{"id":2174,"date":"2016-11-03T18:39:03","date_gmt":"2016-11-03T18:39:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2174"},"modified":"2019-05-30T20:56:33","modified_gmt":"2019-05-30T20:56:33","slug":"introduction-systems-of-linear-equations-two-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/chapter\/introduction-systems-of-linear-equations-two-variables\/","title":{"raw":"Systems of Linear Equations: Two Variables","rendered":"Systems of Linear Equations: Two Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul class=\"ul1\">\r\n \t<li class=\"li2\"><span class=\"s1\">Solve systems of equations by graphing,\u00a0substitution, and\u00a0addition.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Identify inconsistent systems of equations containing two variables.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Express the solution of a system of dependent equations containing two variables using standard notations.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nA skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section we will consider linear equations with two variables to answer these and similar questions.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183558\/CNX_Precalc_Figure_09_01_0012.jpg\" alt=\"Skateboarders at a skating rink by the beach.\" width=\"487\" height=\"252\" \/> (credit: Thomas S\u00f8renes)[\/caption]\r\n<h2>Introduction to Solutions of Systems<\/h2>\r\nIn order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. 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. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.\r\n\r\nIn this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&amp;=15\\\\[1mm] 3x-y&amp;=5\\end{align}[\/latex]<\/p>\r\nThe <em>solution<\/em> to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair [latex](4,7)[\/latex] is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\left(4\\right)+\\left(7\\right)&amp;=15 &amp;&amp;\\text{True} \\\\[1mm] 3\\left(4\\right)-\\left(7\\right)&amp;=5 &amp;&amp;\\text{True} \\end{align}[\/latex]<\/p>\r\nIn addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A <strong>consistent system<\/strong> of equations has at least one solution. A consistent system is considered to be an <strong>independent system<\/strong> if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a <strong>dependent system<\/strong> if the equations have the same slope and the same <em>y<\/em>-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.\r\n\r\nAnother type of system of linear equations is an <strong>inconsistent system<\/strong>, which is one in which the equations represent two parallel lines. The lines have the same slope and different <em>y-<\/em>intercepts. There are no points common to both lines; hence, there is no solution to the system.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Types of Linear Systems<\/h3>\r\nThere are three types of systems of linear equations in two variables, and three types of solutions.\r\n<ul>\r\n \t<li>An <strong>independent system<\/strong> has exactly one solution pair [latex]\\left(x,y\\right)[\/latex]. The point where the two lines intersect is the only solution.<\/li>\r\n \t<li>An <strong>inconsistent system<\/strong> has no solution. Notice that the two lines are parallel and will never intersect.<\/li>\r\n \t<li>A <strong>dependent system<\/strong> has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.<\/li>\r\n<\/ul>\r\nBelow is a comparison of\u00a0graphical representations of each type of system.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183601\/CNX_Precalc_Figure_09_01_002n2.jpg\" alt=\"Graphs of an independent system, an inconsistent system, and a dependent system. The independent system has two lines which cross at the point seven-fifths, negative eleven fifths. The inconsistent system shows two parallel lines. The dependent system shows a single line running through the points negative one, negative two and one, two.\" width=\"945\" height=\"479\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.<\/h3>\r\n<ol>\r\n \t<li>Substitute the ordered pair into each equation in the system.<\/li>\r\n \t<li>Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determining Whether an Ordered Pair Is a Solution to a System of Equations<\/h3>\r\nDetermine whether the ordered pair [latex]\\left(5,1\\right)[\/latex] is a solution to the given system of equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&amp;=8\\\\ 2x-9&amp;=y \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"899056\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"899056\"]\r\n\r\nSubstitute the ordered pair [latex]\\left(5,1\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(5\\right)+3\\left(1\\right)&amp;=8 \\\\[1mm] 8&amp;=8 &amp;&amp;\\text{True} \\\\[3mm] 2\\left(5\\right)-9&amp;=\\left(1\\right) \\\\[1mm] 1&amp;=1 &amp;&amp;\\text{True} \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The ordered pair [latex]\\left(5,1\\right)[\/latex] satisfies both equations, so it is the solution to the system.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183603\/CNX_Precalc_Figure_09_01_0032.jpg\" alt=\"A graph of two lines running through the point five, one. The first line's equation is x plus 3y equals 8. The second line's equation is 2x minus 9 equals y.\" width=\"487\" height=\"365\" \/>\u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nDetermine whether the ordered pair [latex]\\left(8,5\\right)[\/latex] is a solution to the following system.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}5x-4y&amp;=20\\\\ 2x+1&amp;=3y\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"672974\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"672974\"]\r\n\r\nNot a solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Solving Systems of Equations by Graphing<\/h3>\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<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System of Equations in Two Variables by Graphing<\/h3>\r\nSolve the following system of equations by graphing. Identify the type of system.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&amp;=-8\\\\ x-y&amp;=-1\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"8915\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"8915\"]\r\n\r\nSolve the first equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&amp;=-8\\\\ y&amp;=-2x-8\\end{align}[\/latex]<\/p>\r\nSolve the second equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x-y&amp;=-1\\\\ y&amp;=x+1\\end{align}[\/latex]<\/p>\r\nGraph both equations on the same set of axes:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183605\/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{align}2\\left(-3\\right)+\\left(-2\\right)&amp;=-8 \\\\[1mm] -8=-8 &amp;&amp;\\text{True} \\\\[3mm] \\left(-3\\right)-\\left(-2\\right)&amp;=-1 \\\\[1mm] -1&amp;=-1 &amp;&amp;\\text{True} \\end{align}[\/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\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations by graphing.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2x - 5y=-25 \\\\ -4x+5y=35 \\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"141689\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"141689\"]\r\n\r\nThe solution to the system is the ordered pair [latex]\\left(-5,3\\right)[\/latex].\r\n\r\n<img class=\"aligncenter size-full wp-image-3165\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/18222602\/CNX_Precalc_Figure_09_01_0132.jpg\" alt=\"Two lines that cross at the point negative five, three. One line's equation is y equals four-fifths x plus 7. The other line's equation is y equals two-fifths x plus 5.\" width=\"487\" height=\"409\" \/>\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q&amp; A<\/h3>\r\n<h4>Can graphing be used if the system is inconsistent or dependent?<\/h4>\r\n<em>Yes, in 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.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nPlot the three different systems with an online graphing tool. Categorize\u00a0each solution as either consistent or inconsistent. If the system is consistent determine whether it is dependent or independent. You may find it easier to plot each system individually, then clear out your entries before you plot the next.\r\n1)\r\n[latex]5x-3y = -19[\/latex]\r\n[latex]x=2y-1[\/latex]\r\n\r\n2)\r\n[latex]4x+y=11[\/latex]\r\n[latex]-2y=-25+8x[\/latex]\r\n\r\n3)\r\n[latex]y = -3x+6[\/latex]\r\n[latex]-\\frac{1}{3}y+2=x[\/latex]\r\n\r\n[reveal-answer q=\"720375\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720375\"]\r\n<ol>\r\n \t<li>One solution - consistent, independent<\/li>\r\n \t<li>No solutions, inconsistent, neither dependent nor independent<\/li>\r\n \t<li>Many solutions - \u00a0consistent, dependent<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Systems of Equations by Substitution<\/h2>\r\nSolving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a <strong>system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the <strong>substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\r\n<ol>\r\n \t<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\r\n \t<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\r\n \t<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\r\n \t<li>Check the solution in both equations.