{"id":1150,"date":"2023-07-22T08:07:11","date_gmt":"2023-07-22T08:07:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/?post_type=chapter&#038;p=1150"},"modified":"2023-08-13T05:46:31","modified_gmt":"2023-08-13T05:46:31","slug":"solving-rational-inequalities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solving-rational-inequalities\/","title":{"raw":"\u25aa   Solving Rational Inequalities","rendered":"\u25aa   Solving Rational Inequalities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve rational inequalities using boundary value method.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Solving Rational Inequalities using Boundary Value Method<\/h2>\r\nAny inequality that can be put into one of the following forms\r\n<p style=\"padding-left: 30px;\">[latex]f(x)&gt;0, f(x) \\geq 0, f(x)&lt;0, or f(x) \\leq 0[\/latex], where [latex]f[\/latex] is a rational function<\/p>\r\nis called <strong>rational inequality<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>How To: Solve Rational Inequalities<\/h3>\r\n<ol>\r\n \t<li>Find restriction(s) of the given rational expression(s) in the inequality. The restriction(s) is(are) the <strong>boundary point(s)<\/strong>.<\/li>\r\n \t<li>Rewrite the given rational inequality as an equation by replacing the inequality symbol with the equal sign.<\/li>\r\n \t<li>Solve the rational equation. The real solution(s) of the equation is(are) the <strong>boundary point(s)<\/strong>.<\/li>\r\n \t<li>Plot the boundary point(s) from Step 1 &amp; 3 on a number line.\r\n[latex] \\Rightarrow [\/latex] Use an open circle ALL restrictions.\r\n[latex] \\Rightarrow [\/latex] Use an open circle when the given inequality has [latex]&lt;[\/latex] or [latex]&gt;[\/latex]\r\n[latex]\\Rightarrow[\/latex] Use a closed circle when the given inequality has [latex]\\leq[\/latex] or [latex]\\geq[\/latex].<\/li>\r\n \t<li>Choose one number, which is called a <strong>test value<\/strong>, from each interval and test the intervals by evaluating the given inequality at that number.\r\n[latex]\\Rightarrow[\/latex] If the inequality is TRUE, then the interval is a solution of the inequality.\r\n[latex]\\Rightarrow[\/latex] If the inequality is FALSE, then the interval is not a solution of the inequality.<\/li>\r\n \t<li>Write the solution set (usually in interval notation), selecting the interval(s) from Step 5.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Rational Inequality using Boundary Value Method<\/h3>\r\nSolve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{2x-1}{x+2} &gt; 1[\/latex]<\/p>\r\n[reveal-answer q=\"958747\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"958747\"]\r\n\r\nFirst, find restriction(s):\r\n<p style=\"padding-left: 30px;\">[latex]x+2 \\ne 0[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]x \\ne -2[\/latex]<\/p>\r\nSecond, let's rewrite the inequality as an equation and then solve it:\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{2x-1}{x+2} = 1[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{2x-1}{x+2} = \\frac{1}{1}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]2x-1=x+2[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]x=3[\/latex]<\/p>\r\nThird, plot the restriction and the solution on the number line. All restrictions should be an\u00a0<strong>open<\/strong> circle. Since the inequality has \"[latex]&gt;[\/latex]\" (greater than), which doesn't have \"[latex]=[\/latex]\" (equal sign), the solution should be an\u00a0<strong>open<\/strong> circle as well:\r\n<p style=\"padding-left: 30px;\"><img class=\"aligncenter wp-image-1170\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5.png\" alt=\"Number line plotted -2 (open) as restriction, 4 (open) as solution, marked -3, 0, 4 as test values\" width=\"366\" height=\"68\" \/><\/p>\r\n<p style=\"padding-left: 30px;\">Note that now there are three intervals on the number line: [latex](-\\infty, -2), (-2, 3),[\/latex] and [latex](3, \\infty)[\/latex].<\/p>\r\nFourth, choose your choice of test value from each interval:\r\n<p style=\"padding-left: 30px;\">I chose [latex]-3, 0,[\/latex] and [latex]4[\/latex] from those three intervals.