{"id":1758,"date":"2015-11-12T18:30:45","date_gmt":"2015-11-12T18:30:45","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1758"},"modified":"2015-11-12T18:30:45","modified_gmt":"2015-11-12T18:30:45","slug":"solving-a-system-of-nonlinear-equations-using-substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solving-a-system-of-nonlinear-equations-using-substitution\/","title":{"raw":"Solving a System of Nonlinear Equations Using Substitution","rendered":"Solving a System of Nonlinear Equations Using Substitution"},"content":{"raw":"<p>A <strong>system of nonlinear equations<\/strong> is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form [latex]Ax+By+C=0[\/latex]. Any equation that cannot be written in this form in nonlinear. 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.\n<\/p><h2>Intersection of a Parabola and a Line<\/h2>\nThere are three possible types of solutions for a system of nonlinear equations involving a <strong>parabola<\/strong> and a line.\n<div class=\"textbox\">\n<h3>A General Note: Possible Types of Solutions for Points of Intersection of a Parabola and a Line<\/h3>\nFigure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.\n<ul><li>No solution. The line will never intersect the parabola.<\/li>\n\t<li>One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.<\/li>\n\t<li>Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.<\/li>\n<\/ul>\n[caption id=\"\" align=\"aligncenter\" width=\"945\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202119\/CNX_Precalc_Figure_09_03_002n2.jpg\" alt=\"Graphs described in main body.\" width=\"945\" height=\"389\" data-media-type=\"image\/jpg\"\/><b>Figure 2<\/b>[\/caption]\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations containing a line and a parabola, find the solution.<strong>\n<\/strong><\/h3>\n<ol><li>Solve the linear equation for one of the variables.<\/li>\n\t<li>Substitute the expression obtained in step one into the parabola equation.<\/li>\n\t<li>Solve for the remaining variable.<\/li>\n\t<li>Check your solutions in both equations.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Solving a System of Nonlinear Equations Representing a Parabola and a Line<\/h3>\nSolve the system of equations.\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>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nSolve the first equation for [latex]x[\/latex] and then substitute the resulting expression into the second equation.\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}x-y=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }x=y - 1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Solve for }x.\\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }y={x}^{2}+1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }y={\\left(y - 1\\right)}^{2}+1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Substitute expression for }x.\\hfill \\end{array}[\/latex]<\/div>\nExpand the equation and set it equal to zero.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y={\\left(y - 1\\right)}^{2}\\hfill \\\\ \\text{ }=\\left({y}^{2}-2y+1\\right)+1\\hfill \\\\ \\text{ }={y}^{2}-2y+2\\hfill \\\\ 0={y}^{2}-3y+2\\hfill \\\\ \\text{ }=\\left(y - 2\\right)\\left(y - 1\\right)\\hfill \\end{array}[\/latex]<\/div>\nSolving for [latex]y[\/latex] gives [latex]y=2[\/latex] and [latex]y=1[\/latex]. Next, 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.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x-y=-1\\hfill \\\\ x-\\left(2\\right)=-1\\hfill \\\\ \\text{ }x=1\\hfill \\\\ \\hfill \\\\ x-\\left(1\\right)=-1\\hfill \\\\ \\text{ }x=0\\hfill \\end{array}[\/latex]<\/div>\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.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202120\/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\" data-media-type=\"image\/jpg\"\/><b>Figure 3<\/b>[\/caption]\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Could we have substituted values for [latex]y[\/latex] into the second equation to solve for [latex]x[\/latex] in Example 1?<\/h3>\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>\n\n<em>For<\/em> [latex]y=1[\/latex]\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y={x}^{2}+1\\hfill \\\\ y={x}^{2}+1\\hfill \\\\ {x}^{2}=0\\hfill \\\\ x=\\pm \\sqrt{0}=0\\hfill \\end{array}[\/latex]<\/div>\n<em>This gives us the same value as in the solution.<\/em>\n\n<em>For<\/em> [latex]y=2[\/latex]\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y={x}^{2}+1\\hfill \\\\ 2={x}^{2}+1\\hfill \\\\ {x}^{2}=1\\hfill \\\\ x=\\pm \\sqrt{1}=\\pm 1\\hfill \\end{array}[\/latex]<\/div>\n<em>Notice that [latex]-1[\/latex] is an extraneous solution.<\/em>\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\nSolve the given system of equations by substitution.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x-y=-2\\hfill \\\\ 2{x}^{2}-y=0\\hfill \\end{array}[\/latex]<\/div>\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-19\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<h2>Intersection of a Circle and a Line<\/h2>\nJust as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.\n<div class=\"textbox\">\n<h3>A General Note: Possible Types of Solutions for the Points of Intersection of a Circle and a Line<\/h3>\nFigure 4 illustrates possible solution sets for a system of equations involving a <strong>circle<\/strong> and a line.\n<ul><li>No solution. The line does not intersect the circle.