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System of Equations in Two Variables by Substitution<\/h3>\r\nSolve the following system of equations by substitution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ 2x-5y&amp;=1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"786744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"786744\"]\r\n\r\nFirst, we will solve the first equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ y&amp;=x - 5 \\end{align}[\/latex]<\/p>\r\nNow we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 5y&amp;=1 \\\\ 2x - 5\\left(x - 5\\right)&amp;=1 \\\\ 2x - 5x+25&amp;=1 \\\\ -3x&amp;=-24 \\\\ x&amp;=8 \\end{align}[\/latex]<\/p>\r\nNow, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-\\left(8\\right)+y&amp;=-5 \\\\ y&amp;=3 \\end{align}[\/latex]<\/p>\r\nOur solution is [latex]\\left(8,3\\right)[\/latex].\r\n\r\nCheck the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ -\\left(8\\right)+\\left(3\\right)&amp;=-5 &amp;&amp; \\text{True} \\\\[3mm] 2x - 5y&amp;=1 \\\\ 2\\left(8\\right)-5\\left(3\\right)&amp;=1 &amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations by substitution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x&amp;=y+3 \\\\ 4&amp;=3x - 2y \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"783260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"783260\"]\r\n\r\n[latex]\\left(-2,-5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115164&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h4>Can the substitution method be used to solve any linear system in two variables?<\/h4>\r\n<em>Yes, but the method works best if one of the equations contains a coefficient of 1 or \u20131 so that we do not have to deal with fractions.<\/em>\r\n\r\n<\/div>\r\nThe following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations. \u00a0We present three different examples, and also use a graphing tool to help summarize the solution for each example.\r\n\r\nhttps:\/\/youtu.be\/HxhacvH49o8\r\n<h2>Solving Systems of Equations in Two Variables by the Addition Method<\/h2>\r\nA third method of <strong>solving systems of linear equations<\/strong> is the <strong>addition method,\u00a0<\/strong>this method is also called the\u00a0<strong>elimination method<\/strong>. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\r\n<ol>\r\n \t<li>Write both equations with <em>x<\/em>- and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\r\n \t<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\r\n \t<li>Solve the resulting equation for the remaining variable.<\/li>\r\n \t<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\r\n \t<li>Check the solution by substituting the values into the other equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System by the Addition Method<\/h3>\r\nSolve the given system of equations by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"924657\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924657\"]\r\n\r\nBoth equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\\\ \\hline 3y&amp;=2\\end{align}[\/latex]<\/p>\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3y&amp;=2 \\\\ y&amp;=\\dfrac{2}{3} \\end{align}[\/latex]<\/p>\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=3 \\\\ -x+\\frac{2}{3}&amp;=3 \\\\ -x&amp;=3-\\frac{2}{3} \\\\ -x&amp;=\\frac{7}{3} \\\\ x&amp;=-\\frac{7}{3} \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\r\nCheck the solution in the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ \\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)&amp;= \\\\ -\\frac{7}{3}+\\frac{4}{3}&amp;= \\\\ \\-\\frac{3}{3}&amp;= \\\\ -1&amp;=-1&amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try IT<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method When Multiplication of One Equation Is Required<\/h3>\r\nSolve the given system of equations by the <strong>addition method<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11 \\\\ x - 2y&amp;=11 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"883001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"883001\"]\r\n\r\nAdding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11 \\\\ -3\\left(x - 2y\\right)&amp;=-3\\left(11\\right) &amp;&amp; \\text{Multiply both sides by }-3 \\\\ -3x+6y&amp;=-33 &amp;&amp; \\text{Use the distributive property}. \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now, let\u2019s add them.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=\u221211 \\\\ \u22123x+6y&amp;=\u221233 \\\\ \\hline 11y&amp;=\u221244 \\\\ y&amp;=\u22124 \\end{align}[\/latex]<\/p>\r\nFor the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11\\\\ 3x+5\\left(-4\\right)&amp;=-11\\\\ 3x - 20&amp;=-11\\\\ 3x&amp;=9\\\\ x&amp;=3\\end{align}[\/latex]<\/p>\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11 \\\\ \\left(3\\right)-2\\left(-4\\right)&amp;=3+8 \\\\ &amp;=11 &amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 7y&amp;=2\\\\ 3x+y&amp;=-20\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"609174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"609174\"]\r\n\r\n[latex]\\left(-6,-2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115120&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method When Multiplication of Both Equations Is Required<\/h3>\r\nSolve the given system of equations in two variables by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3y&amp;=-16 \\\\ 5x - 10y&amp;=30\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"114755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114755\"]\r\n\r\nOne equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} -5\\left(2x+3y\\right)&amp;=-5\\left(-16\\right) \\\\ -10x - 15y&amp;=80 \\\\[3mm] 2\\left(5x - 10y\\right)&amp;=2\\left(30\\right) \\\\ 10x - 20y&amp;=60 \\end{align}[\/latex]<\/p>\r\nThen, we add the two equations together.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \u221210x\u221215y&amp;=80 \\\\ 10x\u221220y&amp;=60 \\\\ \\hline \u221235y&amp;=140 \\\\ y&amp;=\u22124 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]y=-4[\/latex] into the original first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3\\left(-4\\right)&amp;=-16\\\\ 2x - 12&amp;=-16\\\\ 2x&amp;=-4\\\\ x&amp;=-2\\end{align}[\/latex]<\/p>\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 5x - 10y&amp;=30\\\\ 5\\left(-2\\right)-10\\left(-4\\right)&amp;=30\\\\ -10+40&amp;=30\\\\ 30&amp;=30\\end{align}[\/latex]<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183611\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method in Systems of Equations Containing Fractions<\/h3>\r\nSolve the given system of equations in two variables by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{3}+\\frac{y}{6}&amp;=3 \\\\[1mm] \\frac{x}{2}-\\frac{y}{4}&amp;=1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"359287\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"359287\"]\r\n\r\nFirst clear each equation of fractions by multiplying both sides of the equation by the least common denominator.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)&amp;=6\\left(3\\right) \\\\[1mm] 2x+y&amp;=18 \\\\[3mm] 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)&amp;=4\\left(1\\right) \\\\[1mm] 2x-y&amp;=4 \\end{align}[\/latex]<\/p>\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate\u00a0<em>x<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-1\\left(2x-y\\right)&amp;=-1\\left(4\\right) \\\\[1mm] -2x+y&amp;=-4 \\end{align}[\/latex]<\/p>\r\nAdd the two equations to eliminate\u00a0<em>x<\/em>\u00a0and solve the resulting equation for <em>y<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x+y&amp;=18 \\\\ \u22122x+y&amp;=\u22124 \\\\ \\hline 2y&amp;=14 \\\\ y&amp;=7 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]y=7[\/latex] into the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+\\left(7\\right)&amp;=18 \\\\ 2x&amp;=11 \\\\ x&amp;=\\frac{11}{2} \\\\ &amp;=7.5 \\end{align}[\/latex]<\/p>\r\nThe solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{2}-\\frac{y}{4}&amp;=1\\\\[1mm] \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}&amp;=1\\\\[1mm] \\frac{11}{4}-\\frac{7}{4}&amp;=1\\\\[1mm] \\frac{4}{4}&amp;=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations by addition.\r\n\r\n[latex]\\begin{align}2x+3y&amp;=8\\\\ 3x+5y&amp;=10\\end{align}[\/latex]\r\n\r\n[reveal-answer q=\"326265\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"326265\"]\r\n\r\n[latex]\\left(10,-4\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115110&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nin the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.