{"id":15360,"date":"2021-10-11T23:05:37","date_gmt":"2021-10-11T23:05:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/3-1-1-graphing-systems-of-linear-equations\/"},"modified":"2021-11-23T00:39:09","modified_gmt":"2021-11-23T00:39:09","slug":"graphing-systems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/graphing-systems\/","title":{"raw":"Graphing Systems of Equations","rendered":"Graphing Systems of Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph systems of equations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"title1\">Graph a system of linear equations<\/h2>\r\nThere are multiple methods of solving systems of linear equations. For a <strong>system of linear equations<\/strong> in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.\r\n\r\nWe will practice graphing two equations on the same set of axes,\u00a0and then we will explore the different considerations you need to make when graphing two linear inequalities on the same set of axes. The same techniques are used to graph a system of linear equations as you have used to graph single linear equations. We can use tables of values, slope and [latex]y[\/latex]-intercept, or [latex]x[\/latex]- and [latex]y[\/latex]-intercepts to graph both lines on the same set of axes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the following system of equations by graphing. Identify the type of system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"336799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"336799\"]\r\n\r\nSolve the first equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+y=-8\\\\ y=-2x - 8\\end{array}[\/latex]<\/p>\r\nSolve the second equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x-y=-1\\\\ y=x+1\\end{array}[\/latex]<\/p>\r\nGraph both equations on the same set of axes as seen below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222636\/CNX_Precalc_Figure_09_01_0122.jpg\" alt=\"A graph of two lines running through the point negative 3, negative 2. The first line's equation is y equals minus 2x minus 8. The second line's equation is y equals x+1.\" width=\"487\" height=\"316\" \/>\r\n\r\nThe lines appear to intersect at the point [latex]\\left(-3,-2\\right)[\/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}2\\left(-3\\right)+\\left(-2\\right)=-8\\hfill &amp; \\hfill \\\\ \\text{ }-8=-8\\hfill &amp; \\text{True}\\hfill \\\\ \\text{ }\\left(-3\\right)-\\left(-2\\right)=-1\\hfill &amp; \\hfill \\\\ \\text{ }-1=-1\\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution to the system is the ordered pair [latex]\\left(-3,-2\\right)[\/latex], so the system is independent.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can watch the video below for another example of how to solve a system of equations by first graphing the lines and then identifying the solution the system has.\r\n\r\nhttps:\/\/youtu.be\/Lv832rXAQ5k\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]38339[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<p id=\"video1\" class=\"no-indent\" style=\"text-align: left;\">In the following example, you will be given a system to graph that consists of two parallel lines.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph the system [latex]\\begin{array}{c}y=2x+1\\\\y=2x-3\\end{array}[\/latex] using the slopes and [latex]y[\/latex]-intercepts of the lines.\r\n[reveal-answer q=\"478796\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"478796\"]\r\n\r\nFirst, graph\u00a0[latex]y=2x+1[\/latex] using the slope [latex]m = 2[\/latex] and the [latex]y[\/latex]-intercept [latex](0,1)[\/latex]\r\n\r\n<img class=\" wp-image-4139 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13190852\/Screen-Shot-2016-05-13-at-12.07.19-PM-300x294.png\" alt=\"y=2x+1\" width=\"372\" height=\"365\" \/>\r\n\r\nNext, add\u00a0[latex]y=2x-3[\/latex] using the slope [latex]m = 2[\/latex], and the [latex]y[\/latex]-intercept [latex](0,-3)[\/latex]\r\n\r\n<img class=\" wp-image-4140 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13191012\/Screen-Shot-2016-05-13-at-12.03.10-PM-300x295.png\" alt=\"y = 2x+1 and y = 2x-3\" width=\"355\" height=\"349\" \/>\r\n\r\nNotice how these are parallel lines, and they don't cross. \u00a0In the previous section, we discussed how there are no solutions to a system of equations that\u00a0are parallel lines.\u00a0 We classified this type of system as an inconsistent system.