{"id":5958,"date":"2016-09-07T23:17:37","date_gmt":"2016-09-07T23:17:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/beginalgebra\/?post_type=chapter&#038;p=5958"},"modified":"2018-01-03T23:55:48","modified_gmt":"2018-01-03T23:55:48","slug":"introduction-to-systems-of-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/beginalgebra\/chapter\/introduction-to-systems-of-linear-equations\/","title":{"raw":"Graphs and Solutions to Systems of Linear Equations","rendered":"Graphs and Solutions to Systems of Linear Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph systems of equations\r\n<ul>\r\n \t<li>Graph a system of two linear equations<\/li>\r\n \t<li>Graph a system of two linear\u00a0inequalities<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Evaluate ordered pairs as solutions to systems\r\n<ul>\r\n \t<li>Determine whether an ordered pair is a solution to a system of linear equations<\/li>\r\n \t<li>Determine whether an ordered pair is a solution to a system of linear inequalities<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Classify solutions to systems\r\n<ul>\r\n \t<li>Identify what type of solution a system will have based on its graph<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n&nbsp;\r\n\r\nThe way a river flows depends on many variables including how big the river is, how much water it contains, what sorts of things are floating in the river, whether or not it is raining, and so forth. If you want to best describe its flow, you must take into account these other variables. A system of linear equations can help with that.\r\n\r\nA <strong>system of linear equations<\/strong> consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. You will find systems of equations in every application of mathematics. They are a useful tool for discovering and describing how behaviors or processes are interrelated. It is rare to find, for example, a pattern of traffic flow that that is only affected by weather. Accidents, time of day, and major sporting events are just a few of the other variables that can affect the flow of traffic in a city. In this section, we will explore some basic principles for graphing and describing the intersection of two lines that make up a system of equations.\r\n<h2 id=\"title1\">Graph a system of linear equations<\/h2>\r\nIn this section, we will look at systems of linear equations and inequalities in two variables. \u00a0First, 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 <em>y<\/em>-intercept, or <em>x<\/em>- and <em>y<\/em>-intercepts to graph both lines on the same set of axes.\r\n\r\nFor example, consider the following system of linear equations in two variables.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex]<\/div>\r\nLet's\u00a0graph these using slope-intercept form on the same set of axes. Remember that slope-intercept form looks like [latex]y=mx+b[\/latex], \u00a0so we will want to solve both equations for [latex]y[\/latex].\r\n\r\nFirst, solve for y in [latex]2x+y=-8[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+y=-8\\\\ y=-2x - 8\\end{array}[\/latex]<\/p>\r\n\r\n<div style=\"text-align: center;\"><\/div>\r\n<div style=\"text-align: left;\">Second, solve for y in [latex]x-y=-1[\/latex]<\/div>\r\n<div style=\"text-align: left;\"><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}x-y=-1\\,\\,\\,\\,\\,\\\\ y=x+1\\end{array}[\/latex]<\/div>\r\nThe system is now written as\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y=-2x - 8\\\\y=x+1\\end{array}[\/latex]<\/p>\r\nNow you can graph both equations using their slopes and intercepts on the same set of axes, as seen in the figure below. Note how the graphs share one point in common. This is their point of intersection, a point that lies on both of the lines. \u00a0In the next section we will verify that\u00a0this point is a solution to the system.\r\n\r\n<img class=\"size-full wp-image-5878 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01202809\/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<p id=\"video1\" class=\"no-indent\" style=\"text-align: left;\">\u00a0In 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 y-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 m = 2 and the y-intercept (0,1)\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 m = 2, and the y-intercept (0,-3)\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 next section we will discuss how there are no solutions to a system of equations that\u00a0are parallel lines.\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 x- and y- intercepts of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex]\r\n\r\nThe x-intercept will have a value of 0 for y, so substitute y=0 into the equation, and isolate the variable x.\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 x-intercept of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex] is [latex]\\left(-4,0\\right)[\/latex].\r\n\r\nThe y-intercept is easier to find since this equation is in slope-intercept form. \u00a0The y-intercept is (2,0).\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 y = 0 in to the equation to find the x-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 x-intercept of\u00a0[latex]2y-x=4[\/latex] is [latex]\\left(-4,0\\right)[\/latex].\r\n\r\nNow substitute x = 0 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 y-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 next section, we will see that systems with two of the same equations in them have an infinite number of solutions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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. When you graph a system of linear inequalities on the same set of axes, there are a few more things you will need to consider.\r\n<h2 id=\"title2\">Graph a system of two inequalities<\/h2>\r\nRemember from the module on graphing that the graph of a single linear inequality splits the <b>coordinate plane<\/b> into two regions.\u00a0On one side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality [latex]y&lt;2x+5[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064400\/image012.gif\" alt=\"An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1).\" width=\"346\" height=\"343\" \/>\r\n\r\nThe dashed line is [latex]y=2x+5[\/latex]. Every ordered pair in the shaded\u00a0area below the line is a solution to [latex]y&lt;2x+5[\/latex], as all of the points below the line will make the inequality true. If you doubt that, try substituting the <i>x<\/i> and <i>y<\/i> coordinates of Points A and B into the inequality\u2014you\u2019ll see that they work. So, the shaded area shows all of the solutions for this inequality.\r\n\r\nThe boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don\u2019t satisfy the inequality. If the inequality had been [latex]y\\leq2x+5[\/latex], then the boundary line would have been solid.\r\n\r\nLet\u2019s graph another inequality: [latex]y&gt;\u2212x[\/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064401\/image013.gif\" alt=\"Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1).