\r\n\r\nhttps:\/\/youtu.be\/ova8GSmPV4o\r\n<h2>Classify Solutions to Systems<\/h2>\r\nNow that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different [latex]y[\/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Inconsistent System of Equations<\/h3>\r\nSolve the following system of equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}&amp;x=9 - 2y \\\\ &amp;x+2y=13 \\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"888134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"888134\"]\r\n\r\nWe can approach this problem in two ways. Because one equation is already solved for [latex]x[\/latex], the most obvious step is to use substitution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=13 \\\\ \\left(9 - 2y\\right)+2y&amp;=13 \\\\ 9+0y&amp;=13 \\\\ 9&amp;=13 \\end{align}[\/latex]<\/p>\r\nClearly, this statement is a contradiction because [latex]9\\ne 13[\/latex]. Therefore, the system has no solution.\r\n\r\nThe second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=9 - 2y \\\\ 2y=-x+9 \\\\ y=-\\frac{1}{2}x+\\frac{9}{2} \\end{gathered}[\/latex]<\/p>\r\nWe then convert the second equation expressed to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+2y=13 \\\\ 2y=-x+13 \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\r\nComparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=-\\frac{1}{2}x+\\frac{9}{2} \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWriting the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183613\/CNX_Precalc_Figure_09_01_0072.jpg\" alt=\"A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.\" width=\"488\" height=\"297\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2y - 2x=2\\\\ 2y - 2x=6\\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"681787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"681787\"]\r\n\r\nNo solution. It is an inconsistent system.\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29699&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Expressing the Solution of a System of Dependent Equations Containing Two Variables<\/h3>\r\nRecall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Solution to a Dependent System of Linear Equations<\/h3>\r\nFind a solution to the system of equations using the <strong>addition method<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"390828\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"390828\"]\r\n\r\nWith the addition method, we want to eliminate one of the variables by adding the equations. In this case, let\u2019s focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&amp;=2 \\\\ \\left(-3\\right)\\left(x+3y\\right)&amp;=\\left(-3\\right)\\left(2\\right) \\\\ -3x - 9y&amp;=-6 \\end{align}[\/latex]<\/p>\r\nNow add the equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \u22123x\u22129y&amp;=\u22126 \\\\ +3x+9y&amp;=6 \\\\ \\hline 0&amp;=0 \\end{align}[\/latex]<\/p>\r\nWe can see that there will be an infinite number of solutions that satisfy both equations.\r\n<h4>Analysis of the Solution<\/h4>\r\nIf we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let\u2019s look at what happens when we convert the system to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\begin{gathered}x+3y=2 \\\\ 3y=-x+2 \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered} \\hspace{2cm} \\begin{gathered} 3x+9y=6 \\\\9y=-3x+6 \\\\ y=-\\frac{3}{9}x+\\frac{6}{9} \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered}\\end{align}[\/latex]<\/p>\r\nLook at the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183615\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/>\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Writing the general solution<\/h3>\r\nIn the previous example, we presented an analysis of the solution to the following system of equations:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\r\nAfter a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as\u00a0[latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. \u00a0It tells us that <em>x<\/em> can be anything, <em>x<\/em> is <em>x<\/em>. \u00a0It also tells us that <em>y<\/em> is going to depend on <em>x<\/em>, just like when we write a function rule. \u00a0In this case, depending on what you put in for <em>x<\/em>, <em>y<\/em> will be defined in terms of <em>x<\/em> as [latex]-\\frac{1}{3}x+\\frac{2}{3}[\/latex].\r\n\r\nIn other words, there are infinitely many (<em>x<\/em>,<em>y<\/em>) pairs that will satisfy this system of equations, and they all fall on the line\u00a0[latex]f(x)-\\frac{1}{3}x+\\frac{2}{3}[\/latex].\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y - 2x=5 \\\\ -3y+6x=-15 \\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"218404\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"218404\"]\r\n\r\nThe system is dependent so there are infinitely many solutions of the form [latex]\\left(x,2x+5\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom17\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15665&amp;theme=oea&amp;iframe_resize_id=mom17\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h2><strong>Using Systems of Equations to Investigate Profits<\/strong><\/h2>\r\nUsing what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer\u2019s <strong>revenue function<\/strong> is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue function is shown in orange in the graph below.\r\n\r\nThe <strong>cost function<\/strong> is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The <em>x<\/em>-axis represents quantity in hundreds of units. The <em>y<\/em>-axis represents either cost or revenue in hundreds of dollars.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183617\/CNX_Precalc_Figure_09_01_0092.jpg\" alt=\"A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"488\" height=\"347\" \/>\r\n\r\nThe point at which the two lines intersect is called the <strong>break-even point<\/strong>. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.\r\n\r\nThe shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The <strong>profit function<\/strong> is the revenue function minus the cost function, written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Break-Even Point and the Profit Function Using Substitution<\/h3>\r\nGiven the cost function [latex]C\\left(x\\right)=0.85x+35{,}000[\/latex] and the revenue function [latex]R\\left(x\\right)=1.55x[\/latex], find the break-even point and the profit function.\r\n\r\n[reveal-answer q=\"569292\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"569292\"]\r\n\r\nWrite the system of equations using [latex]y[\/latex] to replace function notation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} y&amp;=0.85x+35{,}000 \\\\ y&amp;=1.55x \\end{align}[\/latex]<\/p>\r\nSubstitute the expression [latex]0.85x+35{,}000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0.85x+35{,}000=1.55x\\\\ 35{,}000=0.7x\\\\ 50{,}000=x\\end{gathered}[\/latex]<\/p>\r\nThen, we substitute [latex]x=50{,}000[\/latex] into either the cost function or the revenue function.\r\n<p style=\"text-align: center;\">[latex]1.55\\left(50{,}000\\right)=77{,}500[\/latex]<\/p>\r\nThe break-even point is [latex]\\left(50{,}000,77{,}500\\right)[\/latex].\r\n\r\nThe profit function is found using the formula [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}P\\left(x\\right)&amp;=1.55x-\\left(0.85x+35{,}000\\right) \\\\ &amp;=0.7x - 35{,}000 \\end{align}[\/latex]<\/p>\r\nThe profit function is [latex]P\\left(x\\right)=0.7x - 35{,}000[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183619\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"487\" height=\"390\" \/>\r\n\r\nWe see from the graph below that the profit function has a negative value until [latex]x=50{,}000[\/latex], when the graph crosses the <em>x<\/em>-axis. Then, the graph emerges into positive <em>y<\/em>-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183621\/CNX_Precalc_Figure_09_01_0112.jpg\" alt=\"A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.\" width=\"731\" height=\"507\" \/>\u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Writing a System of Linear Equations Given a Situation<\/h3>\r\nIt is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.<\/h3>\r\n<ol>\r\n \t<li>Identify the input and output of each linear model.<\/li>\r\n \t<li>Identify the slope and <em>y<\/em>-intercept of each linear model.<\/li>\r\n \t<li>Find the solution by setting the two linear functions equal to another and solving for <em>x<\/em>, or find the point of intersection on a graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\nNow let's practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing and Solving a System of Equations in Two Variables<\/h3>\r\nThe cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?\r\n\r\n[reveal-answer q=\"455809\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455809\"]\r\n\r\nLet <em>c<\/em> = the number of children and <em>a<\/em> = the number of adults in attendance.\r\n\r\nThe total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day.\r\n<p style=\"text-align: center;\">[latex]c+a=2{,}000[\/latex]<\/p>\r\nThe revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[\/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.