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, you will be given a system whose equations look different, but after graphing, turn out to be the same line.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph the system [latex]\\begin{array}{c}y=\\frac{1}{2}x+2\\\\2y-x=4\\end{array}[\/latex] using the x - and y-intercepts.\r\n[reveal-answer q=\"342515\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342515\"]\r\n\r\nFirst, find the [latex]x[\/latex]- and [latex]y[\/latex]- intercepts of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex]\r\n\r\nThe [latex]x[\/latex]-intercept will have a value of [latex]0[\/latex] for [latex]y[\/latex], so substitute [latex]y=0[\/latex] into the equation, and isolate the variable [latex]x[\/latex].\r\n\r\n[latex]\\begin{array}{c}0=\\frac{1}{2}x+2\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,-2}\\\\-2=\\frac{1}{2}x\\\\\\left(2\\right)\\left(-2\\right)=\\left(2\\right)\\frac{1}{2}x\\\\-4=x\\end{array}[\/latex]\r\n\r\nThe [latex]x[\/latex]-intercept of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex] is [latex]\\left(-4,0\\right)[\/latex].\r\n\r\nThe [latex]y[\/latex]-intercept is easier to find since this equation is in slope-intercept form. \u00a0The [latex]y[\/latex]-intercept is [latex](2,0)[\/latex].\r\n\r\nNow we can plot\u00a0[latex]y=\\frac{1}{2}x+2[\/latex] using the intercepts\r\n\r\n<img class=\" wp-image-4144 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13194855\/Screen-Shot-2016-05-13-at-12.47.41-PM-300x295.png\" alt=\"y=1\/2x+2 with intercepts labeled\" width=\"399\" height=\"392\" \/>\r\n\r\nNow find the intercepts of [latex]2y-x=4[\/latex]\r\n\r\nSubstitute [latex]y = 0[\/latex] in to the equation to find the [latex]x[\/latex]-intercept.\r\n\r\n[latex]\\begin{array}{c}2y-x=4\\\\2\\left(0\\right)-x=4\\\\x=-4\\end{array}[\/latex]\r\n\r\nThe [latex]x[\/latex]-intercept of\u00a0[latex]2y-x=4[\/latex] is [latex]\\left(-4,0\\right)[\/latex].\r\n\r\nNow substitute [latex]x = 0[\/latex] into the equation to find the y-intercept.\r\n\r\n[latex]\\begin{array}{c}2y-x=4\\\\2y-0=4\\\\2y=4\\\\y=2\\end{array}[\/latex]\r\n\r\nThe [latex]y[\/latex]-intercept of\u00a0[latex]2y-x=4[\/latex] is [latex]\\left(0,2\\right)[\/latex].\r\n\r\nWAIT, these are the same intercepts as\u00a0[latex]y=\\frac{1}{2}x+2[\/latex]! \u00a0In fact, [latex]y=\\frac{1}{2}x+2[\/latex] and\u00a0[latex]2y-x=4[\/latex] are really the same equation, expressed in different ways. \u00a0If you were to write them both in slope-intercept form you would see that they are the same equation.\r\n\r\nWhen you graph them, they are the same line. In the previous section, we learned that systems with two of the same equations in them have an infinite number of solutions and are classified as a dependent system.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs demonstrated by the examples above, graphing can be used if the system is inconsistent or dependent.\u00a0In both cases, we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.\r\n\r\nThe video below provides more examples of how to graph systems of linear equations.\r\n\r\nhttps:\/\/youtu.be\/BBmB3rFZLXU\r\n\r\nGraphing a system of linear equations consists of choosing which graphing method you want to use and drawing the graphs of both equations on the same set of axes.\r\n<h2>Contribute!<\/h2>\r\n<div style=\"margin-bottom: 8px;\">Did you have an idea for improving this content? We\u2019d love your input.<\/div>\r\n<a style=\"font-size: 10pt; font-weight: 600; color: #077fab; text-decoration: none; border: 2px solid #077fab; border-radius: 7px; padding: 5px 25px; text-align: center; cursor: pointer; line-height: 1.5em;\" href=\"https:\/\/docs.google.com\/document\/d\/1iqDW15Xya6NVOQa0QhjAxr8-lJ5Us2ErB8irnL3GB5U\" target=\"_blank\" rel=\"noopener\">Improve this page<\/a><a style=\"margin-left: 16px;\" href=\"https:\/\/docs.google.com\/document\/d\/1vy-T6DtTF-BbMfpVEI7VP_R7w2A4anzYZLXR8Pk4Fu4\" target=\"_blank\" rel=\"noopener\">Learn More<\/a>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph systems of equations<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"title1\">Graph a system of linear equations<\/h2>\n<p>There are multiple methods of solving systems of linear equations. For a <strong>system of linear equations<\/strong> in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.