<\/p>\r\nfifth, test each interval using the test value:\r\n<p style=\"padding-left: 30px;\">When [latex]x=-3[\/latex], [latex]\\frac{2(-3)-1}{(-3)+2} &gt; 1 \\rightarrow \\frac{-7}{-1}&gt;1 \\rightarrow 7&gt;1[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\r\n<p style=\"padding-left: 30px;\">When [latex]x=-3[\/latex], [latex]\\frac{2(0)-1}{(0)+2} &gt; 1 \\rightarrow \\frac{-1}{2}&gt;1 \\rightarrow -\\frac{1}{2}&gt;1[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\r\n<p style=\"padding-left: 30px;\">When [latex]x=4[\/latex], [latex]\\frac{2(4)-1}{(4)+2} &gt; 1 \\rightarrow \\frac{7}{6}&gt;1[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\r\n<img class=\"aligncenter wp-image-1171\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1.png\" alt=\"Number line plotted -2 (open) as restriction, 4 (open) as solution, marked -3, 0, 4 as test values, first and third intervals are marked in red\" width=\"367\" height=\"68\" \/>\r\n<p style=\"padding-left: 30px;\">Actually, we can check the results of test values from the graph of [latex]f(x)=\\frac{2x-1}{x+2}-1[\/latex] below:<\/p>\r\n<img class=\"aligncenter wp-image-1172\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2.png\" alt=\"Graph of f(x)=(2x-1)\/(x+2)-1, marked above x-axis in red\" width=\"309\" height=\"238\" \/>\r\n\r\nTherefore, the solution of the rational inequality\u00a0[latex]\\frac{2x-1}{x+2} &gt; 1[\/latex] is [latex](-\\infty, -2) \\cup (3, \\infty)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{5x}{x-3} \\leq 2[\/latex]<\/p>\r\n[reveal-answer q=\"431373\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"431373\"]\r\n\r\nFirst, find restriction(s):\r\n<p style=\"padding-left: 30px;\">[latex]x-3 \\ne 0[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]x \\ne 3[\/latex]<\/p>\r\nSecond, let's rewrite the inequality as an equation and then solve it:\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{5x}{x-3}=2[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{5x}{x-3} = \\frac{2}{1}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]5x=2(x-3)[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]5x=2x-6[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]3x=-6[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]x=-2[\/latex]<\/p>\r\nThird, plot the restriction and the solution on the number line. All restrictions should be an\u00a0<strong>open<\/strong> circle. Since the inequality has \"[latex]\\leq[\/latex]\" (less than or equal to), which has \"[latex]=[\/latex]\" (equal sign), the solution should be a\u00a0<strong>closed<\/strong> circle:\r\n<p style=\"padding-left: 30px;\"><img class=\"aligncenter wp-image-1173\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6.png\" alt=\"Number line plotted 3 (open) as restriction, -2 (closed) as solution, marked -3, 0, 4 as test values\" width=\"360\" height=\"70\" \/><\/p>\r\n<p style=\"padding-left: 30px;\">Note that now there are three intervals on the number line: [latex](-\\infty, -2], [-2, 3),[\/latex] and [latex](3, \\infty)[\/latex].<\/p>\r\nFourth, choose your choice of test value from each interval:\r\n<p style=\"padding-left: 30px;\">I chose [latex]-3, 0,[\/latex] and [latex]4[\/latex] from those three intervals.<\/p>\r\nfifth, test each interval using the test value:\r\n<p style=\"padding-left: 30px;\">When [latex]x=-3[\/latex], [latex]\\frac{5(-3)}{(-3)-3} \\leq 2 \\rightarrow \\frac{-15}{-6} \\leq 2 \\rightarrow 2.5 \\leq 2[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\r\n<p style=\"padding-left: 30px;\">When [latex]x=0[\/latex], [latex]\\frac{5(0)}{(0)-3} \\leq 2 \\rightarrow \\frac{0}{-3} \\leq 2 \\rightarrow 0 \\leq 2[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\r\n<p style=\"padding-left: 30px;\">When [latex]x=4[\/latex], [latex]\\frac{5(4)}{(4)-3} \\leq 2 \\rightarrow \\frac{20}{1} \\leq 2 \\rightarrow 20 \\leq 2[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\r\n<img class=\"aligncenter wp-image-1176\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3.png\" alt=\"Number line plotted 3 (open) as restriction, -2 (closed) as solution, marked -3, 0, 4 as test values, middle intervals are marked in red\" width=\"360\" height=\"79\" \/>\r\n<p style=\"padding-left: 30px;\">Actually, we can check the results of test values from the graph of [latex]f(x)=\\frac{5x}{x-3}-2[\/latex] below:<\/p>\r\n<img class=\"aligncenter wp-image-1175\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2.png\" alt=\"Graph of f(x)=5x\/(x-3)-2, marked below x-axis in red\" width=\"282\" height=\"238\" \/>\r\n\r\nTherefore, the solution of the rational inequality\u00a0[latex]\\frac{5x}{x-3} \\leq 2[\/latex] is [latex][-2, 3)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{4x-5}{x+1} \\geq 0[\/latex]<\/p>\r\n[reveal-answer q=\"607532\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"607532\"]\r\n\r\nFirst, find restriction(s):\r\n<p style=\"padding-left: 30px;\">[latex]x+1 \\ne 0[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]x \\ne -1[\/latex]<\/p>\r\nSecond, let's rewrite the inequality as an equation and then solve it:\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{4x-5}{x+1}=0[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]\\frac{4x-5}{x+1} = \\frac{0}{1}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]4x-5=0[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]4x=5[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]x=\\frac{5}{4}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">Third, plot the restriction and the solution on the number line. All restrictions should be an\u00a0<strong>open<\/strong> circle. Since the inequality has \"[latex]\\geq[\/latex]\" (greater than or equal to), which has \"[latex]=[\/latex]\" (equal sign), the solution should be a\u00a0<strong>closed<\/strong> circle:<\/p>\r\n<p style=\"padding-left: 30px;\"><img class=\"aligncenter wp-image-1177\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7.png\" alt=\"Number line plotted -1 (open) as restriction, 5\/4 (closed) as solution, marked -2, 0, 2 as test values\" width=\"381\" height=\"90\" \/><\/p>\r\n<p style=\"padding-left: 30px;\">Note that now there are three intervals on the number line: [latex](-\\infty, -1), (-1, \\frac{5}{4}],[\/latex] and [latex][\\frac{5}{4}, \\infty)[\/latex].<\/p>\r\nFourth, choose your choice of test value from each interval:\r\n<p style=\"padding-left: 30px;\">I chose [latex]-2, 0,[\/latex] and [latex]2[\/latex] from those three intervals.<\/p>\r\nfifth, test each interval using the test value:\r\n<p style=\"padding-left: 30px;\">When [latex]x=-2[\/latex], [latex]\\frac{4(-2)-5}{(-2)+1} \\geq 0 \\rightarrow \\frac{-13}{-1} \\geq 0 \\rightarrow 13 \\geq 0[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\r\n<p style=\"padding-left: 30px;\">When [latex]x=0[\/latex], [latex]\\frac{4(0)-5}{(0)+1} \\geq 0 \\rightarrow \\frac{-5}{1} \\geq 0 \\rightarrow -5 \\geq 0[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\r\n<p style=\"padding-left: 30px;\">When [latex]x=2[\/latex], [latex]\\frac{4(2)-5}{(2)+1} \\geq 0 \\rightarrow \\frac{3}{3} \\geq 0 \\rightarrow 1 \\geq 0[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\r\n<img class=\"aligncenter wp-image-1178\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1.png\" alt=\"Number line plotted -1 (open) as restriction, 5\/4 (closed) as solution, marked -2, 0, 2 as test values, first and third intervals are marked in red\" width=\"382\" height=\"89\" \/>\r\n<p style=\"padding-left: 30px;\">Actually, we can check the results of test values from the graph of [latex]f(x)=\\frac{4x-5}{x+1}[\/latex] below:<\/p>\r\n<img class=\"aligncenter wp-image-1179\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2.png\" alt=\"Graph of f(x)=(4x-5)\/(x+1), marked above x-axis in red\" width=\"282\" height=\"245\" \/>\r\n\r\nTherefore, the solution of the rational inequality\u00a0[latex]\\frac{4x-5}{x+1} \\geq 0[\/latex] is [latex](-\\infty, -1) \\cup [\\frac{5}{4}, \\infty)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve rational inequalities using boundary value method.