\r\n<p style=\"text-align: center;\">[latex]25c+50a=70{,}000[\/latex]<\/p>\r\nWe now have a system of linear equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2,000\\\\ 25c+50a=70{,}000\\end{gathered}[\/latex]<\/p>\r\nIn the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2{,}000\\\\ a=2{,}000-c\\end{gathered}[\/latex]<\/p>\r\nSubstitute the expression [latex]2{,}000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 25c+50\\left(2{,}000-c\\right)&amp;=70{,}000 \\\\ 25c+100{,}000 - 50c&amp;=70{,}000 \\\\ -25c&amp;=-30{,}000 \\\\ c&amp;=1{,}200 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]c=1{,}200[\/latex] into the first equation to solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}1{,}200+a&amp;=2{,}000 \\\\ a&amp;=800 \\end{align}[\/latex]<\/p>\r\nWe find that 1,200 children and 800 adults bought tickets to the circus that day.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nMeal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?\r\n\r\n[reveal-answer q=\"454145\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"454145\"]\r\n\r\n700 children, 950 adults\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23774&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nSometimes, a system of equations can inform a decision. \u00a0In our next example, we help answer the question, \"Which truck rental company will give the best value?\"\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Building a System of Linear Models to Choose a Truck Rental Company<\/h3>\r\nJamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile.[footnote]Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/[\/footnote] When will Keep on Trucking, Inc. be the better choice for Jamal?\r\n\r\n[reveal-answer q=\"869152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"869152\"]\r\n\r\nThe two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.\r\n<table summary=\"Three rows and three columns. In the first column, are the years 1950 and 2000. In the second columns are the house values for Indiana, which are 37700 for 1950 and 94300 for 2000. In the third columns are the house values for Alabama, which are 27100 for 1950 and 85100 for 2000.\">\r\n<tbody>\r\n<tr>\r\n<td>Input<\/td>\r\n<td><em>d<\/em>, distance driven in miles<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Outputs<\/td>\r\n<td><em>K<\/em>(<em>d<\/em>): cost, in dollars, for renting from Keep on Trucking<em>M<\/em>(<em>d<\/em>) cost, in dollars, for renting from Move It Your Way<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Initial Value<\/td>\r\n<td>Up-front fee: <em>K<\/em>(0) = 20 and <em>M<\/em>(0) = 16<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rate of Change<\/td>\r\n<td><em>K<\/em>(<em>d<\/em>) = $0.59\/mile and <em>P<\/em>(<em>d<\/em>) = $0.63\/mile<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nA linear function is of the form [latex]f\\left(x\\right)=mx+b[\/latex]. Using the rates of change and initial charges, we can write the equations\r\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)=0.59d+20\\\\ M\\left(d\\right)=0.63d+16\\end{align}[\/latex]<\/p>\r\nUsing these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\\left(d\\right)&lt;M\\left(d\\right)[\/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\\left(d\\right)[\/latex] function is smaller.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19011929\/CNX_Precalc_Figure_02_03_0072.jpg\" width=\"731\" height=\"340\" \/>\r\n\r\nThese graphs are sketched above, with <em>K<\/em>(<em>d<\/em>)\u00a0in blue.\r\n\r\nTo find the intersection, we set the equations equal and solve:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)&amp;=M\\left(d\\right) \\\\ 0.59d+20&amp;=0.63d+16 \\\\ 4&amp;=0.04d \\\\ 100&amp;=d \\\\ d&amp;=100 \\end{align}[\/latex]<\/p>\r\nThis tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\\left(d\\right)[\/latex]\u00a0is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d&gt;100[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solve a Chemical Mixture Problem<\/h3>\r\nA chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?\r\n[reveal-answer q=\"350379\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350379\"]\r\n\r\nWe will use the following table to help us solve this mixture problem:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Amount<\/td>\r\n<td>Part<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Start<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Final<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe start with 70 mL of solution, and the unknown amount can be <em>x<\/em>. The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Amount<\/td>\r\n<td>Part<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Start<\/td>\r\n<td>70mL<\/td>\r\n<td>0.5<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add<\/td>\r\n<td>[latex]x[\/latex]<\/td>\r\n<td>0.8<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Final<\/td>\r\n<td>[latex]70+x[\/latex]<\/td>\r\n<td>0.6<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdd amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Amount<\/td>\r\n<td>Part<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Start<\/td>\r\n<td>70mL<\/td>\r\n<td>0.5<\/td>\r\n<td>35<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add<\/td>\r\n<td>[latex]x[\/latex]<\/td>\r\n<td>0.8<\/td>\r\n<td>[latex]0.8x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Final<\/td>\r\n<td>[latex]70+x[\/latex]<\/td>\r\n<td>0.6<\/td>\r\n<td>\u00a0[latex]42+0.6x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nMultiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[\/latex].\r\n\r\nIf we add the start and add entries in the Total column, we get the final equation that represents the total amount and it's concentration.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}35+0.8x&amp; = 42+0.6x \\\\ 0.2x&amp;=7 \\\\ \\frac{0.2}{0.2}x&amp;=\\frac{7}{0.2} \\\\ x&amp;=35 \\end{align}[\/latex]<\/p>\r\n35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.\r\n\r\nThe same process can be used if the starting and final amount have a price attached to them, rather than a percentage.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try IT<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8589&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n<iframe id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2239&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.<\/li>\r\n \t<li>The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.<\/li>\r\n \t<li>Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.<\/li>\r\n \t<li>One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes.<\/li>\r\n \t<li>Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation.<\/li>\r\n \t<li>A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables.<\/li>\r\n \t<li>It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together.<\/li>\r\n \t<li>Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect.<\/li>\r\n \t<li>The solution to a system of dependent equations will always be true because both equations describe the same line.<\/li>\r\n \t<li>Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<strong>addition method<\/strong> an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable\r\n\r\n<strong>break-even point<\/strong> the point at which a cost function intersects a revenue function; where profit is zero\r\n\r\n<strong>consistent system<\/strong> a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system\r\n\r\n<strong>cost function<\/strong> the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs\r\n\r\n<strong>dependent system<\/strong> a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system\r\n\r\n<strong>inconsistent system<\/strong> a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common\r\n\r\n<strong>independent system<\/strong> a system of linear equations with exactly one solution pair [latex]\\left(x,y\\right)[\/latex]\r\n\r\n<strong>profit function<\/strong> the profit function is written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex], revenue minus cost\r\n\r\n<strong>revenue function<\/strong> the function that is used to calculate revenue, simply written as [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price\r\n\r\n<strong>substitution method<\/strong> an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable\r\n\r\n<strong>system of linear equations<\/strong> a set of two or more equations in two or more variables that must be considered simultaneously.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n<li class=\"li2\"><span class=\"s1\">Solve systems of equations by graphing,\u00a0substitution, and\u00a0addition.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Identify inconsistent systems of equations containing two variables.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Express the solution of a system of dependent equations containing two variables using standard notations.