\" width=\"327\" height=\"324\" \/>\r\n\r\nTo create a system of inequalities, you need to graph two or more inequalities together. Let\u2019s use\u00a0[latex]y&lt;2x+5[\/latex] and [latex]y&gt;\u2212x[\/latex] since we have already graphed each of them.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014.gif\" alt=\"The two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region.\" width=\"318\" height=\"315\" \/>\r\n\r\nThe purple area shows where the solutions of the two inequalities overlap. This area is the solution to the <i>system of inequalities<\/i>. Any point within this purple region will be true for both [latex]y&gt;\u2212x[\/latex] and [latex]y&lt;2x+5[\/latex].\r\n\r\nIn the next example, you are given a system of two inequalities whose boundary lines are parallel to each other.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nGraph the system\u00a0[latex]\\begin{array}{c}y\\ge2x+1\\\\y\\lt2x-3\\end{array}[\/latex]\r\n[reveal-answer q=\"780322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"780322\"]\r\n\r\nThe boundary lines for this system are the same as the system of\u00a0equations from a previous example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y=2x+1\\\\y=2x-3\\end{array}[\/latex]<\/p>\r\nPlotting the boundary lines will be similar, except that the inequality [latex]y\\lt2x-3[\/latex] requires that we draw a dashed line, while the inequality [latex]y\\ge2x+1[\/latex] will require a solid line. The graphs will look like this:\r\n\r\n<img class=\" wp-image-4148 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13205706\/Screen-Shot-2016-05-13-at-1.56.45-PM-300x300.png\" alt=\"y=2x+1\" width=\"410\" height=\"410\" \/>\r\n\r\nNow we need to add the regions that represent the inequalities. \u00a0For the inequality [latex]y\\ge2x+1[\/latex] we can test a point on either side of the line to see which region to shade. Let's test [latex]\\left(0,0\\right)[\/latex] to make it easy.\r\n\r\nSubstitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\ge2x+1[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y\\ge2x+1\\\\0\\ge2\\left(0\\right)+1\\\\0\\ge{1}\\end{array}[\/latex]<\/p>\r\nThis is not true, so we know that we need to shade the other side of the boundary line for the inequality\u00a0\u00a0[latex]y\\ge2x+1[\/latex]. The graph will now look like this:\r\n\r\n<img class=\" wp-image-4149 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13210312\/Screen-Shot-2016-05-13-at-2.02.49-PM-300x300.png\" alt=\"y=2x+1 \" width=\"355\" height=\"355\" \/>\r\n\r\nNow let's shade the region that shows the solutions to the inequality [latex]y\\lt2x-3[\/latex]. \u00a0Again, we can pick\u00a0[latex]\\left(0,0\\right)[\/latex] to test because it makes easy algebra.\r\n\r\nSubstitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\lt2x-3[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y\\lt2x-3\\\\0\\lt2\\left(0,\\right)x-3\\\\0\\lt{-3}\\end{array}[\/latex]<\/p>\r\nThis is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\\lt2x-3[\/latex]. The graph will now look like this:\r\n\r\n<img class=\" wp-image-4150 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13210724\/Screen-Shot-2016-05-13-at-2.07.01-PM-297x300.png\" alt=\"y=2x+1\" width=\"394\" height=\"398\" \/>\r\n\r\nThis system of inequalities shares no points in common.\r\n\r\nWhat would the graph look like if the system had looked like this?\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y\\ge2x+1\\\\y\\gt2x-3\\end{array}[\/latex].<\/p>\r\nTesting the point [latex]\\left(0,0\\right)[\/latex] would return a positive result for the inequality\u00a0\u00a0[latex]y\\gt2x-3[\/latex], and the graph would then look like this:\r\n\r\n<img class=\" wp-image-4157 aligncenter\" style=\"height: auto; max-width: 100%; display: block; margin-left: auto; margin-right: auto;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13212029\/Screen-Shot-2016-05-13-at-2.19.42-PM-297x300.png\" alt=\"y&gt;2x-3 and y&gt;=2x+1\" width=\"388\" height=\"392\" \/>\r\n\r\nThe purple region is the region of overlap for both inequalities.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/ACTxJv1h2_c\r\n<p class=\"no-indent\" style=\"text-align: left;\">In the next section, we will see that points can be solutions to systems of equations and inequalities. \u00a0We will verify algebraically whether a point is a solution to a linear equation or inequality.<\/p>\r\n\r\n<h2 id=\"title2\">Determine whether an ordered pair is a solution for a system of linear equations<\/h2>\r\n<img class=\"size-medium wp-image-388 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064355\/image009-2-300x300.jpg\" alt=\"image009-2\" width=\"300\" height=\"300\" \/>\r\n\r\nThe lines in the graph above are defined as\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex].<\/p>\r\nThey cross at what appears to be [latex]\\left(-3,-2\\right)[\/latex].\r\n\r\nUsing algebra, we can verify that this shared point is actually [latex]\\left(-3,-2\\right)[\/latex] and not [latex]\\left(-2.999,-1.999\\right)[\/latex]. By substituting the <i>x<\/i>- and <i>y<\/i>-values of the ordered pair into the equation of each line, you can test whether the point is on both lines. If the substitution results in a true statement, then you have found a\u00a0solution to the system of equations!\r\n\r\nSince the solution of the system must be a solution to <i>all<\/i> the equations in the system, you will need to check the point in each equation. In the following example, we will substitute -3 for <i>x<\/i> and -2 for <i>y<\/i> in each equation to test whether it is actually the solution.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs [latex]\\left(-3,-2\\right)[\/latex] a solution of the system\r\n\r\n[latex]\\begin{array}{r}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"919027\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"919027\"]Test\u00a0[latex]2x+y=-8[\/latex] first:\r\n\r\n[latex]\\begin{array}{r}2(-3)+(-2) = -8\\\\-8 = -8\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\nNow test [latex]x-y=-1[\/latex].\r\n\r\n[latex]\\begin{array}{r}(-3)-(-2) = -1\\\\-1 = -1\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\n[latex]\\left(-3,-2\\right)[\/latex] is a solution of [latex]x-y=-1[\/latex]\r\n\r\nSince[latex]\\left(-3,-2\\right)[\/latex] is a solution of each of the equations in the system,[latex]\\left(-3,-2\\right)[\/latex] is a solution of the system.\r\n<h4>Answer<\/h4>\r\n[latex]\\left(-3,-2\\right)[\/latex] is a solution to the system.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs (3, 9) a solution of the system\r\n\r\n[latex]\\begin{array}{r}y=3x\\\\2x\u2013y=6\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"190963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"190963\"]Since the solution of the system must be a solution to <i>all<\/i> the equations in the system, check the point in each equation.\r\n\r\nSubstitute 3 for <i>x<\/i> and 9 for <i>y<\/i> in each equation.\r\n\r\n[latex]\\begin{array}{l}y=3x\\\\9=3\\left(3\\right)\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\n(3, 9) is a solution of [latex]y=3x[\/latex].\r\n\r\n[latex]\\begin{array}{r}2x\u2013y=6\\\\2\\left(3\\right)\u20139=6\\\\6\u20139=6\\\\-3=6\\\\\\text{FALSE}\\end{array}[\/latex]\r\n\r\n(3, 9) is <i>not<\/i> a solution of [latex]2x\u2013y=6[\/latex].