<\/li>\n<\/ul>\n<\/div>\n<h2>Solving Rational Inequalities using Boundary Value Method<\/h2>\n<p>Any inequality that can be put into one of the following forms<\/p>\n<p style=\"padding-left: 30px;\">[latex]f(x)>0, f(x) \\geq 0, f(x)<0, or f(x) \\leq 0[\/latex], where [latex]f[\/latex] is a rational function<\/p>\n<p>is called <strong>rational inequality<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>How To: Solve Rational Inequalities<\/h3>\n<ol>\n<li>Find restriction(s) of the given rational expression(s) in the inequality. The restriction(s) is(are) the <strong>boundary point(s)<\/strong>.<\/li>\n<li>Rewrite the given rational inequality as an equation by replacing the inequality symbol with the equal sign.<\/li>\n<li>Solve the rational equation. The real solution(s) of the equation is(are) the <strong>boundary point(s)<\/strong>.<\/li>\n<li>Plot the boundary point(s) from Step 1 &amp; 3 on a number line.<br \/>\n[latex]\\Rightarrow[\/latex] Use an open circle ALL restrictions.<br \/>\n[latex]\\Rightarrow[\/latex] Use an open circle when the given inequality has [latex]<[\/latex] or [latex]>[\/latex]<br \/>\n[latex]\\Rightarrow[\/latex] Use a closed circle when the given inequality has [latex]\\leq[\/latex] or [latex]\\geq[\/latex].<\/li>\n<li>Choose one number, which is called a <strong>test value<\/strong>, from each interval and test the intervals by evaluating the given inequality at that number.<br \/>\n[latex]\\Rightarrow[\/latex] If the inequality is TRUE, then the interval is a solution of the inequality.<br \/>\n[latex]\\Rightarrow[\/latex] If the inequality is FALSE, then the interval is not a solution of the inequality.<\/li>\n<li>Write the solution set (usually in interval notation), selecting the interval(s) from Step 5.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Rational Inequality using Boundary Value Method<\/h3>\n<p>Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{2x-1}{x+2} > 1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q958747\">Show Answer<\/span><\/p>\n<div id=\"q958747\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find restriction(s):<\/p>\n<p style=\"padding-left: 30px;\">[latex]x+2 \\ne 0[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]x \\ne -2[\/latex]<\/p>\n<p>Second, let&#8217;s rewrite the inequality as an equation and then solve it:<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{2x-1}{x+2} = 1[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{2x-1}{x+2} = \\frac{1}{1}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]2x-1=x+2[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]x=3[\/latex]<\/p>\n<p>Third, plot the restriction and the solution on the number line. All restrictions should be an\u00a0<strong>open<\/strong> circle. Since the inequality has &#8220;[latex]>[\/latex]&#8221; (greater than), which doesn&#8217;t have &#8220;[latex]=[\/latex]&#8221; (equal sign), the solution should be an\u00a0<strong>open<\/strong> circle as well:<\/p>\n<p style=\"padding-left: 30px;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1170\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5.png\" alt=\"Number line plotted -2 (open) as restriction, 4 (open) as solution, marked -3, 0, 4 as test values\" width=\"366\" height=\"68\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5.png 560w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-300x56.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-65x12.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-225x42.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-350x65.png 350w\" sizes=\"auto, (max-width: 366px) 100vw, 366px\" \/><\/p>\n<p style=\"padding-left: 30px;\">Note that now there are three intervals on the number line: [latex](-\\infty, -2), (-2, 3),[\/latex] and [latex](3, \\infty)[\/latex].<\/p>\n<p>Fourth, choose your choice of test value from each interval:<\/p>\n<p style=\"padding-left: 30px;\">I chose [latex]-3, 0,[\/latex] and [latex]4[\/latex] from those three intervals.