<\/span><\/li>\n<\/ul>\n<\/div>\n<p>A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section we will consider linear equations with two variables to answer these and similar questions.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183558\/CNX_Precalc_Figure_09_01_0012.jpg\" alt=\"Skateboarders at a skating rink by the beach.\" width=\"487\" height=\"252\" \/><\/p>\n<p class=\"wp-caption-text\">(credit: Thomas S\u00f8renes)<\/p>\n<\/div>\n<h2>Introduction to Solutions of Systems<\/h2>\n<p>In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. 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. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.<\/p>\n<p>In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&=15\\\\[1mm] 3x-y&=5\\end{align}[\/latex]<\/p>\n<p>The <em>solution<\/em> to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair [latex](4,7)[\/latex] is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2\\left(4\\right)+\\left(7\\right)&=15 &&\\text{True} \\\\[1mm] 3\\left(4\\right)-\\left(7\\right)&=5 &&\\text{True} \\end{align}[\/latex]<\/p>\n<p>In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A <strong>consistent system<\/strong> of equations has at least one solution. A consistent system is considered to be an <strong>independent system<\/strong> if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a <strong>dependent system<\/strong> if the equations have the same slope and the same <em>y<\/em>-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.<\/p>\n<p>Another type of system of linear equations is an <strong>inconsistent system<\/strong>, which is one in which the equations represent two parallel lines. The lines have the same slope and different <em>y-<\/em>intercepts. There are no points common to both lines; hence, there is no solution to the system.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Types of Linear Systems<\/h3>\n<p>There are three types of systems of linear equations in two variables, and three types of solutions.<\/p>\n<ul>\n<li>An <strong>independent system<\/strong> has exactly one solution pair [latex]\\left(x,y\\right)[\/latex]. The point where the two lines intersect is the only solution.<\/li>\n<li>An <strong>inconsistent system<\/strong> has no solution. Notice that the two lines are parallel and will never intersect.<\/li>\n<li>A <strong>dependent system<\/strong> has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.<\/li>\n<\/ul>\n<p>Below is a comparison of\u00a0graphical representations of each type of system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183601\/CNX_Precalc_Figure_09_01_002n2.jpg\" alt=\"Graphs of an independent system, an inconsistent system, and a dependent system. The independent system has two lines which cross at the point seven-fifths, negative eleven fifths. The inconsistent system shows two parallel lines. The dependent system shows a single line running through the points negative one, negative two and one, two.\" width=\"945\" height=\"479\" \/><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.<\/h3>\n<ol>\n<li>Substitute the ordered pair into each equation in the system.<\/li>\n<li>Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether an Ordered Pair Is a Solution to a System of Equations<\/h3>\n<p>Determine whether the ordered pair [latex]\\left(5,1\\right)[\/latex] is a solution to the given system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&=8\\\\ 2x-9&=y \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q899056\">Show Solution<\/span><\/p>\n<div id=\"q899056\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the ordered pair [latex]\\left(5,1\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(5\\right)+3\\left(1\\right)&=8 \\\\[1mm] 8&=8 &&\\text{True} \\\\[3mm] 2\\left(5\\right)-9&=\\left(1\\right) \\\\[1mm] 1&=1 &&\\text{True} \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The ordered pair [latex]\\left(5,1\\right)[\/latex] satisfies both equations, so it is the solution to the system.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183603\/CNX_Precalc_Figure_09_01_0032.jpg\" alt=\"A graph of two lines running through the point five, one. The first line's equation is x plus 3y equals 8. The second line's equation is 2x minus 9 equals y.\" width=\"487\" height=\"365\" \/>\u00a0<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Determine whether the ordered pair [latex]\\left(8,5\\right)[\/latex] is a solution to the following system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}5x-4y&=20\\\\ 2x+1&=3y\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q672974\">Show Solution<\/span><\/p>\n<div id=\"q672974\" class=\"hidden-answer\" style=\"display: none\">\n<p>Not a solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Solving Systems of Equations by Graphing<\/h3>\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<div class=\"textbox exercises\">\n<h3>Example: Solving a System of Equations in Two Variables by Graphing<\/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{align}2x+y&=-8\\\\ x-y&=-1\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8915\">Show Solution<\/span><\/p>\n<div id=\"q8915\" 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{align}2x+y&=-8\\\\ y&=-2x-8\\end{align}[\/latex]<\/p>\n<p>Solve the second equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x-y&=-1\\\\ y&=x+1\\end{align}[\/latex]<\/p>\n<p>Graph both equations on the same set of axes:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183605\/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{align}2\\left(-3\\right)+\\left(-2\\right)&=-8 \\\\[1mm] -8=-8 &&\\text{True} \\\\[3mm] \\left(-3\\right)-\\left(-2\\right)&=-1 \\\\[1mm] -1&=-1 &&\\text{True} \\end{align}[\/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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations by graphing.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2x - 5y=-25 \\\\ -4x+5y=35 \\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q141689\">Show Solution<\/span><\/p>\n<div id=\"q141689\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solution to the system is the ordered pair [latex]\\left(-5,3\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3165\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/18222602\/CNX_Precalc_Figure_09_01_0132.jpg\" alt=\"Two lines that cross at the point negative five, three. One line's equation is y equals four-fifths x plus 7. The other line's equation is y equals two-fifths x plus 5.\" width=\"487\" height=\"409\" \/><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q&amp; A<\/h3>\n<h4>Can graphing be used if the system is inconsistent or dependent?<\/h4>\n<p><em>Yes, in 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.<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Plot the three different systems with an online graphing tool. Categorize\u00a0each solution as either consistent or inconsistent. If the system is consistent determine whether it is dependent or independent. You may find it easier to plot each system individually, then clear out your entries before you plot the next.<br \/>\n1)<br \/>\n[latex]5x-3y = -19[\/latex]<br \/>\n[latex]x=2y-1[\/latex]<\/p>\n<p>2)<br \/>\n[latex]4x+y=11[\/latex]<br \/>\n[latex]-2y=-25+8x[\/latex]<\/p>\n<p>3)<br \/>\n[latex]y = -3x+6[\/latex]<br \/>\n[latex]-\\frac{1}{3}y+2=x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q720375\">Show Solution<\/span><\/p>\n<div id=\"q720375\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>One solution &#8211; consistent, independent<\/li>\n<li>No solutions, inconsistent, neither dependent nor independent<\/li>\n<li>Many solutions &#8211; \u00a0consistent, dependent<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving Systems of Equations by Substitution<\/h2>\n<p>Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a <strong>system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the <strong>substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\n<ol>\n<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\n<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\n<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\n<li>Check the solution in both equations.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a System of Equations in Two Variables by Substitution<\/h3>\n<p>Solve the following system of equations by substitution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ 2x-5y&=1 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786744\">Show Solution<\/span><\/p>\n<div id=\"q786744\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will solve the first equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ y&=x - 5 \\end{align}[\/latex]<\/p>\n<p>Now we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 5y&=1 \\\\ 2x - 5\\left(x - 5\\right)&=1 \\\\ 2x - 5x+25&=1 \\\\ -3x&=-24 \\\\ x&=8 \\end{align}[\/latex]<\/p>\n<p>Now, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-\\left(8\\right)+y&=-5 \\\\ y&=3 \\end{align}[\/latex]<\/p>\n<p>Our solution is [latex]\\left(8,3\\right)[\/latex].