\r\n\r\nSince (3, 9) is not a solution of one of the equations in the system, it cannot be a solution of the system.\r\n<h4>Answer<\/h4>\r\n(3, 9) is not a solution to the system.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs [latex](\u22122,4)[\/latex] a solution for the system\r\n\r\n[latex]\\begin{array}{r}y=2x\\\\3x+2y=1\\end{array}[\/latex]\r\n\r\nBefore you do any calculations, look at the point given and the first equation in the system. \u00a0Can you predict the answer to the question without doing any algebra?\r\n[reveal-answer q=\"598405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"598405\"]\r\n\r\nSubstitute -2 for x, and 4 for y into the first equation:\r\n\r\n[latex]\\begin{array}{l}y=2x\\\\4=2\\left(-2\\right)\\\\4=-4\\\\\\text{FALSE}\\end{array}[\/latex]\r\n\r\nYou can stop testing because a point that is a solution to the system will be a solution to both equations in the system.\r\n\r\n[latex](\u22122,4)[\/latex] is NOT a solution for the system\r\n\r\n[latex]\\begin{array}{r}y=2x\\\\3x+2y=1\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/2IxgKgjX00k\r\n\r\nRemember that in order to be a solution to the system of equations, the values of the point must be a solution for both equations. Once you find one equation for which the point is false, you have determined that it is not a solution for the system.\r\n\r\nWe can use the same method to determine whether a point is a solution to a system of linear inequalities.\r\n<h2 id=\"title4\">Determine whether an ordered pair is a solution to a system of linear inequalities<\/h2>\r\n<img class=\"size-medium wp-image-398 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014-300x297.gif\" alt=\"image014\" width=\"300\" height=\"297\" \/>\r\n\r\nOn the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.\r\n\r\nIn contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y&gt;\u2212x[\/latex] and point A is a solution for the inequality [latex]y&lt;2x+5[\/latex], neither point is a solution for the <i>system<\/i>. The following example shows how to test a point to see whether it is a solution to a system of inequalities.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs the point (2, 1) a solution of the system [latex]x+y&gt;1[\/latex] and [latex]2x+y&lt;8[\/latex]?\r\n\r\n[reveal-answer q=\"84880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"84880\"]Check the point with each of the inequalities. Substitute 2 for <i>x<\/i> and 1 for <i>y<\/i>. Is the point a solution of both inequalities?\r\n\r\n[latex]\\begin{array}{r}x+y&gt;1\\\\2+1&gt;1\\\\3&gt;1\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\n(2, 1) is a solution for [latex]x+y&gt;1[\/latex].\r\n\r\n[latex]\\begin{array}{r}2x+y&lt;8\\\\2\\left(2\\right)+1&lt;8\\\\4+1&lt;8\\\\5&lt;8\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\n(2, 1) is a solution for [latex]2x+y&lt;8.[\/latex]\r\n\r\nSince (2, 1) is a solution of each inequality, it is also a solution of the system.\r\n<h4>Answer<\/h4>\r\nThe point (2, 1) is a solution of the system [latex]x+y&gt;1[\/latex] and [latex]2x+y&lt;8[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is a graph of the system in the example above. Notice that (2, 1) lies in the purple area, which is the overlapping area for the two inequalities.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064404\/image015.gif\" alt=\"Two dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.\" width=\"321\" height=\"317\" \/>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIs the point (2, 1) a solution of the system [latex]x+y&gt;1[\/latex] and [latex]3x+y&lt;4[\/latex]?\r\n\r\n[reveal-answer q=\"833522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"833522\"]\r\n\r\nCheck the point with each of the inequalities. Substitute 2 for <i>x<\/i> and 1 for <i>y<\/i>. Is the point a solution of both inequalities?\r\n\r\n[latex]\\begin{array}{r}x+y&gt;1\\\\2+1&gt;1\\\\3&gt;1\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\n(2, 1) is a solution for [latex]x+y&gt;1[\/latex].\r\n\r\n[latex]\\begin{array}{r}3x+y&lt;4\\\\3\\left(2\\right)+1&lt;4\\\\6+1&lt;4\\\\7&lt;4\\\\\\text{FALSE}\\end{array}[\/latex]\r\n\r\n(2, 1) is <i>not <\/i>a solution for [latex]3x+y&lt;4[\/latex].\r\n\r\nSince (2, 1) is <i>not <\/i>a solution of one of the inequalities, it is not a solution of the system.\r\n<h4>Answer<\/h4>\r\n<span lang=\"X-NONE\">The point (2, 1) is not a solution of the system [latex]x+y&gt;1[\/latex]<\/span>\u00a0and [latex]3x+y&lt;4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is a graph of this system. Notice that (2, 1) is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064405\/image016.gif\" alt=\"A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region.\" width=\"346\" height=\"342\" \/>\r\n\r\nAs shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities. \u00a0The general steps are outlined below:\r\n<ul>\r\n \t<li>Graph each inequality as a line and determine whether it will be solid or dashed<\/li>\r\n \t<li>Determine which side of each boundary line represents solutions to the inequality by testing a point on each side<\/li>\r\n \t<li>Shade the region\u00a0that represents solutions for both inequalities<\/li>\r\n<\/ul>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nShade the region of the graph that represents solutions for both inequalities.\u00a0\u00a0[latex]x+y\\geq1[\/latex] and [latex]y\u2013x\\geq5[\/latex].\r\n\r\n[reveal-answer q=\"873537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"873537\"]Graph one inequality. First graph the boundary line, using a table of values, intercepts, or any other method you prefer. The boundary line for [latex]x+y\\geq1[\/latex] is [latex]x+y=1[\/latex], or [latex]y=\u2212x+1[\/latex]. Since the equal sign is included with the greater than sign, the boundary line is solid.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064406\/image017-2.jpg\" alt=\"A downward-sloping solid line labeled x+y is greater than 1.\" width=\"370\" height=\"370\" \/>\r\n\r\nFind an ordered pair on either side of the boundary line. Insert the <i>x<\/i>- and <i>y<\/i>-values into the inequality [latex]x+y\\geq1[\/latex] and see which ordered pair results in a true statement.\r\n\r\n[latex]\\begin{array}{r}\\text{Test }1:\\left(\u22123,0\\right)\\\\x+y\\geq1\\\\\u22123+0\\geq1\\\\\u22123\\geq1\\\\\\text{FALSE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\x+y\\geq1\\\\4+1\\geq1\\\\5\\geq1\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\nSince (4, 1) results in a true statement, the region that includes (4, 1) should be shaded.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064409\/image018.gif\" alt=\"A solid downward-sloping line with the region above it shaded and labeled x+y is greater than or equal to 1. The point (4,1) is in the shaded region. The point (-3,0) is not.