{"id":1086,"date":"2022-03-12T22:07:50","date_gmt":"2022-03-12T22:07:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1086"},"modified":"2025-11-15T17:25:55","modified_gmt":"2025-11-15T17:25:55","slug":"2-6-2-the-substitution-method-for-solving-systems-of-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/2-6-2-the-substitution-method-for-solving-systems-of-linear-equations\/","title":{"raw":"2.6.2: The Substitution Method","rendered":"2.6.2: The Substitution Method"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-106\" class=\"standard post-106 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Solve a system of linear equations using the substitution method<\/li>\r\n \t<li>Write the general solution of a dependent system as an ordered pair in terms of one variable<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Substitution Method<\/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\u00a0system of linear equations\u00a0that are more precise than graphing. One such method is solving a system of equations by the\u00a0<strong><em>substitution method.<\/em>\u00a0<\/strong>\r\n\r\nLet's assume we have a consistent system of two linear equations in two variables. This means that the lines representing this system intersect. At any point of intersection the [latex]x[\/latex]- and [latex]y[\/latex]- values satisfy both equations. So the [latex]x[\/latex]-value in the first equation equals the\u00a0[latex]x[\/latex]-value in the second equation and the\u00a0[latex]y[\/latex]-value in the first equation equals the\u00a0[latex]y[\/latex]-value in the second equation. The assumption that an intersection point exists allows us to\u00a0solve one of the equations for one variable and then substitute the result into the other equation to end up with a linear equation in one variable that can be solved for the second variable. Once the second variable is known, the first variable can be found by substituting the value of the second variable into either of the original equations in the system.\r\n<div class=\"textbox\">\r\n<h3>solving a system of linear equations: 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 other original equation, and 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 other 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 examples\">\r\n<h3>Example 1<\/h3>\r\nSolve the following system of equations using the substitution method:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nFirst, we will solve the first equation for [latex]y[\/latex] by adding [latex]x[\/latex] to both sides of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-x+y&amp;=-5\\\\y&amp;=x-5\\end{aligned}\\end{equation}[\/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{equation}\\begin{aligned}2x - 5\\color{blue}{y}&amp;=1 \\\\ 2x - 5\\color{blue}{(x - 5)}&amp;=1\\\\ 2x - 5x+25&amp;=1 \\\\ -3x&amp;=-24 \\\\ x=8 \\end{aligned}\\end{equation}[\/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{equation}\\begin{aligned}-\\color{blue}{x}+y&amp;=-5\\\\-\\color{blue}{8}+y=-5 \\\\ y=3 \\end{aligned}\\end{equation}[\/latex]<\/p>\r\nOur solution is [latex]\\left(8,3\\right)[\/latex].\r\n\r\nCheck the solution by substituting [latex](8, 3)[\/latex] into both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-x+y&amp;=-5\\\\ -(8)+(3)&amp;=-5 \\;\\;\\;\\;\\;\\;\\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}2x - 5y&amp;=1\\\\ 2(8)-5(3)&amp;=1\\;\\;\\;\\;\\;\\;\\;\\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIf we had chosen the other equation to start with in the previous example, we would still determine the same solution. It is really a matter of preference which equation to start with. Had we started with the second equation solving for a variable would have resulted in having to work with fractions, so starting with the first equation was, perhaps, the better choice.\u00a0The substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of 1 or [latex]\u20131[\/latex] so that we do not have to deal with fractions.\r\n\r\nThe following video shows an example of solving a system of two equations using the substitution method.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/MIXL35YRzRw?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.2-1.docx\">Transcript-2.6.2-1<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nSolve the following system of equations using the substitution method:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y=5\\hfill \\\\ \\text{ }x - 2y=1\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm721\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm721\"]\r\n\r\nSolution is [latex](x, y)=(3, 1)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nConsider the system of equations:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=9 - 2y\\hfill \\\\ x+2y=13\\hfill \\end{array}[\/latex]<\/p>\r\nThe first equation is already solved for [latex]x[\/latex] so it makes sense to substitute [latex]9-2y[\/latex] for [latex]x[\/latex] in the second equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\color{blue}{x}+2y=13\\hfill \\\\ \\color{blue}{\\left(9 - 2y\\right)}+2y=13\\hfill \\\\ 9+0y=13\\hfill \\\\ 9=13\\hfill \\end{array}[\/latex]<\/p>\r\nClearly, this statement is a <em><strong>contradiction<\/strong><\/em> because [latex]9\\ne 13[\/latex]. But what does this mean?\r\n\r\nWhen we introduced the substitution method, we made an assumption: that the system had a solution.\r\n\r\nSInce we have a contradiction, we are contradicting that assumption, so the system has no solution.\r\n\r\nTo double check, we rearrange the equations so that they are both in slope-intercept form and we will be able to tell whether we have parallel lines (meaning no solution) or coinciding lines (meaning infinite solutions on the line).