{"id":134,"date":"2025-02-13T22:44:10","date_gmt":"2025-02-13T22:44:10","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/systems-of-nonlinear-equations-and-inequalities-two-variables\/"},"modified":"2026-03-20T18:26:00","modified_gmt":"2026-03-20T18:26:00","slug":"systems-of-nonlinear-equations-and-inequalities-two-variables","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/systems-of-nonlinear-equations-and-inequalities-two-variables\/","title":{"raw":"Systems of Nonlinear Equations and Inequalities: Learn It 1","rendered":"Systems of Nonlinear Equations and Inequalities: Learn It 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve a system of nonlinear equations.<\/li>\r\n \t<li>Graph a system of nonlinear inequalities.<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<h2 data-type=\"title\">Solving a System of Nonlinear Equations Using Substitution<\/h2>\r\nA\u00a0<strong><span id=\"term-00004\" data-type=\"term\">system of nonlinear equations<\/span><\/strong> is a system of two or more equations in two or more variables containing at least one equation that is not linear.\u00a0Recall that <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">a linear equation can take the form [latex]Ax+By+C = 0[\/latex].<\/span> Any equation that cannot be written in this form in nonlinear.\r\n\r\nThe substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.\r\n<h3>Intersection of a Parabola and a Line<\/h3>\r\n<p id=\"fs-id1165135367535\">There are three possible types of solutions for a system of nonlinear equations involving a\u00a0<span id=\"term-00005\" class=\"no-emphasis\" data-type=\"term\">parabola<\/span> and a line.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>possible types of solutions for points of intersection of a parabola and a line<\/h3>\r\nThe graphs below illustrate possible solution sets for a system of equations involving a parabola (quadratic function) and a line (linear function).\r\n<ul>\r\n \t<li>No solution. The line will never intersect the parabola.<\/li>\r\n \t<li>One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.<\/li>\r\n \t<li>Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.<\/li>\r\n<\/ul>\r\n<img class=\"size-full wp-image-5686 aligncenter\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/07\/28162507\/178eac8e3b6e24d899b9de23cd5d79940d7592c5.jpg\" alt=\"The image illustrates three scenarios of intersections between a parabola and a line. In the first graph, there are no intersections, indicating no solutions. The second graph shows the line touching the parabola at a single point, representing one solution. In the third graph, the line intersects the parabola at two points, resulting in two solutions.\" width=\"944\" height=\"325\" \/>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p id=\"fs-id1165135694401\"><strong>Given a system of equations containing a line and a parabola, find the solution.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165131884672\" type=\"1\">\r\n \t<li>Solve the linear equation for one of the variables.<\/li>\r\n \t<li>Substitute the expression obtained in step one into the parabola equation.<\/li>\r\n \t<li>Solve for the remaining variable.<\/li>\r\n \t<li>Check your solutions in both equations.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the system of equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x-y=-1\\hfill \\\\ y={x}^{2}+1\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"654580\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"654580\"]\r\n<div style=\"text-align: left;\">Solve the first equation for [latex]x[\/latex] and then substitute the resulting expression into the second equation.<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;x-y=-1 \\\\ &amp;x=y - 1 &amp;&amp; \\text{Solve for }x. \\\\ \\\\ &amp;y={x}^{2}+1 \\\\ &amp;y={\\left(y - 1\\right)}^{2}+1 &amp;&amp; \\text{Substitute expression for }x. \\\\ &amp;y=\\left({y}^{2}-2y+1\\right)+1 &amp;&amp; \\text{Expand} \\\\ &amp;y={y}^{2}-2y+2 \\\\[3mm] &amp;0={y}^{2}-3y+2 &amp;&amp; \\text{Set equal to 0 and solve.} \\\\ &amp;0=\\left(y - 2\\right)\\left(y - 1\\right) \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nSolving for [latex]y[\/latex] gives [latex]y=2[\/latex] and [latex]y=1[\/latex].\r\n[latex]\\\\[\/latex]\r\nNext, substitute each value for [latex]y[\/latex] into the first equation to solve for [latex]x[\/latex]. Always substitute the value into the linear equation to check for extraneous solutions.