\" width=\"345\" height=\"342\" \/>\r\n\r\nDo the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y\u2013x=5\\left(\\text{or }y=x+5\\right)[\/latex] and is solid. Test point (\u22123, 0) is not a solution of [latex]y\u2013x\\geq5[\/latex], and test point (0, 6) is a solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064410\/image019.gif\" alt=\"A solid blue line with the region above it shaded and labeled y-x is greater than or equal to 5. A solid red line with the region above it shaded and labeled x+y is greater than 1. The point (-3,0) is not in any shaded region. The point (0,6) is in the overlapping shaded region.\" width=\"337\" height=\"334\" \/>\r\n<h4>Answer<\/h4>\r\nThe purple region in this graph shows the set of all solutions of the system.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064412\/image020-2.jpg\" alt=\"The previous graph, with the purple overlapping shaded region labeled x+y is greater than or equal to 1 and y-x is greater than or equal to 5.\" width=\"329\" height=\"325\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/o9hTFJEBcXs\r\n\r\nIn this section we have seen that solutions to systems of linear equations and inequalities can be ordered pairs. In the next section, we will work with systems that have no solutions or infinitely many solutions.\r\n<h2 id=\"title1\">Use a graph to\u00a0classify solutions to systems<\/h2>\r\nRecall that a linear equation graphs as a line, which indicates that all of the points on the line are solutions to that linear equation. There are an infinite number of solutions. As we saw in the last section, if you have a system of linear equations that intersect at one point, this point is a solution to the system. \u00a0What happens if the lines never cross, as in the case of parallel lines? \u00a0How would you describe the solutions to that kind of system? In this section, we will explore the three possible outcomes for solutions to a system of linear equations.\r\n<h3>Three possible outcomes for solutions to systems of equations<\/h3>\r\nRecall that the solution for a system of equations is the value or values that are true for <i>all<\/i> equations in the system.\u00a0There are three possible outcomes for solutions to systems of linear equations. \u00a0The graphs of equations within a system can tell you how many solutions exist for that system. Look at the images below. Each shows two lines that make up a system of equations.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>One Solution<\/th>\r\n<th>No Solutions<\/th>\r\n<th>Infinite Solutions<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064348\/image001-1.jpg\" alt=\"Two intersecting lines.\" width=\"206\" height=\"193\" \/><\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064349\/image002-1.jpg\" alt=\"Two parallel lines\" width=\"203\" height=\"190\" \/><\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064349\/image003-2.jpg\" alt=\"Two identical lines that overlap so that they appear to be one line\" width=\"204\" height=\"191\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>If the graphs of the equations intersect, then there is one solution that is true for both equations.<\/td>\r\n<td>If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.<\/td>\r\n<td>If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ul>\r\n \t<li><strong>One Solution:<\/strong>\u00a0When a system of equations intersects at an ordered pair, the system has one solution.<\/li>\r\n \t<li><strong>Infinite Solutions:<\/strong> Sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions.<\/li>\r\n \t<li><strong>No Solution:<\/strong> When the lines that make up a system are parallel, there are no solutions because the two lines share no points in common.<\/li>\r\n<\/ul>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUsing the graph of\u00a0[latex]\\begin{array}{r}y=x\\\\x+2y=6\\end{array}[\/latex], shown below, determine how many solutions the system has.\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\/04064350\/image004-1.gif\" alt=\"A line labeled x+2y=6 and a line labeled y=x.\" width=\"371\" height=\"371\" \/>\r\n\r\n[reveal-answer q=\"896900\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"896900\"]The lines intersect at one point. So the two lines have only one point in common, there is only one solution to the system.\r\n<h4>Answer<\/h4>\r\nThere is one solution to this system.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example (Advanced)<\/h3>\r\nUsing the graph of [latex]\\begin{array}{r}y=3.5x+0.25\\\\14x\u20134y=-4.5\\end{array}[\/latex], shown below, determine how many solutions the system has.\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\/04064351\/image005-1.jpg\" alt=\"Two parallel lines. One line is y=3.5x+0.25. The other line is 14x-4y=-4.5.\" width=\"352\" height=\"349\" \/>\r\n\r\n[reveal-answer q=\"337033\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"337033\"]The lines are parallel, meaning they do not intersect. There are no solutions to the system.\r\n<h4>Answer<\/h4>\r\n<span style=\"line-height: 1.5;\">There are no solutions to the system.<\/span>\r\n\r\n<span style=\"line-height: 1.5;\">[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nHow many solutions does the system\u00a0[latex]\\begin{array}{r}y=2x+1\\\\\u22124x+2y=2\\end{array}[\/latex] have?\r\n[reveal-answer q=\"94971\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"94971\"]\r\nFirst, graph both equations on the same axes.\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\/04064354\/image008.gif\" alt=\"Two lines that overlap each other. One line is y=2x+1. The other line is -4x+2y=2.\" width=\"390\" height=\"390\" \/>\r\n\r\nThe two equations graph as the same line. So every point on that line is a solution for the system of equations.\r\n<h4>Answer<\/h4>\r\nThe system [latex]\\begin{array}{r}y=2x+1\\\\\u22124x+2y=2\\end{array}[\/latex]\u00a0has an infinite number of solutions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/ZolxtOjcEQY\r\n\r\nIn the next section, we will learn some algebraic methods for finding solutions to systems of equations. \u00a0Recall that linear equations in one variable can have one solution, no solution, or many solutions and we can verify this algebraically. \u00a0We will use the same ideas to classify solutions to systems in two variables algebraically.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph systems of equations\n<ul>\n<li>Graph a system of two linear equations<\/li>\n<li>Graph a system of two linear\u00a0inequalities<\/li>\n<\/ul>\n<\/li>\n<li>Evaluate ordered pairs as solutions to systems\n<ul>\n<li>Determine whether an ordered pair is a solution to a system of linear equations<\/li>\n<li>Determine whether an ordered pair is a solution to a system of linear inequalities<\/li>\n<\/ul>\n<\/li>\n<li>Classify solutions to systems\n<ul>\n<li>Identify what type of solution a system will have based on its graph<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The way a river flows depends on many variables including how big the river is, how much water it contains, what sorts of things are floating in the river, whether or not it is raining, and so forth. If you want to best describe its flow, you must take into account these other variables. A system of linear equations can help with that.