<\/p>\n<p>We will practice graphing two equations on the same set of axes,\u00a0and then we will explore the different considerations you need to make when graphing two linear inequalities on the same set of axes. The same techniques are used to graph a system of linear equations as you have used to graph single linear equations. We can use tables of values, slope and [latex]y[\/latex]-intercept, or [latex]x[\/latex]&#8211; and [latex]y[\/latex]-intercepts to graph both lines on the same set of axes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the following system of equations by graphing. Identify the type of system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336799\">Show Solution<\/span><\/p>\n<div id=\"q336799\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve the first equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+y=-8\\\\ y=-2x - 8\\end{array}[\/latex]<\/p>\n<p>Solve the second equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x-y=-1\\\\ y=x+1\\end{array}[\/latex]<\/p>\n<p>Graph both equations on the same set of axes as seen below.<\/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\/07\/01222636\/CNX_Precalc_Figure_09_01_0122.jpg\" alt=\"A graph of two lines running through the point negative 3, negative 2. The first line's equation is y equals minus 2x minus 8. The second line's equation is y equals x+1.\" width=\"487\" height=\"316\" \/><\/p>\n<p>The lines appear to intersect at the point [latex]\\left(-3,-2\\right)[\/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}2\\left(-3\\right)+\\left(-2\\right)=-8\\hfill & \\hfill \\\\ \\text{ }-8=-8\\hfill & \\text{True}\\hfill \\\\ \\text{ }\\left(-3\\right)-\\left(-2\\right)=-1\\hfill & \\hfill \\\\ \\text{ }-1=-1\\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution to the system is the ordered pair [latex]\\left(-3,-2\\right)[\/latex], so the system is independent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You can watch the video below for another example of how to solve a system of equations by first graphing the lines and then identifying the solution the system has.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2:  Solve a System of Equations by Graphing\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Lv832rXAQ5k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm38339\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38339&theme=oea&iframe_resize_id=ohm38339&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"video1\" class=\"no-indent\" style=\"text-align: left;\">In the following example, you will be given a system to graph that consists of two parallel lines.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph the system [latex]\\begin{array}{c}y=2x+1\\\\y=2x-3\\end{array}[\/latex] using the slopes and [latex]y[\/latex]-intercepts of the lines.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q478796\">Show Solution<\/span><\/p>\n<div id=\"q478796\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, graph\u00a0[latex]y=2x+1[\/latex] using the slope [latex]m = 2[\/latex] and the [latex]y[\/latex]-intercept [latex](0,1)[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4139 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13190852\/Screen-Shot-2016-05-13-at-12.07.19-PM-300x294.png\" alt=\"y=2x+1\" width=\"372\" height=\"365\" \/><\/p>\n<p>Next, add\u00a0[latex]y=2x-3[\/latex] using the slope [latex]m = 2[\/latex], and the [latex]y[\/latex]-intercept [latex](0,-3)[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4140 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13191012\/Screen-Shot-2016-05-13-at-12.03.10-PM-300x295.png\" alt=\"y = 2x+1 and y = 2x-3\" width=\"355\" height=\"349\" \/><\/p>\n<p>Notice how these are parallel lines, and they don&#8217;t cross. \u00a0In the previous section, we discussed how there are no solutions to a system of equations that\u00a0are parallel lines.\u00a0 We classified this type of system as an inconsistent system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, you will be given a system whose equations look different, but after graphing, turn out to be the same line.