{"id":1380,"date":"2025-07-24T19:14:43","date_gmt":"2025-07-24T19:14:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1380"},"modified":"2026-03-24T19:22:48","modified_gmt":"2026-03-24T19:22:48","slug":"solving-systems-with-cramers-rule-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-cramers-rule-learn-it-2\/","title":{"raw":"Solving Systems with Cramer's Rule: Learn It 2","rendered":"Solving Systems with Cramer&#8217;s Rule: Learn It 2"},"content":{"raw":"<h2>Using Cramer\u2019s Rule to Solve a System of Two Equations in Two Variables<\/h2>\r\nWe will now introduce a final method for solving systems of equations that uses determinants. Known as <strong>Cramer\u2019s Rule<\/strong>, this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704\u20131752), who introduced it in 1750 in Introduction \u00e0 l'Analyse des lignes Courbes alg\u00e9briques. Cramer\u2019s Rule is a viable and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns.\r\n\r\nCramer\u2019s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used.\r\n\r\nTo understand Cramer\u2019s Rule, let\u2019s look closely at how we solve systems of linear equations using basic row operations.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">Consider a system of two equations in two variables.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}x+{b}_{1}y&amp;={c}_{1}&amp;&amp;R_{1}\\\\ {a}_{2}x+{b}_{2}y&amp;={c}_{2}&amp;&amp;R_2\\end{align}[\/latex]<\/div>\r\nWe eliminate one variable using row operations and solve for the other. Say that we wish to solve for [latex]x[\/latex]. If equation (2) is multiplied by the opposite of the coefficient of [latex]y[\/latex] in equation (1), equation (1) is multiplied by the coefficient of [latex]y[\/latex] in equation (2), and we add the two equations, the variable [latex]y[\/latex] will be eliminated.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}b_{2}a_{1}x+b_{2}b_{1}y&amp;=b_{2}c_{1} \\\\ \u2212b_{1}a_{2}x\u2212b_{1}b_{2}y&amp;=\u2212b_{1}c_{2} \\\\ \\hline b_{2}a_{1}x\u2212b_{1}a_{2}x&amp;=\u2212b_{2}c_{1}\u2212b_{1}c_{2}\\end{align}[\/latex] [latex]\\begin{align}&amp;\\text{Multiply }R_{1}\\text{ by }b_{2} \\\\ &amp;\\text{Multiply }R_{2}\\text{ by }\u2212b_{2} \\\\ \\text{ } \\end{align}[\/latex]<\/p>\r\nNow, solve for [latex]x[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{b}_{2}{a}_{1}x-{b}_{1}{a}_{2}x={b}_{2}{c}_{1}-{b}_{1}{c}_{2}\\\\ \\hfill \\\\ x\\left({b}_{2}{a}_{1}-{b}_{1}{a}_{2}\\right)={b}_{2}{c}_{1}-{b}_{1}{c}_{2}\\\\ \\hfill \\\\ x=\\frac{{b}_{2}{c}_{1}-{b}_{1}{c}_{2}}{{b}_{2}{a}_{1}-{b}_{1}{a}_{2}}=\\frac{\\left\\rvert\\begin{array}{cc}{c}_{1}&amp; {b}_{1}\\\\ {c}_{2}&amp; {b}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right\\rvert}\\hfill \\end{gathered}[\/latex]<\/div>\r\nSimilarly, to solve for [latex]y[\/latex], we will eliminate [latex]x[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}a_{2}a_{1}x+a_{2}b_{1}y&amp;=a_{2}c_{1} \\\\\u2212a_{1}a_{2}x\u2212a_{1}b_{2}y&amp;=\u2212a_{1}c_{2} \\\\ \\hline a_{2}b_{1}y\u2212a_{1}b_{2}y&amp;=a_{2}c_{1}\u2212a_{1}c_{2}\\end{align}[\/latex] [latex]\\begin{align}&amp;\\text{Multiply }R_{1}\\text{ by }a_{2} \\\\&amp;\\text{Multiply }R_{2}\\text{ by }\u2212a_{1} \\\\ \\text{ } \\end{align}[\/latex]<\/div>\r\nSolving for [latex]y[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{a}_{2}{b}_{1}y-{a}_{1}{b}_{2}y={a}_{2}{c}_{1}-{a}_{1}{c}_{2} \\\\ y\\left({a}_{2}{b}_{1}-{a}_{1}{b}_{2}\\right)={a}_{2}{c}_{1}-{a}_{1}{c}_{2} \\\\ y=\\frac{{a}_{2}{c}_{1}-{a}_{1}{c}_{2}}{{a}_{2}{b}_{1}-{a}_{1}{b}_{2}}=\\frac{{a}_{1}{c}_{2}-{a}_{2}{c}_{1}}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}=\\frac{\\left\\rvert\\begin{array}{cc}{a}_{1}&amp; {c}_{1}\\\\ {a}_{2}&amp; {c}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right\\rvert} \\end{gathered}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/section>Notice that the denominator for both [latex]x[\/latex] and [latex]y[\/latex] is the determinant of the coefficient matrix.