<\/p>\n<p>A <strong>system of linear equations<\/strong> consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. You will find systems of equations in every application of mathematics. They are a useful tool for discovering and describing how behaviors or processes are interrelated. It is rare to find, for example, a pattern of traffic flow that that is only affected by weather. Accidents, time of day, and major sporting events are just a few of the other variables that can affect the flow of traffic in a city. In this section, we will explore some basic principles for graphing and describing the intersection of two lines that make up a system of equations.<\/p>\n<h2 id=\"title1\">Graph a system of linear equations<\/h2>\n<p>In this section, we will look at systems of linear equations and inequalities in two variables. \u00a0First, 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 <em>y<\/em>-intercept, or <em>x<\/em>&#8211; and <em>y<\/em>-intercepts to graph both lines on the same set of axes.<\/p>\n<p>For example, consider the following system of linear equations in two variables.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex]<\/div>\n<p>Let&#8217;s\u00a0graph these using slope-intercept form on the same set of axes. Remember that slope-intercept form looks like [latex]y=mx+b[\/latex], \u00a0so we will want to solve both equations for [latex]y[\/latex].<\/p>\n<p>First, solve for y in [latex]2x+y=-8[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+y=-8\\\\ y=-2x - 8\\end{array}[\/latex]<\/p>\n<div style=\"text-align: center;\"><\/div>\n<div style=\"text-align: left;\">Second, solve for y in [latex]x-y=-1[\/latex]<\/div>\n<div style=\"text-align: left;\"><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}x-y=-1\\,\\,\\,\\,\\,\\\\ y=x+1\\end{array}[\/latex]<\/div>\n<p>The system is now written as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y=-2x - 8\\\\y=x+1\\end{array}[\/latex]<\/p>\n<p>Now you can graph both equations using their slopes and intercepts on the same set of axes, as seen in the figure below. Note how the graphs share one point in common. This is their point of intersection, a point that lies on both of the lines. \u00a0In the next section we will verify that\u00a0this point is a solution to the system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5878 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01202809\/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 id=\"video1\" class=\"no-indent\" style=\"text-align: left;\">\u00a0In 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 y-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 m = 2 and the y-intercept (0,1)<\/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 m = 2, and the y-intercept (0,-3)<\/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 next section we will discuss how there are no solutions to a system of equations that\u00a0are parallel lines.<\/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 x- and y- intercepts of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex]<\/p>\n<p>The x-intercept will have a value of 0 for y, so substitute y=0 into the equation, and isolate the variable x.<\/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 x-intercept of\u00a0[latex]y=\\frac{1}{2}x+2[\/latex] is [latex]\\left(-4,0\\right)[\/latex].<\/p>\n<p>The y-intercept is easier to find since this equation is in slope-intercept form. \u00a0The y-intercept is (2,0).<\/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 y = 0 in to the equation to find the x-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 x-intercept of\u00a0[latex]2y-x=4[\/latex] is [latex]\\left(-4,0\\right)[\/latex].<\/p>\n<p>Now substitute x = 0 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 y-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 next section, we will see that systems with two of the same equations in them have an infinite number of solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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. When you graph a system of linear inequalities on the same set of axes, there are a few more things you will need to consider.<\/p>\n<h2 id=\"title2\">Graph a system of two inequalities<\/h2>\n<p>Remember from the module on graphing that the graph of a single linear inequality splits the <b>coordinate plane<\/b> into two regions.\u00a0On one side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality [latex]y<2x+5[\/latex].\n\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064400\/image012.gif\" alt=\"An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1).\" width=\"346\" height=\"343\" \/><\/p>\n<p>The dashed line is [latex]y=2x+5[\/latex]. Every ordered pair in the shaded\u00a0area below the line is a solution to [latex]y<2x+5[\/latex], as all of the points below the line will make the inequality true. If you doubt that, try substituting the <i>x<\/i> and <i>y<\/i> coordinates of Points A and B into the inequality\u2014you\u2019ll see that they work. So, the shaded area shows all of the solutions for this inequality.<\/p>\n<p>The boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don\u2019t satisfy the inequality. If the inequality had been [latex]y\\leq2x+5[\/latex], then the boundary line would have been solid.<\/p>\n<p>Let\u2019s graph another inequality: [latex]y>\u2212x[\/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064401\/image013.gif\" alt=\"Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1).\" width=\"327\" height=\"324\" \/><\/p>\n<p>To create a system of inequalities, you need to graph two or more inequalities together. Let\u2019s use\u00a0[latex]y<2x+5[\/latex] and [latex]y>\u2212x[\/latex] since we have already graphed each of them.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014.gif\" alt=\"The two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region.\" width=\"318\" height=\"315\" \/><\/p>\n<p>The purple area shows where the solutions of the two inequalities overlap. This area is the solution to the <i>system of inequalities<\/i>. Any point within this purple region will be true for both [latex]y>\u2212x[\/latex] and [latex]y<2x+5[\/latex].\n\nIn the next example, you are given a system of two inequalities whose boundary lines are parallel to each other.\n\n\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>Graph the system\u00a0[latex]\\begin{array}{c}y\\ge2x+1\\\\y\\lt2x-3\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q780322\">Show Solution<\/span><\/p>\n<div id=\"q780322\" class=\"hidden-answer\" style=\"display: none\">\n<p>The boundary lines for this system are the same as the system of\u00a0equations from a previous example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y=2x+1\\\\y=2x-3\\end{array}[\/latex]<\/p>\n<p>Plotting the boundary lines will be similar, except that the inequality [latex]y\\lt2x-3[\/latex] requires that we draw a dashed line, while the inequality [latex]y\\ge2x+1[\/latex] will require a solid line. The graphs will look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4148 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13205706\/Screen-Shot-2016-05-13-at-1.56.45-PM-300x300.png\" alt=\"y=2x+1\" width=\"410\" height=\"410\" \/><\/p>\n<p>Now we need to add the regions that represent the inequalities. \u00a0For the inequality [latex]y\\ge2x+1[\/latex] we can test a point on either side of the line to see which region to shade. Let&#8217;s test [latex]\\left(0,0\\right)[\/latex] to make it easy.