\r\n\r\nFirst equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ 2y=-x+9\\hfill \\\\ \\text{ }y=-\\dfrac{1}{2}x+\\dfrac{9}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nSecond equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\text{ }2y=-x+13\\hfill \\\\ \\text{ }y=-\\dfrac{1}{2}x+\\dfrac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nComparing the equations, we see that the lines they represent have the same slope of [latex]-\\frac{1}{2}[\/latex] but different [latex]y[\/latex]-intercepts. Therefore, the lines are parallel and do not intersect.\r\n<div class=\"textbox shaded\">\r\n<h3>Contradictions<\/h3>\r\nWhen we get a contradiction in an equation using the substitution method, it means the system has no solution.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nSolve the system of equations\u00a0using the substitution method:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=-x+4\\hfill \\\\ x+y=10\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nThe first equation is already solved for [latex]y[\/latex] so it makes sense to substitute [latex]-x+4[\/latex] for [latex]y[\/latex] in the second equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+\\color{blue}{y}=10\\hfill \\\\ x+\\color{blue}{\\left(-x+4\\right)}=10\\hfill \\\\ 4=10\\hfill \\end{array}[\/latex]<\/p>\r\nClearly, this statement is a contradiction because [latex]4\\ne 10[\/latex], so the system has no solution.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe next video shows another example of using the substitution method to solve a system that has no solution.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/kTtKfh5gFUc?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.2-2.docx\">Transcript-2.6.2-2<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nSolve the system of equations using the substitution method:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}4x=1 - 2y\\hfill \\\\ 4x+2y=3\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm245\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm245\"]The system has no solution.[\/hidden-answer]\r\n\r\n<\/div>\r\nWe have seen that the substitution method results in solving a linear equation in one variable. We have also seen that we can get a single solution when this equation leads to a numerical answer, or no solution when this equation leads to a contradiction (an equation that is never true). The only remaining option occurs when this equation turns into an <em><strong>identity<\/strong><\/em>\u00a0(an equation that is always true) with the result of an infinite number of solutions that lie on the line of the original equation.\r\n\r\nConsider the system of equations:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x=6 - y\\hfill \\\\ 3x+y=6\\hfill \\end{array}[\/latex]<\/p>\r\nTo solve this system we would first\u00a0rearrange one of the equations for a single variable. Let's solve the 2nd equation for [latex]y[\/latex] by subtracting [latex]3x[\/latex] from both sides:\u00a0 [latex]y=-3x+6[\/latex]\r\n\r\nNow we can substitute\u00a0[latex]-3x+6[\/latex] for\u00a0[latex]y[\/latex] in the 1st equation:\r\n\r\n[latex]\\begin{equation}\\begin{aligned}3x&amp;=6-\\color{blue}{y}\\\\3x&amp;=6-\\color{blue}{(-3x+6)}\\\\ 3x&amp;=6+3x-6\\\\3x&amp;=3x\\end{aligned}\\end{equation}[\/latex]\r\n\r\nWe have an <em><strong>identity<\/strong><\/em>. This means that the lines coincide. Consequently, there are infinite solutions that lie on the line with equation [latex]3x+y=6[\/latex].\r\n\r\nThe next video shows that a system can have an infinite number of solutions.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/Pcqb109yK5Q?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.2-3.docx\">Transcript-2.6.2-3<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nSolve the system of equations using the substitution method:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-7x=-5 + 2y\\hfill \\\\ 14x+4y=10\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm806\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm806\"]\r\n\r\nThere are infinite solutions that lie on the line [latex]2y+7x=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox tryit\">\r\n<h3>TRY IT 4<\/h3>\r\n[ohm_question]247796[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>TRY IT 5<\/h3>\r\n[ohm_question]247798[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>TRY iT 6<\/h3>\r\n[ohm_question]247799[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\nThe substitution method works well when one of the equations in the system can be solved relatively easily for a single variable. However, it can get a bit tricky if fractions are involved. In the next section, we will consider an alternative method that makes dealing with fractions in the equations a little easier.\r\n\r\n<\/div>\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-106\" class=\"standard post-106 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Solve a system of linear equations using the substitution method<\/li>\n<li>Write the general solution of a dependent system as an ordered pair in terms of one variable<\/li>\n<\/ul>\n<\/div>\n<h2>The Substitution Method<\/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\u00a0system of linear equations\u00a0that are more precise than graphing. One such method is solving a system of equations by the\u00a0<strong><em>substitution method.<\/em>\u00a0<\/strong><\/p>\n<p>Let&#8217;s assume we have a consistent system of two linear equations in two variables. This means that the lines representing this system intersect. At any point of intersection the [latex]x[\/latex]&#8211; and [latex]y[\/latex]&#8211; values satisfy both equations. So the [latex]x[\/latex]-value in the first equation equals the\u00a0[latex]x[\/latex]-value in the second equation and the\u00a0[latex]y[\/latex]-value in the first equation equals the\u00a0[latex]y[\/latex]-value in the second equation. The assumption that an intersection point exists allows us to\u00a0solve one of the equations for one variable and then substitute the result into the other equation to end up with a linear equation in one variable that can be solved for the second variable. Once the second variable is known, the first variable can be found by substituting the value of the second variable into either of the original equations in the system.