<\/p>\n<p>Check the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ -\\left(8\\right)+\\left(3\\right)&=-5 && \\text{True} \\\\[3mm] 2x - 5y&=1 \\\\ 2\\left(8\\right)-5\\left(3\\right)&=1 && \\text{True} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations by substitution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x&=y+3 \\\\ 4&=3x - 2y \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q783260\">Show Solution<\/span><\/p>\n<div id=\"q783260\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-2,-5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115164&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Can the substitution method be used to solve any linear system in two variables?<\/h4>\n<p><em>Yes, but the method works best if one of the equations contains a coefficient of 1 or \u20131 so that we do not have to deal with fractions.<\/em><\/p>\n<\/div>\n<p>The following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations. \u00a0We present three different examples, and also use a graphing tool to help summarize the solution for each example.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solving Systems of Equations using Substitution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/HxhacvH49o8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Solving Systems of Equations in Two Variables by the Addition Method<\/h2>\n<p>A third method of <strong>solving systems of linear equations<\/strong> is the <strong>addition method,\u00a0<\/strong>this method is also called the\u00a0<strong>elimination method<\/strong>. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\n<ol>\n<li>Write both equations with <em>x<\/em>&#8211; and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\n<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\n<li>Solve the resulting equation for the remaining variable.<\/li>\n<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\n<li>Check the solution by substituting the values into the other equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a System by the Addition Method<\/h3>\n<p>Solve the given system of equations by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ -x+y&=3 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924657\">Show Solution<\/span><\/p>\n<div id=\"q924657\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} x+2y&=-1 \\\\ -x+y&=3 \\\\ \\hline 3y&=2\\end{align}[\/latex]<\/p>\n<p>Now that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3y&=2 \\\\ y&=\\dfrac{2}{3} \\end{align}[\/latex]<\/p>\n<p>Then, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=3 \\\\ -x+\\frac{2}{3}&=3 \\\\ -x&=3-\\frac{2}{3} \\\\ -x&=\\frac{7}{3} \\\\ x&=-\\frac{7}{3} \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ \\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)&= \\\\ -\\frac{7}{3}+\\frac{4}{3}&= \\\\ \\-\\frac{3}{3}&= \\\\ -1&=-1&& \\text{True} \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method When Multiplication of One Equation Is Required<\/h3>\n<p>Solve the given system of equations by the <strong>addition method<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11 \\\\ x - 2y&=11 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883001\">Show Solution<\/span><\/p>\n<div id=\"q883001\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11 \\\\ -3\\left(x - 2y\\right)&=-3\\left(11\\right) && \\text{Multiply both sides by }-3 \\\\ -3x+6y&=-33 && \\text{Use the distributive property}. \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now, let\u2019s add them.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=\u221211 \\\\ \u22123x+6y&=\u221233 \\\\ \\hline 11y&=\u221244 \\\\ y&=\u22124 \\end{align}[\/latex]<\/p>\n<p>For the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11\\\\ 3x+5\\left(-4\\right)&=-11\\\\ 3x - 20&=-11\\\\ 3x&=9\\\\ x&=3\\end{align}[\/latex]<\/p>\n<p>Our solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11 \\\\ \\left(3\\right)-2\\left(-4\\right)&=3+8 \\\\ &=11 && \\text{True} \\end{align}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 7y&=2\\\\ 3x+y&=-20\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q609174\">Show Solution<\/span><\/p>\n<div id=\"q609174\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-6,-2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115120&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method When Multiplication of Both Equations Is Required<\/h3>\n<p>Solve the given system of equations in two variables by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3y&=-16 \\\\ 5x - 10y&=30\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114755\">Show Solution<\/span><\/p>\n<div id=\"q114755\" class=\"hidden-answer\" style=\"display: none\">\n<p>One equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} -5\\left(2x+3y\\right)&=-5\\left(-16\\right) \\\\ -10x - 15y&=80 \\\\[3mm] 2\\left(5x - 10y\\right)&=2\\left(30\\right) \\\\ 10x - 20y&=60 \\end{align}[\/latex]<\/p>\n<p>Then, we add the two equations together.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \u221210x\u221215y&=80 \\\\ 10x\u221220y&=60 \\\\ \\hline \u221235y&=140 \\\\ y&=\u22124 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]y=-4[\/latex] into the original first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3\\left(-4\\right)&=-16\\\\ 2x - 12&=-16\\\\ 2x&=-4\\\\ x&=-2\\end{align}[\/latex]<\/p>\n<p>The solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 5x - 10y&=30\\\\ 5\\left(-2\\right)-10\\left(-4\\right)&=30\\\\ -10+40&=30\\\\ 30&=30\\end{align}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183611\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method in Systems of Equations Containing Fractions<\/h3>\n<p>Solve the given system of equations in two variables by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{3}+\\frac{y}{6}&=3 \\\\[1mm] \\frac{x}{2}-\\frac{y}{4}&=1 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q359287\">Show Solution<\/span><\/p>\n<div id=\"q359287\" class=\"hidden-answer\" style=\"display: none\">\n<p>First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)&=6\\left(3\\right) \\\\[1mm] 2x+y&=18 \\\\[3mm] 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)&=4\\left(1\\right) \\\\[1mm] 2x-y&=4 \\end{align}[\/latex]<\/p>\n<p>Now multiply the second equation by [latex]-1[\/latex] so that we can eliminate\u00a0<em>x<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-1\\left(2x-y\\right)&=-1\\left(4\\right) \\\\[1mm] -2x+y&=-4 \\end{align}[\/latex]<\/p>\n<p>Add the two equations to eliminate\u00a0<em>x<\/em>\u00a0and solve the resulting equation for <em>y<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x+y&=18 \\\\ \u22122x+y&=\u22124 \\\\ \\hline 2y&=14 \\\\ y&=7 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]y=7[\/latex] into the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+\\left(7\\right)&=18 \\\\ 2x&=11 \\\\ x&=\\frac{11}{2} \\\\ &=7.5 \\end{align}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{2}-\\frac{y}{4}&=1\\\\[1mm] \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}&=1\\\\[1mm] \\frac{11}{4}-\\frac{7}{4}&=1\\\\[1mm] \\frac{4}{4}&=1\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations by addition.<\/p>\n<p>[latex]\\begin{align}2x+3y&=8\\\\ 3x+5y&=10\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q326265\">Show Solution<\/span><\/p>\n<div id=\"q326265\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(10,-4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115110&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>in the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Solving Systems of Equations using Elimination\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ova8GSmPV4o?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Classify Solutions to Systems<\/h2>\n<p>Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different [latex]y[\/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Inconsistent System of Equations<\/h3>\n<p>Solve the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}&x=9 - 2y \\\\ &x+2y=13 \\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888134\">Show Solution<\/span><\/p>\n<div id=\"q888134\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can approach this problem in two ways. Because one equation is already solved for [latex]x[\/latex], the most obvious step is to use substitution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=13 \\\\ \\left(9 - 2y\\right)+2y&=13 \\\\ 9+0y&=13 \\\\ 9&=13 \\end{align}[\/latex]<\/p>\n<p>Clearly, this statement is a contradiction because [latex]9\\ne 13[\/latex]. Therefore, the system has no solution.<\/p>\n<p>The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=9 - 2y \\\\ 2y=-x+9 \\\\ y=-\\frac{1}{2}x+\\frac{9}{2} \\end{gathered}[\/latex]<\/p>\n<p>We then convert the second equation expressed to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+2y=13 \\\\ 2y=-x+13 \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\n<p>Comparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=-\\frac{1}{2}x+\\frac{9}{2} \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown 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\/896\/2016\/11\/03183613\/CNX_Precalc_Figure_09_01_0072.jpg\" alt=\"A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.