<\/p>\n<p>Substitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\ge2x+1[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y\\ge2x+1\\\\0\\ge2\\left(0\\right)+1\\\\0\\ge{1}\\end{array}[\/latex]<\/p>\n<p>This is not true, so we know that we need to shade the other side of the boundary line for the inequality\u00a0\u00a0[latex]y\\ge2x+1[\/latex]. The graph will now look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4149 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13210312\/Screen-Shot-2016-05-13-at-2.02.49-PM-300x300.png\" alt=\"y=2x+1\" width=\"355\" height=\"355\" \/><\/p>\n<p>Now let&#8217;s shade the region that shows the solutions to the inequality [latex]y\\lt2x-3[\/latex]. \u00a0Again, we can pick\u00a0[latex]\\left(0,0\\right)[\/latex] to test because it makes easy algebra.<\/p>\n<p>Substitute\u00a0[latex]\\left(0,0\\right)[\/latex] into\u00a0[latex]y\\lt2x-3[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y\\lt2x-3\\\\0\\lt2\\left(0,\\right)x-3\\\\0\\lt{-3}\\end{array}[\/latex]<\/p>\n<p>This is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\\lt2x-3[\/latex]. The graph will now look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4150 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13210724\/Screen-Shot-2016-05-13-at-2.07.01-PM-297x300.png\" alt=\"y=2x+1\" width=\"394\" height=\"398\" \/><\/p>\n<p>This system of inequalities shares no points in common.<\/p>\n<p>What would the graph look like if the system had looked like this?<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}y\\ge2x+1\\\\y\\gt2x-3\\end{array}[\/latex].<\/p>\n<p>Testing the point [latex]\\left(0,0\\right)[\/latex] would return a positive result for the inequality\u00a0\u00a0[latex]y\\gt2x-3[\/latex], and the graph would then look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4157 aligncenter\" style=\"height: auto; max-width: 100%; display: block; margin-left: auto; margin-right: auto;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/13212029\/Screen-Shot-2016-05-13-at-2.19.42-PM-297x300.png\" alt=\"y&gt;2x-3 and y&gt;=2x+1\" width=\"388\" height=\"392\" \/><\/p>\n<p>The purple region is the region of overlap for both inequalities.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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 class=\"no-indent\" style=\"text-align: left;\">In the next section, we will see that points can be solutions to systems of equations and inequalities. \u00a0We will verify algebraically whether a point is a solution to a linear equation or inequality.<\/p>\n<h2 id=\"title2\">Determine whether an ordered pair is a solution for a system of linear equations<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-388 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064355\/image009-2-300x300.jpg\" alt=\"image009-2\" width=\"300\" height=\"300\" \/><\/p>\n<p>The lines in the graph above are defined as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=-8\\\\ x-y=-1\\end{array}[\/latex].<\/p>\n<p>They cross at what appears to be [latex]\\left(-3,-2\\right)[\/latex].<\/p>\n<p>Using algebra, we can verify that this shared point is actually [latex]\\left(-3,-2\\right)[\/latex] and not [latex]\\left(-2.999,-1.999\\right)[\/latex]. By substituting the <i>x<\/i>&#8211; and <i>y<\/i>-values of the ordered pair into the equation of each line, you can test whether the point is on both lines. If the substitution results in a true statement, then you have found a\u00a0solution to the system of equations!<\/p>\n<p>Since the solution of the system must be a solution to <i>all<\/i> the equations in the system, you will need to check the point in each equation. In the following example, we will substitute -3 for <i>x<\/i> and -2 for <i>y<\/i> in each equation to test whether it is actually the solution.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is [latex]\\left(-3,-2\\right)[\/latex] a solution of the system<\/p>\n<p>[latex]\\begin{array}{r}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=\"q919027\">Show Solution<\/span><\/p>\n<div id=\"q919027\" class=\"hidden-answer\" style=\"display: none\">Test\u00a0[latex]2x+y=-8[\/latex] first:<\/p>\n<p>[latex]\\begin{array}{r}2(-3)+(-2) = -8\\\\-8 = -8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>Now test [latex]x-y=-1[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}(-3)-(-2) = -1\\\\-1 = -1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>[latex]\\left(-3,-2\\right)[\/latex] is a solution of [latex]x-y=-1[\/latex]<\/p>\n<p>Since[latex]\\left(-3,-2\\right)[\/latex] is a solution of each of the equations in the system,[latex]\\left(-3,-2\\right)[\/latex] is a solution of the system.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(-3,-2\\right)[\/latex] is a solution to the system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is (3, 9) a solution of the system<\/p>\n<p>[latex]\\begin{array}{r}y=3x\\\\2x\u2013y=6\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q190963\">Show Solution<\/span><\/p>\n<div id=\"q190963\" class=\"hidden-answer\" style=\"display: none\">Since the solution of the system must be a solution to <i>all<\/i> the equations in the system, check the point in each equation.<\/p>\n<p>Substitute 3 for <i>x<\/i> and 9 for <i>y<\/i> in each equation.<\/p>\n<p>[latex]\\begin{array}{l}y=3x\\\\9=3\\left(3\\right)\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>(3, 9) is a solution of [latex]y=3x[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}2x\u2013y=6\\\\2\\left(3\\right)\u20139=6\\\\6\u20139=6\\\\-3=6\\\\\\text{FALSE}\\end{array}[\/latex]<\/p>\n<p>(3, 9) is <i>not<\/i> a solution of [latex]2x\u2013y=6[\/latex].<\/p>\n<p>Since (3, 9) is not a solution of one of the equations in the system, it cannot be a solution of the system.<\/p>\n<h4>Answer<\/h4>\n<p>(3, 9) is not a solution to the system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is [latex](\u22122,4)[\/latex] a solution for the system<\/p>\n<p>[latex]\\begin{array}{r}y=2x\\\\3x+2y=1\\end{array}[\/latex]<\/p>\n<p>Before you do any calculations, look at the point given and the first equation in the system. \u00a0Can you predict the answer to the question without doing any algebra?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q598405\">Show Solution<\/span><\/p>\n<div id=\"q598405\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute -2 for x, and 4 for y into the first equation:<\/p>\n<p>[latex]\\begin{array}{l}y=2x\\\\4=2\\left(-2\\right)\\\\4=-4\\\\\\text{FALSE}\\end{array}[\/latex]<\/p>\n<p>You can stop testing because a point that is a solution to the system will be a solution to both equations in the system.