<\/p>\n<div class=\"textbox\">\n<h3>solving a system of linear equations: 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 other original equation, and then solve for the remaining variable.<\/li>\n<li>Substitute that solution into either of the original equations to find the value of the other 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 examples\">\n<h3>Example 1<\/h3>\n<p>Solve the following system of equations using the substitution method:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>First, we will solve the first equation for [latex]y[\/latex] by adding [latex]x[\/latex] to both sides of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-x+y&=-5\\\\y&=x-5\\end{aligned}\\end{equation}[\/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{equation}\\begin{aligned}2x - 5\\color{blue}{y}&=1 \\\\ 2x - 5\\color{blue}{(x - 5)}&=1\\\\ 2x - 5x+25&=1 \\\\ -3x&=-24 \\\\ x=8 \\end{aligned}\\end{equation}[\/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{equation}\\begin{aligned}-\\color{blue}{x}+y&=-5\\\\-\\color{blue}{8}+y=-5 \\\\ y=3 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Our solution is [latex]\\left(8,3\\right)[\/latex].<\/p>\n<p>Check the solution by substituting [latex](8, 3)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-x+y&=-5\\\\ -(8)+(3)&=-5 \\;\\;\\;\\;\\;\\;\\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}2x - 5y&=1\\\\ 2(8)-5(3)&=1\\;\\;\\;\\;\\;\\;\\;\\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>If we had chosen the other equation to start with in the previous example, we would still determine the same solution. It is really a matter of preference which equation to start with. Had we started with the second equation solving for a variable would have resulted in having to work with fractions, so starting with the first equation was, perhaps, the better choice.\u00a0The substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of 1 or [latex]\u20131[\/latex] so that we do not have to deal with fractions.<\/p>\n<p>The following video shows an example of solving a system of two equations using the substitution method.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/MIXL35YRzRw?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.2-1.docx\">Transcript-2.6.2-1<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Solve the following system of equations using the substitution method:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y=5\\hfill \\\\ \\text{ }x - 2y=1\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm721\">Show Answer<\/span><\/p>\n<div id=\"qhjm721\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution is [latex](x, y)=(3, 1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Consider the system of equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=9 - 2y\\hfill \\\\ x+2y=13\\hfill \\end{array}[\/latex]<\/p>\n<p>The first equation is already solved for [latex]x[\/latex] so it makes sense to substitute [latex]9-2y[\/latex] for [latex]x[\/latex] in the second equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\color{blue}{x}+2y=13\\hfill \\\\ \\color{blue}{\\left(9 - 2y\\right)}+2y=13\\hfill \\\\ 9+0y=13\\hfill \\\\ 9=13\\hfill \\end{array}[\/latex]<\/p>\n<p>Clearly, this statement is a <em><strong>contradiction<\/strong><\/em> because [latex]9\\ne 13[\/latex]. But what does this mean?<\/p>\n<p>When we introduced the substitution method, we made an assumption: that the system had a solution.<\/p>\n<p>SInce we have a contradiction, we are contradicting that assumption, so the system has no solution.<\/p>\n<p>To double check, we rearrange the equations so that they are both in slope-intercept form and we will be able to tell whether we have parallel lines (meaning no solution) or coinciding lines (meaning infinite solutions on the line).<\/p>\n<p>First equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ 2y=-x+9\\hfill \\\\ \\text{ }y=-\\dfrac{1}{2}x+\\dfrac{9}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Second equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\text{ }2y=-x+13\\hfill \\\\ \\text{ }y=-\\dfrac{1}{2}x+\\dfrac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Comparing the equations, we see that the lines they represent have the same slope of [latex]-\\frac{1}{2}[\/latex] but different [latex]y[\/latex]-intercepts. Therefore, the lines are parallel and do not intersect.<\/p>\n<div class=\"textbox shaded\">\n<h3>Contradictions<\/h3>\n<p>When we get a contradiction in an equation using the substitution method, it means the system has no solution.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Solve the system of equations\u00a0using the substitution method:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=-x+4\\hfill \\\\ x+y=10\\hfill \\end{array}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>The first equation is already solved for [latex]y[\/latex] so it makes sense to substitute [latex]-x+4[\/latex] for [latex]y[\/latex] in the second equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+\\color{blue}{y}=10\\hfill \\\\ x+\\color{blue}{\\left(-x+4\\right)}=10\\hfill \\\\ 4=10\\hfill \\end{array}[\/latex]<\/p>\n<p>Clearly, this statement is a contradiction because [latex]4\\ne 10[\/latex], so the system has no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The next video shows another example of using the substitution method to solve a system that has no solution.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/kTtKfh5gFUc?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.2-2.docx\">Transcript-2.6.2-2<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Solve the system of equations using the substitution method:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}4x=1 - 2y\\hfill \\\\ 4x+2y=3\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm245\">Show Answer<\/span><\/p>\n<div id=\"qhjm245\" class=\"hidden-answer\" style=\"display: none\">The system has no solution.