<\/li>\n\t<li>One solution. The line is tangent to the circle and intersects the circle at exactly one point.<\/li>\n\t<li>Two solutions. The line crosses the circle and intersects it at two points.<\/li>\n<\/ul>\n[caption id=\"\" align=\"aligncenter\" width=\"945\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202122\/CNX_Precalc_Figure_09_03_004n2.jpg\" alt=\"Image described in main body\" width=\"945\" height=\"337\" data-media-type=\"image\/jpg\"\/><b>Figure 4<\/b>[\/caption]\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations containing a line and a circle, find the solution.<strong>\n<\/strong><\/h3>\n<ol><li>Solve the linear equation for one of the variables.<\/li>\n\t<li>Substitute the expression obtained in step one into the equation for the circle.<\/li>\n\t<li>Solve for the remaining variable.<\/li>\n\t<li>Check your solutions in both equations.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding the Intersection of a Circle and a Line by Substitution<\/h3>\nFind the intersection of the given circle and the given line by substitution.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+{y}^{2}=5\\hfill \\\\ y=3x - 5\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nOne of the equations has already been solved for [latex]y[\/latex]. We will substitute [latex]y=3x - 5[\/latex] into the equation for the circle.\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{x}^{2}+{\\left(3x - 5\\right)}^{2}=5\\\\ {x}^{2}+9{x}^{2}-30x+25=5\\\\ 10{x}^{2}-30x+20=0\\end{array}[\/latex]<\/div>\nNow, we factor and solve for [latex]x[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}10\\left({x}^{2}-3x+2\\right)=0\\hfill \\\\ 10\\left(x - 2\\right)\\left(x - 1\\right)=0\\hfill \\\\ x=2\\hfill \\\\ x=1\\hfill \\end{array}[\/latex]<\/div>\nSubstitute the two <em>x<\/em>-values into the original linear equation to solve for [latex]y[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y=3\\left(2\\right)-5\\hfill \\\\ =1\\hfill \\\\ y=3\\left(1\\right)-5\\hfill \\\\ =-2\\hfill \\end{array}[\/latex]<\/div>\nThe line intersects the circle at [latex]\\left(2,1\\right)[\/latex] and [latex]\\left(1,-2\\right)[\/latex], which can be verified by substituting these [latex]\\left(x,y\\right)[\/latex] values into both of the original equations.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202123\/CNX_Precalc_Figure_09_03_0052.jpg\" alt=\"Line y equals 3x minus 5 crosses circle x squared plus y squared equals five at the points 2,1 and 1, negative 2.\" width=\"487\" height=\"367\" data-media-type=\"image\/jpg\"\/><b>Figure 5<\/b>[\/caption]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\nSolve the system of nonlinear equations.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+{y}^{2}=10\\hfill \\\\ x - 3y=-10\\hfill \\end{array}[\/latex]<\/div>\n<div><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-19\/\" target=\"_blank\">Solution<\/a><\/div>\n<\/div>","rendered":"<p>A <strong>system of nonlinear equations<\/strong> is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form [latex]Ax+By+C=0[\/latex]. Any equation that cannot be written in this form in nonlinear. 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.\n<\/p>\n<h2>Intersection of a Parabola and a Line<\/h2>\n<p>There are three possible types of solutions for a system of nonlinear equations involving a <strong>parabola<\/strong> and a line.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Possible Types of Solutions for Points of Intersection of a Parabola and a Line<\/h3>\n<p>Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.<\/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<div style=\"width: 955px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202119\/CNX_Precalc_Figure_09_03_002n2.jpg\" alt=\"Graphs described in main body.\" width=\"945\" height=\"389\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations containing a line and a parabola, find the solution.<strong><br \/>\n<\/strong><\/h3>\n<ol>\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<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Solving a System of Nonlinear Equations Representing a Parabola and a Line<\/h3>\n<p>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>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>Solve the first equation for [latex]x[\/latex] and then substitute the resulting expression into the second equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}x-y=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }x=y - 1\\hfill & \\hfill & \\hfill & \\text{Solve for }x.\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }y={x}^{2}+1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }y={\\left(y - 1\\right)}^{2}+1\\hfill & \\hfill & \\hfill & \\text{Substitute expression for }x.\\hfill \\end{array}[\/latex]<\/div>\n<p>Expand the equation and set it equal to zero.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y={\\left(y - 1\\right)}^{2}\\hfill \\\\ \\text{ }=\\left({y}^{2}-2y+1\\right)+1\\hfill \\\\ \\text{ }={y}^{2}-2y+2\\hfill \\\\ 0={y}^{2}-3y+2\\hfill \\\\ \\text{ }=\\left(y - 2\\right)\\left(y - 1\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>Solving for [latex]y[\/latex] gives [latex]y=2[\/latex] and [latex]y=1[\/latex]. Next, 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{array}{l}\\text{ }x-y=-1\\hfill \\\\ x-\\left(2\\right)=-1\\hfill \\\\ \\text{ }x=1\\hfill \\\\ \\hfill \\\\ x-\\left(1\\right)=-1\\hfill \\\\ \\text{ }x=0\\hfill \\end{array}[\/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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202120\/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\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Could we have substituted values for [latex]y[\/latex] into the second equation to solve for [latex]x[\/latex] in Example 1?