\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}x-y=-1 \\\\ x-\\left(2\\right)=-1 \\\\ x=1 \\\\[3mm] x-\\left(1\\right)=-1 \\\\ x=0 \\end{gathered}[\/latex]<\/div>\r\nThe solutions are [latex]\\left(1,2\\right)[\/latex] and [latex]\\left(0,1\\right),\\text{}[\/latex], which can be verified by substituting these [latex]\\left(x,y\\right)[\/latex] values into both of the original equations.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03190507\/CNX_Precalc_Figure_09_03_0032.jpg\" alt=\"Line x minus y equals negative one crosses parabola y equals x squared plus one at the points zero, one and one, two.\" width=\"487\" height=\"292\" \/>\r\n\r\nCould we have substituted values for [latex]y[\/latex] into the second equation to solve for [latex]x[\/latex]?\r\n\r\n<em>Yes, but because [latex]x[\/latex] is squared in the second equation this could give us extraneous solutions for [latex]x[\/latex]. <\/em>\r\n\r\n<em>For<\/em> [latex]y=1[\/latex]\r\n<div style=\"text-align: center;\">\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;y={x}^{2}+1 \\\\ &amp;y={x}^{2}+1 \\\\ &amp;{x}^{2}=0 \\\\ &amp;x=\\pm \\sqrt{0}=0 \\end{align}[\/latex]<\/div>\r\n<em>This gives us the same value as in the solution.<\/em>\r\n\r\n<em>For<\/em> [latex]y=2[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;y={x}^{2}+1 \\\\ &amp;2={x}^{2}+1 \\\\ &amp;{x}^{2}=1 \\\\ &amp;x=\\pm \\sqrt{1}=\\pm 1 \\end{align}[\/latex]<\/div>\r\n<\/div>\r\n<em>Notice that [latex]-1[\/latex] is an extraneous solution.<\/em>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\n[ohm_question hide_question_numbers=1]321625[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321626[\/ohm_question]<\/section><section class=\"textbox interact\" aria-label=\"Interact\">Use an online graphing calculator to graph the parabola [latex]y\\ =\\ x^2+2x-3[\/latex]. Now graph the line [latex]ax+by+c\\ =\\ 0[\/latex] and adjust the values of [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] to find equations for lines that produce\u00a0systems with the following types of solutions:\r\n<ul>\r\n \t<li>One solution<\/li>\r\n \t<li>Two solutions<\/li>\r\n \t<li>No solutions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<dl id=\"fs-id1165137681185\" class=\"definition\">\r\n \t<dd id=\"fs-id1165137681190\"><\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve a system of nonlinear equations.<\/li>\n<li>Graph a system of nonlinear inequalities.<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<h2 data-type=\"title\">Solving a System of Nonlinear Equations Using Substitution<\/h2>\n<p>A\u00a0<strong><span id=\"term-00004\" data-type=\"term\">system of nonlinear equations<\/span><\/strong> is a system of two or more equations in two or more variables containing at least one equation that is not linear.\u00a0Recall that <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">a linear equation can take the form [latex]Ax+By+C = 0[\/latex].<\/span> Any equation that cannot be written in this form in nonlinear.<\/p>\n<p>The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.<\/p>\n<h3>Intersection of a Parabola and a Line<\/h3>\n<p id=\"fs-id1165135367535\">There are three possible types of solutions for a system of nonlinear equations involving a\u00a0<span id=\"term-00005\" class=\"no-emphasis\" data-type=\"term\">parabola<\/span> and a line.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>possible types of solutions for points of intersection of a parabola and a line<\/h3>\n<p>The graphs below illustrate possible solution sets for a system of equations involving a parabola (quadratic function) and a line (linear function).<\/p>\n<ul>\n<li>No solution. The line will never intersect the parabola.<\/li>\n<li>One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.<\/li>\n<li>Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5686 aligncenter\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/07\/28162507\/178eac8e3b6e24d899b9de23cd5d79940d7592c5.jpg\" alt=\"The image illustrates three scenarios of intersections between a parabola and a line. In the first graph, there are no intersections, indicating no solutions. The second graph shows the line touching the parabola at a single point, representing one solution. In the third graph, the line intersects the parabola at two points, resulting in two solutions.\" width=\"944\" height=\"325\" \/><\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p id=\"fs-id1165135694401\"><strong>Given a system of equations containing a line and a parabola, find the solution.<\/strong><\/p>\n<ol id=\"fs-id1165131884672\" type=\"1\">\n<li>Solve the linear equation for one of the variables.<\/li>\n<li>Substitute the expression obtained in step one into the parabola equation.<\/li>\n<li>Solve for the remaining variable.