\r\n\r\nWe can use these formulas to solve for [latex]x[\/latex] and [latex]y[\/latex], but Cramer\u2019s Rule also introduces new notation:\r\n<ul>\r\n \t<li>[latex]D:[\/latex] determinant of the coefficient matrix<\/li>\r\n \t<li>[latex]{D}_{x}:[\/latex] determinant of the numerator in the solution of [latex]x[\/latex]\r\n<div style=\"text-align: center;\">[latex]x=\\frac{{D}_{x}}{D}[\/latex]<\/div><\/li>\r\n \t<li>[latex]{D}_{y}:[\/latex] determinant of the numerator in the solution of [latex]y[\/latex]\r\n<div style=\"text-align: center;\">[latex]y=\\frac{{D}_{y}}{D}[\/latex]<\/div><\/li>\r\n<\/ul>\r\nThe key to Cramer\u2019s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express [latex]x[\/latex] and [latex]y[\/latex] as a quotient of two determinants.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Cramer's Rule for 2x2 systems<\/h3>\r\n<strong>Cramer\u2019s Rule<\/strong> is a method that uses determinants to solve systems of equations that have the same number of equations as variables.\r\n\r\nConsider a system of two linear equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/p>\r\nThe solution using Cramer\u2019s Rule is given as\r\n<p style=\"text-align: center;\">[latex]x=\\frac{{D}_{x}}{D}=\\frac{\\left\\rvert\\begin{array}{cc}{c}_{1}&amp; {b}_{1}\\\\ {c}_{2}&amp; {b}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right\\rvert},D\\ne 0;\\text{ }\\text{ }y=\\frac{{D}_{y}}{D}=\\frac{\\left\\rvert\\begin{array}{cc}{a}_{1}&amp; {c}_{1}\\\\ {a}_{2}&amp; {c}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}&amp; {b}_{1}\\\\ {a}_{2}&amp; {b}_{2}\\end{array}\\right\\rvert},D\\ne 0[\/latex].<\/p>\r\nIf we are solving for [latex]x[\/latex], the [latex]x[\/latex] column is replaced with the constant column. If we are solving for [latex]y[\/latex], the [latex]y[\/latex] column is replaced with the constant column.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the following [latex]2\\text{ }\\times \\text{ }2[\/latex] system using Cramer\u2019s Rule.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}12x+3y&amp;=15\\\\ 2x - 3y&amp;=13\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"472105\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"472105\"]\r\n\r\nSolve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]x=\\frac{{D}_{x}}{D}=\\frac{\\left\\rvert\\begin{array}{rr}\\hfill 15&amp; \\hfill 3\\\\ \\hfill 13&amp; \\hfill -3\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{rr}\\hfill 12&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill -3\\end{array}\\right\\rvert}=\\frac{-45 - 39}{-36 - 6}=\\frac{-84}{-42}=2[\/latex]<\/p>\r\nSolve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]y=\\frac{{D}_{y}}{D}=\\frac{\\left\\rvert\\begin{array}{rr}\\hfill 12&amp; \\hfill 15\\\\ \\hfill 2&amp; \\hfill 13\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{rr}\\hfill 12&amp; \\hfill 3\\\\ \\hfill 2&amp; \\hfill -3\\end{array}\\right\\rvert}=\\frac{156 - 30}{-36 - 6}=-\\frac{126}{42}=-3[\/latex]<\/p>\r\nThe solution is [latex]\\left(2,-3\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321768[\/ohm_question]<\/section>","rendered":"<h2>Using Cramer\u2019s Rule to Solve a System of Two Equations in Two Variables<\/h2>\n<p>We will now introduce a final method for solving systems of equations that uses determinants. Known as <strong>Cramer\u2019s Rule<\/strong>, this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704\u20131752), who introduced it in 1750 in Introduction \u00e0 l&#8217;Analyse des lignes Courbes alg\u00e9briques. Cramer\u2019s Rule is a viable and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns.<\/p>\n<p>Cramer\u2019s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used.<\/p>\n<p>To understand Cramer\u2019s Rule, let\u2019s look closely at how we solve systems of linear equations using basic row operations.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">Consider a system of two equations in two variables.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}x+{b}_{1}y&={c}_{1}&&R_{1}\\\\ {a}_{2}x+{b}_{2}y&={c}_{2}&&R_2\\end{align}[\/latex]<\/div>\n<p>We eliminate one variable using row operations and solve for the other. Say that we wish to solve for [latex]x[\/latex]. If equation (2) is multiplied by the opposite of the coefficient of [latex]y[\/latex] in equation (1), equation (1) is multiplied by the coefficient of [latex]y[\/latex] in equation (2), and we add the two equations, the variable [latex]y[\/latex] will be eliminated.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}b_{2}a_{1}x+b_{2}b_{1}y&=b_{2}c_{1} \\\\ \u2212b_{1}a_{2}x\u2212b_{1}b_{2}y&=\u2212b_{1}c_{2} \\\\ \\hline b_{2}a_{1}x\u2212b_{1}a_{2}x&=\u2212b_{2}c_{1}\u2212b_{1}c_{2}\\end{align}[\/latex] [latex]\\begin{align}&\\text{Multiply }R_{1}\\text{ by }b_{2} \\\\ &\\text{Multiply }R_{2}\\text{ by }\u2212b_{2} \\\\ \\text{ } \\end{align}[\/latex]<\/p>\n<p>Now, solve for [latex]x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{b}_{2}{a}_{1}x-{b}_{1}{a}_{2}x={b}_{2}{c}_{1}-{b}_{1}{c}_{2}\\\\ \\hfill \\\\ x\\left({b}_{2}{a}_{1}-{b}_{1}{a}_{2}\\right)={b}_{2}{c}_{1}-{b}_{1}{c}_{2}\\\\ \\hfill \\\\ x=\\frac{{b}_{2}{c}_{1}-{b}_{1}{c}_{2}}{{b}_{2}{a}_{1}-{b}_{1}{a}_{2}}=\\frac{\\left\\rvert\\begin{array}{cc}{c}_{1}& {b}_{1}\\\\ {c}_{2}& {b}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right\\rvert}\\hfill \\end{gathered}[\/latex]<\/div>\n<p>Similarly, to solve for [latex]y[\/latex], we will eliminate [latex]x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}a_{2}a_{1}x+a_{2}b_{1}y&=a_{2}c_{1} \\\\\u2212a_{1}a_{2}x\u2212a_{1}b_{2}y&=\u2212a_{1}c_{2} \\\\ \\hline a_{2}b_{1}y\u2212a_{1}b_{2}y&=a_{2}c_{1}\u2212a_{1}c_{2}\\end{align}[\/latex] [latex]\\begin{align}&\\text{Multiply }R_{1}\\text{ by }a_{2} \\\\&\\text{Multiply }R_{2}\\text{ by }\u2212a_{1} \\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<p>Solving for [latex]y[\/latex] gives<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{a}_{2}{b}_{1}y-{a}_{1}{b}_{2}y={a}_{2}{c}_{1}-{a}_{1}{c}_{2} \\\\ y\\left({a}_{2}{b}_{1}-{a}_{1}{b}_{2}\\right)={a}_{2}{c}_{1}-{a}_{1}{c}_{2} \\\\ y=\\frac{{a}_{2}{c}_{1}-{a}_{1}{c}_{2}}{{a}_{2}{b}_{1}-{a}_{1}{b}_{2}}=\\frac{{a}_{1}{c}_{2}-{a}_{2}{c}_{1}}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}=\\frac{\\left\\rvert\\begin{array}{cc}{a}_{1}& {c}_{1}\\\\ {a}_{2}& {c}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right\\rvert} \\end{gathered}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/section>\n<p>Notice that the denominator for both [latex]x[\/latex] and [latex]y[\/latex] is the determinant of the coefficient matrix.