{"id":518,"date":"2025-10-09T15:17:41","date_gmt":"2025-10-09T15:17:41","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=518"},"modified":"2025-11-20T14:51:16","modified_gmt":"2025-11-20T14:51:16","slug":"advanced-vector-operations-and-applications-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/advanced-vector-operations-and-applications-background-youll-need-2\/","title":{"raw":"Advanced Vector Operations and Applications: Background You'll Need 2","rendered":"Advanced Vector Operations and Applications: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Use substitution and the addition\/elimination method to solve systems of equations with two variables<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Systems of Equations by Substitution<\/h2>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When solving systems of linear equations, we need methods that work reliably for all types of solutions\u2014whether they involve integers, decimals, or fractions. The <strong>substitution method<\/strong> provides an algebraic approach that gives us exact answers.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The substitution method works in two main steps: solve one equation for one variable, then substitute that expression into the other equation. This allows us to solve for one variable at a time, which simplifies the process and ensures accuracy.<\/p>\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\nAnother 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>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Use substitution and the addition\/elimination method to solve systems of equations with two variables<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Solving Systems of Equations by Substitution<\/h2>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When solving systems of linear equations, we need methods that work reliably for all types of solutions\u2014whether they involve integers, decimals, or fractions. The <strong>substitution method<\/strong> provides an algebraic approach that gives us exact answers.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The substitution method works in two main steps: solve one equation for one variable, then substitute that expression into the other equation. This allows us to solve for one variable at a time, which simplifies the process and ensures accuracy.<\/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>Another 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","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":[],"class_list":["post-518","chapter","type-chapter","status-publish","hentry"],"part":422,"module-header":"background_you_need","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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/518","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/518\/revisions"}],"predecessor-version":[{"id":2610,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/518\/revisions\/2610"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/422"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/518\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=518"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=518"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=518"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=518"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}