{"id":2080,"date":"2016-07-01T23:08:21","date_gmt":"2016-07-01T23:08:21","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2080"},"modified":"2019-08-06T18:02:38","modified_gmt":"2019-08-06T18:02:38","slug":"read-solve-systems-of-equations-in-two-variables-by-the-addition-method","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-solve-systems-of-equations-in-two-variables-by-the-addition-method\/","title":{"raw":"Solve Systems by Elimination","rendered":"Solve Systems by Elimination"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve systems of equations by addition<\/li>\r\n \t<li>Express the solution of a system of dependent equations containing two variables<\/li>\r\n<\/ul>\r\n<\/div>\r\nA third method of <strong>solving systems of linear equations<\/strong> is the <strong>elimination\u00a0method<\/strong>. In 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 elimination.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by elimination.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"522070\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522070\"]\r\n\r\nBoth equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation,\u00a0[latex]\u20131[\/latex], is the opposite of the coefficient of [latex]x[\/latex] in the first equation,\u00a0[latex]1[\/latex]. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{\\begin{array}{l}\\hfill \\\\ \\:\\:x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}\\:\\:\\:\\:\\:\\:3y=2}[\/latex]<\/div>\r\n&nbsp;\r\n\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{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\dfrac{2}{3}\\hfill \\end{array}[\/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{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\dfrac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\dfrac{2}{3}\\hfill \\\\ \\text{ }-x=\\dfrac{7}{3}\\hfill \\\\ \\text{ }\\:\\:\\:\\:\\:x=-\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].\r\n\r\nCheck the solution in the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }\\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\dfrac{7}{3}+\\dfrac{4}{3}=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\dfrac{3}{3}=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-1=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/p>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. In the graph below, you will see 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<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222638\/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=\"487\" height=\"291\" \/>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see another example of how to use the method of elimination to solve a system of linear equations.\r\n\r\nhttps:\/\/youtu.be\/M4IEmwcqR3c\r\n\r\nSometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system and solve it. In the next example, you will see a technique where we multiply one of the equations in the system by a number that will allow us to eliminate one of the variables.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by the <strong>elimination method.<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"843118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843118\"]\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<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ -3\\left(x - 2y\\right)=-3\\left(11\\right)\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Multiply both sides by }-3.\\hfill \\\\ -3x+6y=-33\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<\/div>\r\nNow, let us add them.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\end{array}[\/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{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/latex]<\/p>\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\left(3\\right)-2\\left(-4\\right)=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ 11=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222640\/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=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBelow is another video example of using the elimination method to solve a system of linear equations.\r\n\r\nhttps:\/\/youtu.be\/_liDhKops2w\r\n\r\nIn the next example, we will see that sometimes both equations need to be multiplied by different numbers in order for one variable to be eliminated.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations in two variables by elimination.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"245990\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245990\"]\r\n\r\nOne equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex], so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} -5\\left(2x+3y\\right)=-5\\left(-16\\right)\\hfill \\\\ \\text{ }-10x - 15y=80\\hfill \\\\ \\text{ }2\\left(5x - 10y\\right)=2\\left(30\\right)\\hfill \\\\ \\text{ }10x - 20y=60\\hfill \\end{array}[\/latex]<\/div>\r\nThen, we add the two equations together.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ \\:\\:10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\u221235y=140 \\\\ \\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:y=\u22124 \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y=-4[\/latex] into the original first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]<\/p>\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the second original equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]<\/div>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222643\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/> [\/hidden-answer]\r\n\r\n<\/div>\r\nBelow is a summary of the general steps for using the elimination method to solve a system of equations.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve using the elimination\u00a0method<\/h3>\r\n<ol>\r\n \t<li>Write both equations with <em>x\u00a0<\/em>and <em>y<\/em>-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. 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. 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 \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<\/div>\r\nIn the next\u00a0example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations in two variables by elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{x}{3}+\\dfrac{y}{6}=3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"288325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"288325\"]\r\n\r\nFirst clear each equation of fractions by multiplying both sides of the equation by the least common denominator\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}6\\left(\\dfrac{x}{3}+\\dfrac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\dfrac{x}{2}-\\dfrac{y}{4}\\right)=4\\left(1\\right)\\hfill \\\\ \\text{ }2x-y=4\\hfill \\end{array}[\/latex]<\/p>\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]<\/p>\r\nAdd the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y=7[\/latex] into the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\dfrac{11}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution is [latex]\\left(\\dfrac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x-y=4\\\\ 2(\\dfrac{11}{2})-7=4\\\\ 11-7=4 \\\\ 4=4\\end{array}[\/latex]<\/p>\r\n\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\nIn the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.