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph the system [latex]\\begin{array}{c}y=\\frac{1}{2}x+2\\\\2y-x=4\\end{array}[\/latex] using the x &#8211; and y-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342515\">Show Solution<\/span><\/p>\n<div id=\"q342515\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the [latex]x[\/latex]&#8211; and [latex]y[\/latex]&#8211; intercepts of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex]<\/p>\n<p>The [latex]x[\/latex]-intercept will have a value of [latex]0[\/latex] for [latex]y[\/latex], so substitute [latex]y=0[\/latex] into the equation, and isolate the variable [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{array}{c}0=\\frac{1}{2}x+2\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,-2}\\\\-2=\\frac{1}{2}x\\\\\\left(2\\right)\\left(-2\\right)=\\left(2\\right)\\frac{1}{2}x\\\\-4=x\\end{array}[\/latex]<\/p>\n<p>The [latex]x[\/latex]-intercept of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex] is [latex]\\left(-4,0\\right)[\/latex].<\/p>\n<p>The [latex]y[\/latex]-intercept is easier to find since this equation is in slope-intercept form. \u00a0The [latex]y[\/latex]-intercept is [latex](2,0)[\/latex].<\/p>\n<p>Now we can plot\u00a0[latex]y=\\frac{1}{2}x+2[\/latex] using the intercepts<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4144 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13194855\/Screen-Shot-2016-05-13-at-12.47.41-PM-300x295.png\" alt=\"y=1\/2x+2 with intercepts labeled\" width=\"399\" height=\"392\" \/><\/p>\n<p>Now find the intercepts of [latex]2y-x=4[\/latex]<\/p>\n<p>Substitute [latex]y = 0[\/latex] in to the equation to find the [latex]x[\/latex]-intercept.<\/p>\n<p>[latex]\\begin{array}{c}2y-x=4\\\\2\\left(0\\right)-x=4\\\\x=-4\\end{array}[\/latex]<\/p>\n<p>The [latex]x[\/latex]-intercept of\u00a0[latex]2y-x=4[\/latex] is [latex]\\left(-4,0\\right)[\/latex].<\/p>\n<p>Now substitute [latex]x = 0[\/latex] into the equation to find the y-intercept.<\/p>\n<p>[latex]\\begin{array}{c}2y-x=4\\\\2y-0=4\\\\2y=4\\\\y=2\\end{array}[\/latex]<\/p>\n<p>The [latex]y[\/latex]-intercept of\u00a0[latex]2y-x=4[\/latex] is [latex]\\left(0,2\\right)[\/latex].<\/p>\n<p>WAIT, these are the same intercepts as\u00a0[latex]y=\\frac{1}{2}x+2[\/latex]! \u00a0In fact, [latex]y=\\frac{1}{2}x+2[\/latex] and\u00a0[latex]2y-x=4[\/latex] are really the same equation, expressed in different ways. \u00a0If you were to write them both in slope-intercept form you would see that they are the same equation.<\/p>\n<p>When you graph them, they are the same line. In the previous section, we learned that systems with two of the same equations in them have an infinite number of solutions and are classified as a dependent system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>As demonstrated by the examples above, graphing can be used if the system is inconsistent or dependent.\u00a0In both cases, we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.<\/p>\n<p>The video below provides more examples of how to graph systems of linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Graphing a System of Linear  Equations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BBmB3rFZLXU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Graphing a system of linear equations consists of choosing which graphing method you want to use and drawing the graphs of both equations on the same set of axes.<\/p>\n<h2>Contribute!<\/h2>\n<div style=\"margin-bottom: 8px;\">Did you have an idea for improving this content? We\u2019d love your input.<\/div>\n<p><a style=\"font-size: 10pt; font-weight: 600; color: #077fab; text-decoration: none; border: 2px solid #077fab; border-radius: 7px; padding: 5px 25px; text-align: center; cursor: pointer; line-height: 1.5em;\" href=\"https:\/\/docs.google.com\/document\/d\/1iqDW15Xya6NVOQa0QhjAxr8-lJ5Us2ErB8irnL3GB5U\" target=\"_blank\" rel=\"noopener\">Improve this page<\/a><a style=\"margin-left: 16px;\" href=\"https:\/\/docs.google.com\/document\/d\/1vy-T6DtTF-BbMfpVEI7VP_R7w2A4anzYZLXR8Pk4Fu4\" target=\"_blank\" rel=\"noopener\">Learn More<\/a><\/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-15360\">\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>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><li>Graphing a System of Linear Equation. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BBmB3rFZLXU\">https:\/\/youtu.be\/BBmB3rFZLXU<\/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>Unit 14: Systems of Equations and Inequalities, 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\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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 Mathispower4u  Mathispower4u. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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