<\/h3>\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;\">[latex]\\begin{array}{l}y={x}^{2}+1\\hfill \\\\ y={x}^{2}+1\\hfill \\\\ {x}^{2}=0\\hfill \\\\ x=\\pm \\sqrt{0}=0\\hfill \\end{array}[\/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{array}{l}y={x}^{2}+1\\hfill \\\\ 2={x}^{2}+1\\hfill \\\\ {x}^{2}=1\\hfill \\\\ x=\\pm \\sqrt{1}=\\pm 1\\hfill \\end{array}[\/latex]<\/div>\n<p><em>Notice that [latex]-1[\/latex] is an extraneous solution.<\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\n<p>Solve the given system of equations by substitution.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3x-y=-2\\hfill \\\\ 2{x}^{2}-y=0\\hfill \\end{array}[\/latex]<\/div>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-19\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<h2>Intersection of a Circle and a Line<\/h2>\n<p>Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Possible Types of Solutions for the Points of Intersection of a Circle and a Line<\/h3>\n<p>Figure 4 illustrates possible solution sets for a system of equations involving a <strong>circle<\/strong> and a line.<\/p>\n<ul>\n<li>No solution. The line does not intersect the circle.<\/li>\n<li>One solution. The line is tangent to the circle and intersects the circle at exactly one point.<\/li>\n<li>Two solutions. The line crosses the circle and intersects it at two points.<\/li>\n<\/ul>\n<div style=\"width: 955px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202122\/CNX_Precalc_Figure_09_03_004n2.jpg\" alt=\"Image described in main body\" width=\"945\" height=\"337\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations containing a line and a circle, find the solution.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Solve the linear equation for one of the variables.<\/li>\n<li>Substitute the expression obtained in step one into the equation for the circle.<\/li>\n<li>Solve for the remaining variable.<\/li>\n<li>Check your solutions in both equations.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding the Intersection of a Circle and a Line by Substitution<\/h3>\n<p>Find the intersection of the given circle and the given line by substitution.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+{y}^{2}=5\\hfill \\\\ y=3x - 5\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>One of the equations has already been solved for [latex]y[\/latex]. We will substitute [latex]y=3x - 5[\/latex] into the equation for the circle.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{x}^{2}+{\\left(3x - 5\\right)}^{2}=5\\\\ {x}^{2}+9{x}^{2}-30x+25=5\\\\ 10{x}^{2}-30x+20=0\\end{array}[\/latex]<\/div>\n<p>Now, we factor and solve for [latex]x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}10\\left({x}^{2}-3x+2\\right)=0\\hfill \\\\ 10\\left(x - 2\\right)\\left(x - 1\\right)=0\\hfill \\\\ x=2\\hfill \\\\ x=1\\hfill \\end{array}[\/latex]<\/div>\n<p>Substitute the two <em>x<\/em>-values into the original linear equation to solve for [latex]y[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y=3\\left(2\\right)-5\\hfill \\\\ =1\\hfill \\\\ y=3\\left(1\\right)-5\\hfill \\\\ =-2\\hfill \\end{array}[\/latex]<\/div>\n<p>The line intersects the circle at [latex]\\left(2,1\\right)[\/latex] and [latex]\\left(1,-2\\right)[\/latex], which can be verified by substituting these [latex]\\left(x,y\\right)[\/latex] values into both of the original equations.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202123\/CNX_Precalc_Figure_09_03_0052.jpg\" alt=\"Line y equals 3x minus 5 crosses circle x squared plus y squared equals five at the points 2,1 and 1, negative 2.\" width=\"487\" height=\"367\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\n<p>Solve the system of nonlinear equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+{y}^{2}=10\\hfill \\\\ x - 3y=-10\\hfill \\end{array}[\/latex]<\/div>\n<div><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-19\/\" target=\"_blank\">Solution<\/a><\/div>\n<\/div>\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-1758\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><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":276,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1758","chapter","type-chapter","status-publish","hentry"],"part":1751,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1758","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1758\/revisions"}],"predecessor-version":[{"id":2272,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1758\/revisions\/2272"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1751"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1758\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=1758"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1758"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1758"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=1758"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}