\r\n\r\nhttps:\/\/youtu.be\/s3S64b1DrtQ\r\n\r\nRecall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or elimination, the resulting equation will be an identity such as [latex]0=0[\/latex]. The last example includes two equations that represent the same line and are therefore dependent.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind a solution to the system of equations using the <strong>elimination method<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"10390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"10390\"]\r\n\r\nWith the elimination method, we want to eliminate one of the variables by adding the equations. In this case, focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\left(-3\\right)\\left(x+3y\\right)=\\left(-3\\right)\\left(2\\right)\\hfill \\\\ \\text{ }-3x - 9y=-6\\hfill \\end{array}[\/latex]<\/p>\r\nNow add the equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array} \\hfill\u22123x\u22129y=\u22126 \\\\ \\hfill3x+9y=6 \\\\ \\hfill \\text{_____________} \\\\ \\hfill 0=0 \\end{array}[\/latex]<\/p>\r\nWe can see that there will be an infinite number of solutions that satisfy both equations.\r\n\r\nIf we rewrote both equations in slope-intercept form, we might know what the solution would look like before adding. Look at what happens when we convert the system to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\text{ }3y=-x+2\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\\\ 3x+9y=6\\hfill \\\\ \\text{ }9y=-3x+6\\hfill \\\\ \\text{ }y=-\\dfrac{3}{9}x+\\dfrac{6}{9}\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nSee the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\dfrac{1}{3}x+\\dfrac{2}{3}\\right)[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222647\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of solving a system that is dependent using elimination.\r\n\r\nhttps:\/\/youtu.be\/NRxh9Q16Ulk\r\n\r\nIn our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other.\r\n\r\nhttps:\/\/youtu.be\/z5_ACYtzW98","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve systems of equations by addition<\/li>\n<li>Express the solution of a system of dependent equations containing two variables<\/li>\n<\/ul>\n<\/div>\n<p>A third method of <strong>solving systems of linear equations<\/strong> is the <strong>elimination\u00a0method<\/strong>. In 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 elimination.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by elimination.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522070\">Show Solution<\/span><\/p>\n<div id=\"q522070\" 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,\u00a0[latex]\u20131[\/latex], is the opposite of the coefficient of [latex]x[\/latex] in the first equation,\u00a0[latex]1[\/latex]. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{\\begin{array}{l}\\hfill \\\\ \\:\\:x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}\\:\\:\\:\\:\\:\\:3y=2}[\/latex]<\/div>\n<p>&nbsp;<\/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{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\dfrac{2}{3}\\hfill \\end{array}[\/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{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\dfrac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\dfrac{2}{3}\\hfill \\\\ \\text{ }-x=\\dfrac{7}{3}\\hfill \\\\ \\text{ }\\:\\:\\:\\:\\:x=-\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }\\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\dfrac{7}{3}+\\dfrac{4}{3}=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\dfrac{3}{3}=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-1=-1\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p>We gain an important perspective on systems of equations by looking at the graphical representation. In the graph below, you will see 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<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222638\/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=\"487\" height=\"291\" \/>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see another example of how to use the method of elimination to solve a system of linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Solve a System of Equations Using the Elimination Method\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/M4IEmwcqR3c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system and solve it. In the next example, you will see a technique where we multiply one of the equations in the system by a number that will allow us to eliminate one of the variables.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by the <strong>elimination method.<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843118\">Show Solution<\/span><\/p>\n<div id=\"q843118\" 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<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ -3\\left(x - 2y\\right)=-3\\left(11\\right)\\hfill & \\hfill & \\hfill & \\text{Multiply both sides by }-3.\\hfill \\\\ -3x+6y=-33\\hfill & \\hfill & \\hfill & \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, let us add them.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\end{array}[\/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{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/latex]<\/p>\n<p>Our solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\left(3\\right)-2\\left(-4\\right)=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ 11=11\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222640\/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=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Below is another video example of using the elimination method to solve a system of linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 2:  Solve a System of Equations Using the Elimination Method\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_liDhKops2w?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we will see that sometimes both equations need to be multiplied by different numbers in order for one variable to be eliminated.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations in two variables by elimination.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q245990\">Show Solution<\/span><\/p>\n<div id=\"q245990\" class=\"hidden-answer\" style=\"display: none\">\n<p>One equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex], so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} -5\\left(2x+3y\\right)=-5\\left(-16\\right)\\hfill \\\\ \\text{ }-10x - 15y=80\\hfill \\\\ \\text{ }2\\left(5x - 10y\\right)=2\\left(30\\right)\\hfill \\\\ \\text{ }10x - 20y=60\\hfill \\end{array}[\/latex]<\/div>\n<p>Then, we add the two equations together.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ \\:\\:10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\u221235y=140 \\\\ \\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:\\:y=\u22124 \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=-4[\/latex] into the original first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the second original equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222643\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/> <\/div>\n<\/div>\n<\/div>\n<p>Below is a summary of the general steps for using the elimination method to solve a system of equations.