<\/li>\n<li>Check your solutions in both equations.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the system of equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x-y=-1\\hfill \\\\ y={x}^{2}+1\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q654580\">Show Solution<\/button><\/p>\n<div id=\"q654580\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: left;\">Solve the first equation for [latex]x[\/latex] and then substitute the resulting expression into the second equation.<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{align}&x-y=-1 \\\\ &x=y - 1 && \\text{Solve for }x. \\\\ \\\\ &y={x}^{2}+1 \\\\ &y={\\left(y - 1\\right)}^{2}+1 && \\text{Substitute expression for }x. \\\\ &y=\\left({y}^{2}-2y+1\\right)+1 && \\text{Expand} \\\\ &y={y}^{2}-2y+2 \\\\[3mm] &0={y}^{2}-3y+2 && \\text{Set equal to 0 and solve.} \\\\ &0=\\left(y - 2\\right)\\left(y - 1\\right) \\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>Solving for [latex]y[\/latex] gives [latex]y=2[\/latex] and [latex]y=1[\/latex].<br \/>\n[latex]\\\\[\/latex]<br \/>\nNext, substitute each value for [latex]y[\/latex] into the first equation to solve for [latex]x[\/latex]. Always substitute the value into the linear equation to check for extraneous solutions.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}x-y=-1 \\\\ x-\\left(2\\right)=-1 \\\\ x=1 \\\\[3mm] x-\\left(1\\right)=-1 \\\\ x=0 \\end{gathered}[\/latex]<\/div>\n<p>The solutions are [latex]\\left(1,2\\right)[\/latex] and [latex]\\left(0,1\\right),\\text{}[\/latex], which can be verified by substituting these [latex]\\left(x,y\\right)[\/latex] values into both of the original equations.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03190507\/CNX_Precalc_Figure_09_03_0032.jpg\" alt=\"Line x minus y equals negative one crosses parabola y equals x squared plus one at the points zero, one and one, two.\" width=\"487\" height=\"292\" \/><\/p>\n<p>Could we have substituted values for [latex]y[\/latex] into the second equation to solve for [latex]x[\/latex]?<\/p>\n<p><em>Yes, but because [latex]x[\/latex] is squared in the second equation this could give us extraneous solutions for [latex]x[\/latex]. <\/em><\/p>\n<p><em>For<\/em> [latex]y=1[\/latex]<\/p>\n<div style=\"text-align: center;\">\n<div style=\"text-align: center;\">[latex]\\begin{align}&y={x}^{2}+1 \\\\ &y={x}^{2}+1 \\\\ &{x}^{2}=0 \\\\ &x=\\pm \\sqrt{0}=0 \\end{align}[\/latex]<\/div>\n<p><em>This gives us the same value as in the solution.<\/em><\/p>\n<p><em>For<\/em> [latex]y=2[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&y={x}^{2}+1 \\\\ &2={x}^{2}+1 \\\\ &{x}^{2}=1 \\\\ &x=\\pm \\sqrt{1}=\\pm 1 \\end{align}[\/latex]<\/div>\n<\/div>\n<p><em>Notice that [latex]-1[\/latex] is an extraneous solution.<\/em><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p><iframe loading=\"lazy\" id=\"ohm321625\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321625&theme=lumen&iframe_resize_id=ohm321625&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321626\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321626&theme=lumen&iframe_resize_id=ohm321626&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox interact\" aria-label=\"Interact\">Use an online graphing calculator to graph the parabola [latex]y\\ =\\ x^2+2x-3[\/latex]. Now graph the line [latex]ax+by+c\\ =\\ 0[\/latex] and adjust the values of [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] to find equations for lines that produce\u00a0systems with the following types of solutions:<\/p>\n<ul>\n<li>One solution<\/li>\n<li>Two solutions<\/li>\n<li>No solutions<\/li>\n<\/ul>\n<\/section>\n<dl id=\"fs-id1165137681185\" class=\"definition\">\n<dd id=\"fs-id1165137681190\"><\/dd>\n<\/dl>\n","protected":false},"author":6,"menu_order":20,"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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":131,"module-header":"learn_it","content_attributions":[{"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":""}],"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/134"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/134\/revisions"}],"predecessor-version":[{"id":5945,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/134\/revisions\/5945"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/131"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/134\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=134"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=134"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=134"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=134"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}