\" width=\"488\" height=\"297\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2y - 2x=2\\\\ 2y - 2x=6\\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q681787\">Show Solution<\/span><\/p>\n<div id=\"q681787\" class=\"hidden-answer\" style=\"display: none\">\n<p>No solution. It is an inconsistent system.<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29699&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h3>Expressing the Solution of a System of Dependent Equations Containing Two Variables<\/h3>\n<p>Recall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Solution to a Dependent System of Linear Equations<\/h3>\n<p>Find a solution to the system of equations using the <strong>addition method<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390828\">Show Solution<\/span><\/p>\n<div id=\"q390828\" class=\"hidden-answer\" style=\"display: none\">\n<p>With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let\u2019s focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&=2 \\\\ \\left(-3\\right)\\left(x+3y\\right)&=\\left(-3\\right)\\left(2\\right) \\\\ -3x - 9y&=-6 \\end{align}[\/latex]<\/p>\n<p>Now add the equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \u22123x\u22129y&=\u22126 \\\\ +3x+9y&=6 \\\\ \\hline 0&=0 \\end{align}[\/latex]<\/p>\n<p>We can see that there will be an infinite number of solutions that satisfy both equations.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let\u2019s look at what happens when we convert the system to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\begin{gathered}x+3y=2 \\\\ 3y=-x+2 \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered} \\hspace{2cm} \\begin{gathered} 3x+9y=6 \\\\9y=-3x+6 \\\\ y=-\\frac{3}{9}x+\\frac{6}{9} \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered}\\end{align}[\/latex]<\/p>\n<p>Look at the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183615\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Writing the general solution<\/h3>\n<p>In the previous example, we presented an analysis of the solution to the following system of equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\n<p>After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as\u00a0[latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. \u00a0It tells us that <em>x<\/em> can be anything, <em>x<\/em> is <em>x<\/em>. \u00a0It also tells us that <em>y<\/em> is going to depend on <em>x<\/em>, just like when we write a function rule. \u00a0In this case, depending on what you put in for <em>x<\/em>, <em>y<\/em> will be defined in terms of <em>x<\/em> as [latex]-\\frac{1}{3}x+\\frac{2}{3}[\/latex].<\/p>\n<p>In other words, there are infinitely many (<em>x<\/em>,<em>y<\/em>) pairs that will satisfy this system of equations, and they all fall on the line\u00a0[latex]f(x)-\\frac{1}{3}x+\\frac{2}{3}[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y - 2x=5 \\\\ -3y+6x=-15 \\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q218404\">Show Solution<\/span><\/p>\n<div id=\"q218404\" class=\"hidden-answer\" style=\"display: none\">\n<p>The system is dependent so there are infinitely many solutions of the form [latex]\\left(x,2x+5\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom17\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15665&amp;theme=oea&amp;iframe_resize_id=mom17\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h2><strong>Using Systems of Equations to Investigate Profits<\/strong><\/h2>\n<p>Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer\u2019s <strong>revenue function<\/strong> is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue function is shown in orange in the graph below.<\/p>\n<p>The <strong>cost function<\/strong> is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The <em>x<\/em>-axis represents quantity in hundreds of units. The <em>y<\/em>-axis represents either cost or revenue in hundreds of dollars.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183617\/CNX_Precalc_Figure_09_01_0092.jpg\" alt=\"A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"488\" height=\"347\" \/><\/p>\n<p>The point at which the two lines intersect is called the <strong>break-even point<\/strong>. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.<\/p>\n<p>The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The <strong>profit function<\/strong> is the revenue function minus the cost function, written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Break-Even Point and the Profit Function Using Substitution<\/h3>\n<p>Given the cost function [latex]C\\left(x\\right)=0.85x+35{,}000[\/latex] and the revenue function [latex]R\\left(x\\right)=1.55x[\/latex], find the break-even point and the profit function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q569292\">Show Solution<\/span><\/p>\n<div id=\"q569292\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the system of equations using [latex]y[\/latex] to replace function notation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} y&=0.85x+35{,}000 \\\\ y&=1.55x \\end{align}[\/latex]<\/p>\n<p>Substitute the expression [latex]0.85x+35{,}000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0.85x+35{,}000=1.55x\\\\ 35{,}000=0.7x\\\\ 50{,}000=x\\end{gathered}[\/latex]<\/p>\n<p>Then, we substitute [latex]x=50{,}000[\/latex] into either the cost function or the revenue function.<\/p>\n<p style=\"text-align: center;\">[latex]1.55\\left(50{,}000\\right)=77{,}500[\/latex]<\/p>\n<p>The break-even point is [latex]\\left(50{,}000,77{,}500\\right)[\/latex].<\/p>\n<p>The profit function is found using the formula [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}P\\left(x\\right)&=1.55x-\\left(0.85x+35{,}000\\right) \\\\ &=0.7x - 35{,}000 \\end{align}[\/latex]<\/p>\n<p>The profit function is [latex]P\\left(x\\right)=0.7x - 35{,}000[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183619\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"487\" height=\"390\" \/><\/p>\n<p>We see from the graph below that the profit function has a negative value until [latex]x=50{,}000[\/latex], when the graph crosses the <em>x<\/em>-axis. Then, the graph emerges into positive <em>y<\/em>-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183621\/CNX_Precalc_Figure_09_01_0112.jpg\" alt=\"A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.\" width=\"731\" height=\"507\" \/>\u00a0<\/div>\n<\/div>\n<\/div>\n<h3>Writing a System of Linear Equations Given a Situation<\/h3>\n<p>It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.<\/h3>\n<ol>\n<li>Identify the input and output of each linear model.<\/li>\n<li>Identify the slope and <em>y<\/em>-intercept of each linear model.<\/li>\n<li>Find the solution by setting the two linear functions equal to another and solving for <em>x<\/em>, or find the point of intersection on a graph.<\/li>\n<\/ol>\n<\/div>\n<p>Now let&#8217;s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing and Solving a System of Equations in Two Variables<\/h3>\n<p>The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455809\">Show Solution<\/span><\/p>\n<div id=\"q455809\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em>c<\/em> = the number of children and <em>a<\/em> = the number of adults in attendance.<\/p>\n<p>The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day.<\/p>\n<p style=\"text-align: center;\">[latex]c+a=2{,}000[\/latex]<\/p>\n<p>The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[\/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.<\/p>\n<p style=\"text-align: center;\">[latex]25c+50a=70{,}000[\/latex]<\/p>\n<p>We now have a system of linear equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2,000\\\\ 25c+50a=70{,}000\\end{gathered}[\/latex]<\/p>\n<p>In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2{,}000\\\\ a=2{,}000-c\\end{gathered}[\/latex]<\/p>\n<p>Substitute the expression [latex]2{,}000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 25c+50\\left(2{,}000-c\\right)&=70{,}000 \\\\ 25c+100{,}000 - 50c&=70{,}000 \\\\ -25c&=-30{,}000 \\\\ c&=1{,}200 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]c=1{,}200[\/latex] into the first equation to solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}1{,}200+a&=2{,}000 \\\\ a&=800 \\end{align}[\/latex]<\/p>\n<p>We find that 1,200 children and 800 adults bought tickets to the circus that day.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q454145\">Show Solution<\/span><\/p>\n<div id=\"q454145\" class=\"hidden-answer\" style=\"display: none\">\n<p>700 children, 950 adults<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23774&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>Sometimes, a system of equations can inform a decision. \u00a0In our next example, we help answer the question, &#8220;Which truck rental company will give the best value?