{"id":791,"date":"2025-06-20T17:13:05","date_gmt":"2025-06-20T17:13:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=791"},"modified":"2025-10-31T16:38:00","modified_gmt":"2025-10-31T16:38:00","slug":"introduction-to-differential-equations-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/introduction-to-differential-equations-background-youll-need-2\/","title":{"raw":"Introduction to Differential Equations: Background You'll Need 2","rendered":"Introduction to Differential Equations: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use graphing, substitution, and addition methods to solve systems of equations with two variables.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Systems of Equations by Graphing<\/h2>\r\nThere are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"os-subtitle\" data-type=\"title\"><strong><span class=\"os-subtitle-label\">How to: Solve a system of linear equations by graphing<\/span><\/strong><\/p>\r\n\r\n<ol id=\"eip-idm264452624\" class=\"os-stepwise\" type=\"1\">\r\n \t<li><span class=\"os-stepwise-content\">Graph the first equation.<\/span><\/li>\r\n \t<li><span class=\"os-stepwise-content\">Graph the second equation on the same rectangular coordinate system.<\/span><\/li>\r\n \t<li><span class=\"os-stepwise-content\">Determine whether the lines intersect, are parallel, or are the same line.<\/span><\/li>\r\n \t<li><span class=\"os-stepwise-content\">Identify the solution to the system.<\/span><\/li>\r\n \t<li><span class=\"os-stepwise-content\">Check the solution in both equations.<\/span><\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Making a quick sketch of any mathematical situation is often a good idea to help you visualize it. Recall the techniques for graphing linear equations include using the y-intercept and slope to plot two points as well as using the intercepts. With practice, you'll get a feel for which technique to use in a given situation.<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve 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\r\n\r\n[caption id=\"attachment_2320\" align=\"alignright\" width=\"400\"]<img class=\"wp-image-2320\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/07\/22211620\/Screenshot-2024-07-22-at-2.16.16%E2%80%AFPM.png\" alt=\"Graph of two lines, y = -2x - 8 and y = x + 1, intersecting at (-3, -2).\" width=\"400\" height=\"306\" \/> Graph of two lines intersecting at the point (-3,-2)[\/caption]\r\n\r\nTo find the solution, we want to graph both equations on the same set of axes:\r\n<ul>\r\n \t<li>Solve the first equation for [latex]y[\/latex].<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&amp;=-8\\\\ y&amp;=-2x-8\\end{align}[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>Solve the second equation for [latex]y[\/latex].<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x-y&amp;=-1\\\\ y&amp;=x+1\\end{align}[\/latex]<\/p>\r\nThe lines appear to intersect at the point [latex]\\left(-3,-2\\right)[\/latex].\r\n\r\nYou can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.\r\n\r\n<center>[latex]\\begin{align*} 2(-3) + (-2) &amp;= -8 &amp; \\text{} \\\\ -8 &amp;= -8 &amp; \\text{True} \\\\ (-3) - (-2) &amp;= -1 &amp; \\text{} \\\\ -1 &amp;= -1 &amp; \\text{True} \\end{align*}[\/latex]<\/center>The solution to the system is the ordered pair [latex]\\left(-3,-2\\right)[\/latex], so the<span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">\u00a0system is independent.<\/span>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24797[\/ohm2_question]<\/section><section aria-label=\"Try It\">\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 system of linear equations that are more precise than graphing.\r\n\r\nOne 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\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a system of two equations in two variables, solve using the substitution method.<\/strong>\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<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve 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<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve 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\nBecause one equation is already solved for [latex]x[\/latex], we can substitute\u00a0 [latex]x=9 - 2y[\/latex] into the second equation:\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 <strong>no solution<\/strong>.\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"301\"]<img 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=\"301\" height=\"183\" \/> Graph demonstrating an inconsistent system[\/caption]\r\n\r\n<strong>Analysis<\/strong>\r\n\r\nLet's graph the equations to confirm that the system has no solution. Writing the equations in slope-intercept form confirms that the system is <strong>inconsistent<\/strong> because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24799[\/ohm2_question]<\/section><section aria-label=\"Try It\">\r\n<h2>Solving Systems of Equations by the Addition Method<\/h2>\r\nA third method of solving systems of linear equations is the <strong>addition method<\/strong>, this method is also called the <strong>elimination method<\/strong>. \u00a0In 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\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a system of equations, solve using the addition method.<\/strong>\r\n<ol>\r\n \t<li>Write both equations with [latex]x[\/latex]- and [latex]y[\/latex]-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.\r\n<ul>\r\n \t<li>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.<\/li>\r\n \t<li>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<\/ul>\r\n<\/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<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve 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\n<strong>Analysis<\/strong>\r\n\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\r\n[caption id=\"\" align=\"alignright\" width=\"350\"]<img 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=\"350\" height=\"209\" \/> A graph of two lines that intersect at the point negative seven-thirds, two-thirds[\/caption]\r\n\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. See the graph to 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[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the given system of equations by the addition method.\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].\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"350\"]<img 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=\"350\" height=\"235\" \/> A graph of two lines that cross at the point 3, negative 4[\/caption]\r\n\r\n<strong>Analysis<\/strong>\r\n\r\nWe can check the solution in the original second equation and also by graphing:\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[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24801[\/ohm2_question]<\/section><\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use graphing, substitution, and addition methods to solve systems of equations with two variables.