/p>\n<div class=\"examplesentence\">\\(0\\leq0+8\\)<\/div>\n<p>\n&nbsp;<br \/>\nOr:<\/p>\n<div class=\"examplesentence\">\\(0\\leq 8\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe know this is true.<\/p>\n<p>Based on our work, the origin point should not be shaded for \\(y>2x+4\\) but should be shaded for \\(y\\leq 3x+8\\). In other words, the dashed line is shaded to its left, and the solid line is shaded to its right. <\/p>\n<p>In this case, the solution set is the area where both shaded portions overlap. Great job! <\/p>\n<hr>\n<p>\nI hope this video about graphing solutions to linear inequalities was helpful. 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