<\/p>\n<p>We can use these formulas to solve for [latex]x[\/latex] and [latex]y[\/latex], but Cramer\u2019s Rule also introduces new notation:<\/p>\n<ul>\n<li>[latex]D:[\/latex] determinant of the coefficient matrix<\/li>\n<li>[latex]{D}_{x}:[\/latex] determinant of the numerator in the solution of [latex]x[\/latex]\n<div style=\"text-align: center;\">[latex]x=\\frac{{D}_{x}}{D}[\/latex]<\/div>\n<\/li>\n<li>[latex]{D}_{y}:[\/latex] determinant of the numerator in the solution of [latex]y[\/latex]\n<div style=\"text-align: center;\">[latex]y=\\frac{{D}_{y}}{D}[\/latex]<\/div>\n<\/li>\n<\/ul>\n<p>The key to Cramer\u2019s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express [latex]x[\/latex] and [latex]y[\/latex] as a quotient of two determinants.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Cramer&#8217;s Rule for 2&#215;2 systems<\/h3>\n<p><strong>Cramer\u2019s Rule<\/strong> is a method that uses determinants to solve systems of equations that have the same number of equations as variables.<\/p>\n<p>Consider a system of two linear equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\\\ {a}_{2}x+{b}_{2}y={c}_{2}\\end{array}[\/latex]<\/p>\n<p>The solution using Cramer\u2019s Rule is given as<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{{D}_{x}}{D}=\\frac{\\left\\rvert\\begin{array}{cc}{c}_{1}& {b}_{1}\\\\ {c}_{2}& {b}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right\\rvert},D\\ne 0;\\text{ }\\text{ }y=\\frac{{D}_{y}}{D}=\\frac{\\left\\rvert\\begin{array}{cc}{a}_{1}& {c}_{1}\\\\ {a}_{2}& {c}_{2}\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{cc}{a}_{1}& {b}_{1}\\\\ {a}_{2}& {b}_{2}\\end{array}\\right\\rvert},D\\ne 0[\/latex].<\/p>\n<p>If we are solving for [latex]x[\/latex], the [latex]x[\/latex] column is replaced with the constant column. If we are solving for [latex]y[\/latex], the [latex]y[\/latex] column is replaced with the constant column.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the following [latex]2\\text{ }\\times \\text{ }2[\/latex] system using Cramer\u2019s Rule.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}12x+3y&=15\\\\ 2x - 3y&=13\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q472105\">Show Solution<\/button><\/p>\n<div id=\"q472105\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{{D}_{x}}{D}=\\frac{\\left\\rvert\\begin{array}{rr}\\hfill 15& \\hfill 3\\\\ \\hfill 13& \\hfill -3\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{rr}\\hfill 12& \\hfill 3\\\\ \\hfill 2& \\hfill -3\\end{array}\\right\\rvert}=\\frac{-45 - 39}{-36 - 6}=\\frac{-84}{-42}=2[\/latex]<\/p>\n<p>Solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]y=\\frac{{D}_{y}}{D}=\\frac{\\left\\rvert\\begin{array}{rr}\\hfill 12& \\hfill 15\\\\ \\hfill 2& \\hfill 13\\end{array}\\right\\rvert}{\\left\\rvert\\begin{array}{rr}\\hfill 12& \\hfill 3\\\\ \\hfill 2& \\hfill -3\\end{array}\\right\\rvert}=\\frac{156 - 30}{-36 - 6}=-\\frac{126}{42}=-3[\/latex]<\/p>\n<p>The solution is [latex]\\left(2,-3\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321768\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321768&theme=lumen&iframe_resize_id=ohm321768&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":22,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":514,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1380"}],"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\/13"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1380\/revisions"}],"predecessor-version":[{"id":5999,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1380\/revisions\/5999"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/514"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1380\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1380"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1380"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1380"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1380"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}