<\/div>\n<\/div>\n<\/div>\n<p>We have seen that the substitution method results in solving a linear equation in one variable. We have also seen that we can get a single solution when this equation leads to a numerical answer, or no solution when this equation leads to a contradiction (an equation that is never true). The only remaining option occurs when this equation turns into an <em><strong>identity<\/strong><\/em>\u00a0(an equation that is always true) with the result of an infinite number of solutions that lie on the line of the original equation.<\/p>\n<p>Consider the system of equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x=6 - y\\hfill \\\\ 3x+y=6\\hfill \\end{array}[\/latex]<\/p>\n<p>To solve this system we would first\u00a0rearrange one of the equations for a single variable. Let&#8217;s solve the 2nd equation for [latex]y[\/latex] by subtracting [latex]3x[\/latex] from both sides:\u00a0 [latex]y=-3x+6[\/latex]<\/p>\n<p>Now we can substitute\u00a0[latex]-3x+6[\/latex] for\u00a0[latex]y[\/latex] in the 1st equation:<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}3x&=6-\\color{blue}{y}\\\\3x&=6-\\color{blue}{(-3x+6)}\\\\ 3x&=6+3x-6\\\\3x&=3x\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>We have an <em><strong>identity<\/strong><\/em>. This means that the lines coincide. Consequently, there are infinite solutions that lie on the line with equation [latex]3x+y=6[\/latex].<\/p>\n<p>The next video shows that a system can have an infinite number of solutions.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/Pcqb109yK5Q?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.2-3.docx\">Transcript-2.6.2-3<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Solve the system of equations using the substitution method:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-7x=-5 + 2y\\hfill \\\\ 14x+4y=10\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm806\">Show Answer<\/span><\/p>\n<div id=\"qhjm806\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are infinite solutions that lie on the line [latex]2y+7x=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox tryit\">\n<h3>TRY IT 4<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm247796\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=247796&theme=oea&iframe_resize_id=ohm247796&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>TRY IT 5<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm247798\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=247798&theme=oea&iframe_resize_id=ohm247798&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>TRY iT 6<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm247799\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=247799&theme=oea&iframe_resize_id=ohm247799&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<p>The substitution method works well when one of the equations in the system can be solved relatively easily for a single variable. However, it can get a bit tricky if fractions are involved. In the next section, we will consider an alternative method that makes dealing with fractions in the equations a little easier.<\/p>\n<\/div>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/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-1086\">\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>Revised and adapted: The Substitution Method. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Try It hjm721; hjm806; hjm245. . <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <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>Ex 2: Solve a System of Equations Using Substitution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/MIXL35YRzRw\">https:\/\/youtu.be\/MIXL35YRzRw<\/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>Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve a System of Equations Using Substitution - No Solution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kTtKfh5gFUc\">https:\/\/youtu.be\/kTtKfh5gFUc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve a System of Equations Using Substitution - Infinite Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Pcqb109yK5Q\">https:\/\/youtu.be\/Pcqb109yK5Q<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, 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>","protected":false},"author":422608,"menu_order":9,"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\":\"cc\",\"description\":\"Ex 2: Solve a System of Equations Using Substitution\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/MIXL35YRzRw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve a System of Equations Using Substitution - No Solution\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/kTtKfh5gFUc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve a System of Equations Using Substitution - Infinite Solutions\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Pcqb109yK5Q\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revised and adapted: The Substitution Method\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Try It hjm721; hjm806; hjm245. \",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1086","chapter","type-chapter","status-publish","hentry"],"part":963,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1086","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":31,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1086\/revisions"}],"predecessor-version":[{"id":4648,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1086\/revisions\/4648"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/963"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1086\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=1086"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1086"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=1086"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=1086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}