<\/p>\n<p>[latex](\u22122,4)[\/latex] is NOT a solution for the system<\/p>\n<p>[latex]\\begin{array}{r}y=2x\\\\3x+2y=1\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Determine if an Ordered Pair is a Solution to a System of Linear Equations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/2IxgKgjX00k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Remember that in order to be a solution to the system of equations, the values of the point must be a solution for both equations. Once you find one equation for which the point is false, you have determined that it is not a solution for the system.<\/p>\n<p>We can use the same method to determine whether a point is a solution to a system of linear inequalities.<\/p>\n<h2 id=\"title4\">Determine whether an ordered pair is a solution to a system of linear inequalities<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-398 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064403\/image014-300x297.gif\" alt=\"image014\" width=\"300\" height=\"297\" \/><\/p>\n<p>On the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.<\/p>\n<p>In contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y>\u2212x[\/latex] and point A is a solution for the inequality [latex]y<2x+5[\/latex], neither point is a solution for the <i>system<\/i>. The following example shows how to test a point to see whether it is a solution to a system of inequalities.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is the point (2, 1) a solution of the system [latex]x+y>1[\/latex] and [latex]2x+y<8[\/latex]?\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q84880\">Show Solution<\/span><\/p>\n<div id=\"q84880\" class=\"hidden-answer\" style=\"display: none\">Check the point with each of the inequalities. Substitute 2 for <i>x<\/i> and 1 for <i>y<\/i>. Is the point a solution of both inequalities?<\/p>\n<p>[latex]\\begin{array}{r}x+y>1\\\\2+1>1\\\\3>1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>(2, 1) is a solution for [latex]x+y>1[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}2x+y<8\\\\2\\left(2\\right)+1<8\\\\4+1<8\\\\5<8\\\\\\text{TRUE}\\end{array}[\/latex]\n\n(2, 1) is a solution for [latex]2x+y<8.[\/latex]\n\nSince (2, 1) is a solution of each inequality, it is also a solution of the system.\n\n\n<h4>Answer<\/h4>\n<p>The point (2, 1) is a solution of the system [latex]x+y>1[\/latex] and [latex]2x+y<8[\/latex].\n\n<\/div>\n<\/div>\n<\/div>\n<p>Here is a graph of the system in the example above. Notice that (2, 1) lies in the purple area, which is the overlapping area for the two inequalities.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064404\/image015.gif\" alt=\"Two dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.\" width=\"321\" height=\"317\" \/><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Is the point (2, 1) a solution of the system [latex]x+y>1[\/latex] and [latex]3x+y<4[\/latex]?\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q833522\">Show Solution<\/span><\/p>\n<div id=\"q833522\" class=\"hidden-answer\" style=\"display: none\">\n<p>Check the point with each of the inequalities. Substitute 2 for <i>x<\/i> and 1 for <i>y<\/i>. Is the point a solution of both inequalities?<\/p>\n<p>[latex]\\begin{array}{r}x+y>1\\\\2+1>1\\\\3>1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>(2, 1) is a solution for [latex]x+y>1[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}3x+y<4\\\\3\\left(2\\right)+1<4\\\\6+1<4\\\\7<4\\\\\\text{FALSE}\\end{array}[\/latex]\n\n(2, 1) is <i>not <\/i>a solution for [latex]3x+y<4[\/latex].\n\nSince (2, 1) is <i>not <\/i>a solution of one of the inequalities, it is not a solution of the system.<\/p>\n<h4>Answer<\/h4>\n<p><span lang=\"X-NONE\">The point (2, 1) is not a solution of the system [latex]x+y>1[\/latex]<\/span>\u00a0and [latex]3x+y<4[\/latex].\n\n<\/div>\n<\/div>\n<\/div>\n<p>Here is a graph of this system. Notice that (2, 1) is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064405\/image016.gif\" alt=\"A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region.\" width=\"346\" height=\"342\" \/><\/p>\n<p>As shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities. \u00a0The general steps are outlined below:<\/p>\n<ul>\n<li>Graph each inequality as a line and determine whether it will be solid or dashed<\/li>\n<li>Determine which side of each boundary line represents solutions to the inequality by testing a point on each side<\/li>\n<li>Shade the region\u00a0that represents solutions for both inequalities<\/li>\n<\/ul>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Shade the region of the graph that represents solutions for both inequalities.\u00a0\u00a0[latex]x+y\\geq1[\/latex] and [latex]y\u2013x\\geq5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q873537\">Show Solution<\/span><\/p>\n<div id=\"q873537\" class=\"hidden-answer\" style=\"display: none\">Graph one inequality. First graph the boundary line, using a table of values, intercepts, or any other method you prefer. The boundary line for [latex]x+y\\geq1[\/latex] is [latex]x+y=1[\/latex], or [latex]y=\u2212x+1[\/latex]. Since the equal sign is included with the greater than sign, the boundary line is solid.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064406\/image017-2.jpg\" alt=\"A downward-sloping solid line labeled x+y is greater than 1.\" width=\"370\" height=\"370\" \/><\/p>\n<p>Find an ordered pair on either side of the boundary line. Insert the <i>x<\/i>&#8211; and <i>y<\/i>-values into the inequality [latex]x+y\\geq1[\/latex] and see which ordered pair results in a true statement.<\/p>\n<p>[latex]\\begin{array}{r}\\text{Test }1:\\left(\u22123,0\\right)\\\\x+y\\geq1\\\\\u22123+0\\geq1\\\\\u22123\\geq1\\\\\\text{FALSE}\\\\\\\\\\text{Test }2:\\left(4,1\\right)\\\\x+y\\geq1\\\\4+1\\geq1\\\\5\\geq1\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>Since (4, 1) results in a true statement, the region that includes (4, 1) should be shaded.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064409\/image018.gif\" alt=\"A solid downward-sloping line with the region above it shaded and labeled x+y is greater than or equal to 1. The point (4,1) is in the shaded region. The point (-3,0) is not.\" width=\"345\" height=\"342\" \/><\/p>\n<p>Do the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y\u2013x=5\\left(\\text{or }y=x+5\\right)[\/latex] and is solid. Test point (\u22123, 0) is not a solution of [latex]y\u2013x\\geq5[\/latex], and test point (0, 6) is a solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064410\/image019.gif\" alt=\"A solid blue line with the region above it shaded and labeled y-x is greater than or equal to 5. A solid red line with the region above it shaded and labeled x+y is greater than 1. The point (-3,0) is not in any shaded region. The point (0,6) is in the overlapping shaded region.\" width=\"337\" height=\"334\" \/><\/p>\n<h4>Answer<\/h4>\n<p>The purple region in this graph shows the set of all solutions of the system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064412\/image020-2.jpg\" alt=\"The previous graph, with the purple overlapping shaded region labeled x+y is greater than or equal to 1 and y-x is greater than or equal to 5.