&#8221;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Building a System of Linear Models to Choose a Truck Rental Company<\/h3>\n<p>Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile.<a class=\"footnote\" title=\"Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/\" id=\"return-footnote-2174-1\" href=\"#footnote-2174-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> When will Keep on Trucking, Inc. be the better choice for Jamal?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q869152\">Show Solution<\/span><\/p>\n<div id=\"q869152\" class=\"hidden-answer\" style=\"display: none\">\n<p>The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.<\/p>\n<table summary=\"Three rows and three columns. In the first column, are the years 1950 and 2000. In the second columns are the house values for Indiana, which are 37700 for 1950 and 94300 for 2000. In the third columns are the house values for Alabama, which are 27100 for 1950 and 85100 for 2000.\">\n<tbody>\n<tr>\n<td>Input<\/td>\n<td><em>d<\/em>, distance driven in miles<\/td>\n<\/tr>\n<tr>\n<td>Outputs<\/td>\n<td><em>K<\/em>(<em>d<\/em>): cost, in dollars, for renting from Keep on Trucking<em>M<\/em>(<em>d<\/em>) cost, in dollars, for renting from Move It Your Way<\/td>\n<\/tr>\n<tr>\n<td>Initial Value<\/td>\n<td>Up-front fee: <em>K<\/em>(0) = 20 and <em>M<\/em>(0) = 16<\/td>\n<\/tr>\n<tr>\n<td>Rate of Change<\/td>\n<td><em>K<\/em>(<em>d<\/em>) = $0.59\/mile and <em>P<\/em>(<em>d<\/em>) = $0.63\/mile<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>A linear function is of the form [latex]f\\left(x\\right)=mx+b[\/latex]. Using the rates of change and initial charges, we can write the equations<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)=0.59d+20\\\\ M\\left(d\\right)=0.63d+16\\end{align}[\/latex]<\/p>\n<p>Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\\left(d\\right)<M\\left(d\\right)[\/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\\left(d\\right)[\/latex] function is smaller.\n\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19011929\/CNX_Precalc_Figure_02_03_0072.jpg\" width=\"731\" height=\"340\" alt=\"image\" \/><\/p>\n<p>These graphs are sketched above, with <em>K<\/em>(<em>d<\/em>)\u00a0in blue.<\/p>\n<p>To find the intersection, we set the equations equal and solve:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)&=M\\left(d\\right) \\\\ 0.59d+20&=0.63d+16 \\\\ 4&=0.04d \\\\ 100&=d \\\\ d&=100 \\end{align}[\/latex]<\/p>\n<p>This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\\left(d\\right)[\/latex]\u00a0is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solve a Chemical Mixture Problem<\/h3>\n<p>A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q350379\">Show Solution<\/span><\/p>\n<div id=\"q350379\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will use the following table to help us solve this mixture problem:<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Amount<\/td>\n<td>Part<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Start<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Add<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Final<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We start with 70 mL of solution, and the unknown amount can be <em>x<\/em>. The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Amount<\/td>\n<td>Part<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Start<\/td>\n<td>70mL<\/td>\n<td>0.5<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Add<\/td>\n<td>[latex]x[\/latex]<\/td>\n<td>0.8<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Final<\/td>\n<td>[latex]70+x[\/latex]<\/td>\n<td>0.6<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Amount<\/td>\n<td>Part<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Start<\/td>\n<td>70mL<\/td>\n<td>0.5<\/td>\n<td>35<\/td>\n<\/tr>\n<tr>\n<td>Add<\/td>\n<td>[latex]x[\/latex]<\/td>\n<td>0.8<\/td>\n<td>[latex]0.8x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Final<\/td>\n<td>[latex]70+x[\/latex]<\/td>\n<td>0.6<\/td>\n<td>\u00a0[latex]42+0.6x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Multiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[\/latex].<\/p>\n<p>If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it&#8217;s concentration.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}35+0.8x& = 42+0.6x \\\\ 0.2x&=7 \\\\ \\frac{0.2}{0.2}x&=\\frac{7}{0.2} \\\\ x&=35 \\end{align}[\/latex]<\/p>\n<p>35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.<\/p>\n<p>The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8589&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2239&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.<\/li>\n<li>The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.<\/li>\n<li>Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.<\/li>\n<li>One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes.<\/li>\n<li>Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation.<\/li>\n<li>A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables.<\/li>\n<li>It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together.<\/li>\n<li>Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect.<\/li>\n<li>The solution to a system of dependent equations will always be true because both equations describe the same line.<\/li>\n<li>Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<p><strong>addition method<\/strong> an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable<\/p>\n<p><strong>break-even point<\/strong> the point at which a cost function intersects a revenue function; where profit is zero<\/p>\n<p><strong>consistent system<\/strong> a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system<\/p>\n<p><strong>cost function<\/strong> the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs<\/p>\n<p><strong>dependent system<\/strong> a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system<\/p>\n<p><strong>inconsistent system<\/strong> a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common<\/p>\n<p><strong>independent system<\/strong> a system of linear equations with exactly one solution pair [latex]\\left(x,y\\right)[\/latex]<\/p>\n<p><strong>profit function<\/strong> the profit function is written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex], revenue minus cost<\/p>\n<p><strong>revenue function<\/strong> the function that is used to calculate revenue, simply written as [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price<\/p>\n<p><strong>substitution method<\/strong> an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable<\/p>\n<p><strong>system of linear equations<\/strong> a set of two or more equations in two or more variables that must be considered simultaneously.<\/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-2174\">\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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Solving Systems of Equations using Elimination. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ova8GSmPV4o\">https:\/\/youtu.be\/ova8GSmPV4o<\/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>Question ID 115164, 115120, 115110. <strong>Authored by<\/strong>: Shabazian, Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Beginning and Intermediate Algebra. <strong>Authored by<\/strong>: Wallace, Tyler. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\">http:\/\/www.wallace.ccfaculty.org\/book\/book.html<\/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>Question ID 29699. <strong>Authored by<\/strong>: McClure, Caren. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 23774. <strong>Authored by<\/strong>: Roy Shahbazian. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 8589. <strong>Authored by<\/strong>: Greg Harbaugh. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 2239. <strong>Authored by<\/strong>: Morales, Lawrence. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/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><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-2174-1\">Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/ <a href=\"#return-footnote-2174-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Solving Systems of Equations using Elimination\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ova8GSmPV4o\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 115164, 115120, 115110\",\"author\":\"Shabazian, Roy\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Beginning and Intermediate Algebra\",\"author\":\"Wallace, Tyler\",\"organization\":\"\",\"url\":\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 29699\",\"author\":\"McClure, Caren\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 23774\",\"author\":\"Roy Shahbazian\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 8589\",\"author\":\"Greg Harbaugh\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 2239\",\"author\":\"Morales, Lawrence\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + 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