<\/li>\n<\/ul>\n<\/section>\n<h2>Solving Systems of Equations by Graphing<\/h2>\n<p>There are multiple methods of solving systems of linear equations. For a system of linear equations 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<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"os-subtitle\" data-type=\"title\"><strong><span class=\"os-subtitle-label\">How to: Solve a system of linear equations by graphing<\/span><\/strong><\/p>\n<ol id=\"eip-idm264452624\" class=\"os-stepwise\" type=\"1\">\n<li><span class=\"os-stepwise-content\">Graph the first equation.<\/span><\/li>\n<li><span class=\"os-stepwise-content\">Graph the second equation on the same rectangular coordinate system.<\/span><\/li>\n<li><span class=\"os-stepwise-content\">Determine whether the lines intersect, are parallel, or are the same line.<\/span><\/li>\n<li><span class=\"os-stepwise-content\">Identify the solution to the system.<\/span><\/li>\n<li><span class=\"os-stepwise-content\">Check the solution in both equations.<\/span><\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Making a quick sketch of any mathematical situation is often a good idea to help you visualize it. Recall the techniques for graphing linear equations include using the y-intercept and slope to plot two points as well as using the intercepts. With practice, you&#8217;ll get a feel for which technique to use in a given situation.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<figure id=\"attachment_2320\" aria-describedby=\"caption-attachment-2320\" style=\"width: 400px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2320\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/07\/22211620\/Screenshot-2024-07-22-at-2.16.16%E2%80%AFPM.png\" alt=\"Graph of two lines, y = -2x - 8 and y = x + 1, intersecting at (-3, -2).\" width=\"400\" height=\"306\" \/><figcaption id=\"caption-attachment-2320\" class=\"wp-caption-text\">Graph of two lines intersecting at the point (-3,-2)<\/figcaption><\/figure>\n<p>To find the solution, we want to graph both equations on the same set of axes:<\/p>\n<ul>\n<li>Solve the first equation for [latex]y[\/latex].<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&=-8\\\\ y&=-2x-8\\end{align}[\/latex]<\/p>\n<ul>\n<li>Solve the second equation for [latex]y[\/latex].<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]\\begin{align}x-y&=-1\\\\ y&=x+1\\end{align}[\/latex]<\/p>\n<p>The lines appear to intersect at the point [latex]\\left(-3,-2\\right)[\/latex].<\/p>\n<p>You can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} 2(-3) + (-2) &= -8 & \\text{} \\\\ -8 &= -8 & \\text{True} \\\\ (-3) - (-2) &= -1 & \\text{} \\\\ -1 &= -1 & \\text{True} \\end{align*}[\/latex]<\/div>\n<p>The solution to the system is the ordered pair [latex]\\left(-3,-2\\right)[\/latex], so the<span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">\u00a0system is independent.<\/span><\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24797\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24797&theme=lumen&iframe_resize_id=ohm24797&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\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 system of linear equations that are more precise than graphing.<\/p>\n<p>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<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a system of two equations in two variables, solve using the substitution method.<\/strong><\/p>\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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q786744\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q888134\">Show Solution<\/button><\/p>\n<div id=\"q888134\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because one equation is already solved for [latex]x[\/latex], we can substitute\u00a0 [latex]x=9 - 2y[\/latex] into the second equation:<\/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 <strong>no solution<\/strong>.<\/p>\n<figure style=\"width: 301px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" 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=\"301\" height=\"183\" \/><figcaption class=\"wp-caption-text\">Graph demonstrating an inconsistent system<\/figcaption><\/figure>\n<p><strong>Analysis<\/strong><\/p>\n<p>Let&#8217;s graph the equations to confirm that the system has no solution. Writing the equations in slope-intercept form confirms that the system is <strong>inconsistent<\/strong> because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24799\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24799&theme=lumen&iframe_resize_id=ohm24799&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<h2>Solving Systems of Equations by the Addition Method<\/h2>\n<p>A third method of solving systems of linear equations is the <strong>addition method<\/strong>, this method is also called the <strong>elimination method<\/strong>. \u00a0In 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<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a system of equations, solve using the addition method.<\/strong><\/p>\n<ol>\n<li>Write both equations with [latex]x[\/latex]&#8211; and [latex]y[\/latex]-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.\n<ul>\n<li>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.<\/li>\n<li>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<\/ul>\n<\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q924657\">Show Solution<\/button><\/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><strong>Analysis<\/strong><\/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<figure style=\"width: 350px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" 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=\"350\" height=\"209\" \/><figcaption class=\"wp-caption-text\">A graph of two lines that intersect at the point negative seven-thirds, two-thirds<\/figcaption><\/figure>\n<p>We gain an important perspective on systems of equations by looking at the graphical representation. See the graph to 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<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the given system of equations by the addition method.<\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q883001\">Show Solution<\/button><\/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].<\/p>\n<figure style=\"width: 350px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" 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=\"350\" height=\"235\" \/><figcaption class=\"wp-caption-text\">A graph of two lines that cross at the point 3, negative 4<\/figcaption><\/figure>\n<p><strong>Analysis<\/strong><\/p>\n<p>We can check the solution in the original second equation and also by graphing:<\/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<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24801\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24801&theme=lumen&iframe_resize_id=ohm24801&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":669,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/791"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/791\/revisions"}],"predecessor-version":[{"id":2478,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/791\/revisions\/2478"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/669"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/791\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=791"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=791"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=791"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=791"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}