\" width=\"329\" height=\"325\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Determine if an Ordered Pair is a Solution to a System of Linear Inequalities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/o9hTFJEBcXs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In this section we have seen that solutions to systems of linear equations and inequalities can be ordered pairs. In the next section, we will work with systems that have no solutions or infinitely many solutions.<\/p>\n<h2 id=\"title1\">Use a graph to\u00a0classify solutions to systems<\/h2>\n<p>Recall that a linear equation graphs as a line, which indicates that all of the points on the line are solutions to that linear equation. There are an infinite number of solutions. As we saw in the last section, if you have a system of linear equations that intersect at one point, this point is a solution to the system. \u00a0What happens if the lines never cross, as in the case of parallel lines? \u00a0How would you describe the solutions to that kind of system? In this section, we will explore the three possible outcomes for solutions to a system of linear equations.<\/p>\n<h3>Three possible outcomes for solutions to systems of equations<\/h3>\n<p>Recall that the solution for a system of equations is the value or values that are true for <i>all<\/i> equations in the system.\u00a0There are three possible outcomes for solutions to systems of linear equations. \u00a0The graphs of equations within a system can tell you how many solutions exist for that system. Look at the images below. Each shows two lines that make up a system of equations.<\/p>\n<table>\n<thead>\n<tr>\n<th>One Solution<\/th>\n<th>No Solutions<\/th>\n<th>Infinite Solutions<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064348\/image001-1.jpg\" alt=\"Two intersecting lines.\" width=\"206\" height=\"193\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064349\/image002-1.jpg\" alt=\"Two parallel lines\" width=\"203\" height=\"190\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064349\/image003-2.jpg\" alt=\"Two identical lines that overlap so that they appear to be one line\" width=\"204\" height=\"191\" \/><\/td>\n<\/tr>\n<tr>\n<td>If the graphs of the equations intersect, then there is one solution that is true for both equations.<\/td>\n<td>If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.<\/td>\n<td>If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><strong>One Solution:<\/strong>\u00a0When a system of equations intersects at an ordered pair, the system has one solution.<\/li>\n<li><strong>Infinite Solutions:<\/strong> Sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions.<\/li>\n<li><strong>No Solution:<\/strong> When the lines that make up a system are parallel, there are no solutions because the two lines share no points in common.<\/li>\n<\/ul>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Using the graph of\u00a0[latex]\\begin{array}{r}y=x\\\\x+2y=6\\end{array}[\/latex], shown below, determine how many solutions the system has.<\/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\/04064350\/image004-1.gif\" alt=\"A line labeled x+2y=6 and a line labeled y=x.\" width=\"371\" height=\"371\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q896900\">Show Solution<\/span><\/p>\n<div id=\"q896900\" class=\"hidden-answer\" style=\"display: none\">The lines intersect at one point. So the two lines have only one point in common, there is only one solution to the system.<\/p>\n<h4>Answer<\/h4>\n<p>There is one solution to this system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example (Advanced)<\/h3>\n<p>Using the graph of [latex]\\begin{array}{r}y=3.5x+0.25\\\\14x\u20134y=-4.5\\end{array}[\/latex], shown below, determine how many solutions the system has.<\/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\/04064351\/image005-1.jpg\" alt=\"Two parallel lines. One line is y=3.5x+0.25. The other line is 14x-4y=-4.5.\" width=\"352\" height=\"349\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q337033\">Show Solution<\/span><\/p>\n<div id=\"q337033\" class=\"hidden-answer\" style=\"display: none\">The lines are parallel, meaning they do not intersect. There are no solutions to the system.<\/p>\n<h4>Answer<\/h4>\n<p><span style=\"line-height: 1.5;\">There are no solutions to the system.<\/span><\/p>\n<p><span style=\"line-height: 1.5;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>How many solutions does the system\u00a0[latex]\\begin{array}{r}y=2x+1\\\\\u22124x+2y=2\\end{array}[\/latex] have?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q94971\">Show Solution<\/span><\/p>\n<div id=\"q94971\" class=\"hidden-answer\" style=\"display: none\">\nFirst, graph both equations on the same axes.<\/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\/04064354\/image008.gif\" alt=\"Two lines that overlap each other. One line is y=2x+1. The other line is -4x+2y=2.\" width=\"390\" height=\"390\" \/><\/p>\n<p>The two equations graph as the same line. So every point on that line is a solution for the system of equations.<\/p>\n<h4>Answer<\/h4>\n<p>The system [latex]\\begin{array}{r}y=2x+1\\\\\u22124x+2y=2\\end{array}[\/latex]\u00a0has an infinite number of solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Determine the Number of Solutions to a System of Linear Equations From a Graph\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ZolxtOjcEQY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next section, we will learn some algebraic methods for finding solutions to systems of equations. \u00a0Recall that linear equations in one variable can have one solution, no solution, or many solutions and we can verify this algebraically. \u00a0We will use the same ideas to classify solutions to systems in two variables algebraically.<\/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-5958\">\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><li>Ex 1: Graph a System of Linear Inequalities. <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><li>Determine if an Ordered Pair is a Solution to a System of Linear Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/2IxgKgjX00k\">https:\/\/youtu.be\/2IxgKgjX00k<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Determine if an Ordered Pair is a Solution to a System of Linear Inequalities. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/o9hTFJEBcXs\">https:\/\/youtu.be\/o9hTFJEBcXs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Determine the Number of Solutions to a System of Linear Equations From a Graph. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ZolxtOjcEQY\">https:\/\/youtu.be\/ZolxtOjcEQY<\/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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Graphing a System of Linear Equation\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/BBmB3rFZLXU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex 1: Graph a System of Linear Inequalities\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ACTxJv1h2_c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Determine if an Ordered Pair is a Solution to a System of Linear Equations\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen 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