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve using the elimination\u00a0method<\/h3>\n<ol>\n<li>Write both equations with <em>x\u00a0<\/em>and <em>y<\/em>-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. 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. 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<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<\/div>\n<p>In the next\u00a0example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations in two variables by elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{x}{3}+\\dfrac{y}{6}=3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q288325\">Show Solution<\/span><\/p>\n<div id=\"q288325\" class=\"hidden-answer\" style=\"display: none\">\n<p>First clear each equation of fractions by multiplying both sides of the equation by the least common denominator<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}6\\left(\\dfrac{x}{3}+\\dfrac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\dfrac{x}{2}-\\dfrac{y}{4}\\right)=4\\left(1\\right)\\hfill \\\\ \\text{ }2x-y=4\\hfill \\end{array}[\/latex]<\/p>\n<p>Now multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]<\/p>\n<p>Add the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=7[\/latex] into the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\dfrac{11}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\dfrac{11}{2},7\\right)[\/latex]. Check it in the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x-y=4\\\\ 2(\\dfrac{11}{2})-7=4\\\\ 11-7=4 \\\\ 4=4\\end{array}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex: Solve a System of Equations Using Eliminations (Fractions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/s3S64b1DrtQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Recall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or elimination, the resulting equation will be an identity such as [latex]0=0[\/latex]. The last example includes two equations that represent the same line and are therefore dependent.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find a solution to the system of equations using the <strong>elimination method<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q10390\">Show Solution<\/span><\/p>\n<div id=\"q10390\" class=\"hidden-answer\" style=\"display: none\">\n<p>With the elimination method, we want to eliminate one of the variables by adding the equations. In this case, focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\left(-3\\right)\\left(x+3y\\right)=\\left(-3\\right)\\left(2\\right)\\hfill \\\\ \\text{ }-3x - 9y=-6\\hfill \\end{array}[\/latex]<\/p>\n<p>Now add the equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array} \\hfill\u22123x\u22129y=\u22126 \\\\ \\hfill3x+9y=6 \\\\ \\hfill \\text{_____________} \\\\ \\hfill 0=0 \\end{array}[\/latex]<\/p>\n<p>We can see that there will be an infinite number of solutions that satisfy both equations.<\/p>\n<p>If we rewrote both equations in slope-intercept form, we might know what the solution would look like before adding. Look at what happens when we convert the system to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\text{ }3y=-x+2\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\\\ 3x+9y=6\\hfill \\\\ \\text{ }9y=-3x+6\\hfill \\\\ \\text{ }y=-\\dfrac{3}{9}x+\\dfrac{6}{9}\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>See the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\dfrac{1}{3}x+\\dfrac{2}{3}\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222647\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of solving a system that is dependent using elimination.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  System of Equations Using Elimination (Infinite Solutions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/NRxh9Q16Ulk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex:  System of Equations Using Elimination (No Solution)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/z5_ACYtzW98?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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-2080\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Solve a System of Equations Using the Elimination Method. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/M4IEmwcqR3c\">https:\/\/youtu.be\/M4IEmwcqR3c<\/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 2: Solve a System of Equations Using the Elimination Method. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/_liDhKops2w\">https:\/\/youtu.be\/_liDhKops2w<\/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 Eliminations (Fractions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/s3S64b1DrtQ\">https:\/\/youtu.be\/s3S64b1DrtQ<\/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: System of Equations Using Elimination (Infinite Solutions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/NRxh9Q16Ulk\">https:\/\/youtu.be\/NRxh9Q16Ulk<\/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: System of Equations Using Elimination (No Solution). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/z5_ACYtzW98\">https:\/\/youtu.be\/z5_ACYtzW98<\/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":21,"menu_order":5,"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 1: Solve a System of Equations Using the Elimination Method\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/M4IEmwcqR3c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solve a System of Equations Using the Elimination Method\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/_liDhKops2w\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve a System of Equations Using Eliminations (Fractions)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/s3S64b1DrtQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Using Elimination (Infinite Solutions)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/NRxh9Q16Ulk\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Using Elimination (No Solution)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/z5_ACYtzW98\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"128ab918-da6f-4cb7-a53a-5e59438b5923","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2080","chapter","type-chapter","status-publish","hentry"],"part":2069,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2080","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":26,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2080\/revisions"}],"predecessor-version":[{"id":5524,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2080\/revisions\/5524"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/parts\/2069"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2080\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/media?parent=2080"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2080"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/contributor?post=2080"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/license?post=2080"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}