<\/p>\n<p>fifth, test each interval using the test value:<\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=-3[\/latex], [latex]\\frac{2(-3)-1}{(-3)+2} > 1 \\rightarrow \\frac{-7}{-1}>1 \\rightarrow 7>1[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=-3[\/latex], [latex]\\frac{2(0)-1}{(0)+2} > 1 \\rightarrow \\frac{-1}{2}>1 \\rightarrow -\\frac{1}{2}>1[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=4[\/latex], [latex]\\frac{2(4)-1}{(4)+2} > 1 \\rightarrow \\frac{7}{6}>1[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1171\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1.png\" alt=\"Number line plotted -2 (open) as restriction, 4 (open) as solution, marked -3, 0, 4 as test values, first and third intervals are marked in red\" width=\"367\" height=\"68\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1.png 556w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1-300x56.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1-65x12.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1-225x42.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-1-350x65.png 350w\" sizes=\"auto, (max-width: 367px) 100vw, 367px\" \/><\/p>\n<p style=\"padding-left: 30px;\">Actually, we can check the results of test values from the graph of [latex]f(x)=\\frac{2x-1}{x+2}-1[\/latex] below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1172\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2.png\" alt=\"Graph of f(x)=(2x-1)\/(x+2)-1, marked above x-axis in red\" width=\"309\" height=\"238\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2.png 415w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2-300x231.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2-65x50.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2-225x173.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-5-2-350x269.png 350w\" sizes=\"auto, (max-width: 309px) 100vw, 309px\" \/><\/p>\n<p>Therefore, the solution of the rational inequality\u00a0[latex]\\frac{2x-1}{x+2} > 1[\/latex] is [latex](-\\infty, -2) \\cup (3, \\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{5x}{x-3} \\leq 2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q431373\">Show Answer<\/span><\/p>\n<div id=\"q431373\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find restriction(s):<\/p>\n<p style=\"padding-left: 30px;\">[latex]x-3 \\ne 0[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]x \\ne 3[\/latex]<\/p>\n<p>Second, let&#8217;s rewrite the inequality as an equation and then solve it:<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{5x}{x-3}=2[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{5x}{x-3} = \\frac{2}{1}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]5x=2(x-3)[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]5x=2x-6[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]3x=-6[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]x=-2[\/latex]<\/p>\n<p>Third, plot the restriction and the solution on the number line. All restrictions should be an\u00a0<strong>open<\/strong> circle. Since the inequality has &#8220;[latex]\\leq[\/latex]&#8221; (less than or equal to), which has &#8220;[latex]=[\/latex]&#8221; (equal sign), the solution should be a\u00a0<strong>closed<\/strong> circle:<\/p>\n<p style=\"padding-left: 30px;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1173\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6.png\" alt=\"Number line plotted 3 (open) as restriction, -2 (closed) as solution, marked -3, 0, 4 as test values\" width=\"360\" height=\"70\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6.png 560w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-300x58.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-65x13.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-225x44.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-350x68.png 350w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/p>\n<p style=\"padding-left: 30px;\">Note that now there are three intervals on the number line: [latex](-\\infty, -2], [-2, 3),[\/latex] and [latex](3, \\infty)[\/latex].<\/p>\n<p>Fourth, choose your choice of test value from each interval:<\/p>\n<p style=\"padding-left: 30px;\">I chose [latex]-3, 0,[\/latex] and [latex]4[\/latex] from those three intervals.<\/p>\n<p>fifth, test each interval using the test value:<\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=-3[\/latex], [latex]\\frac{5(-3)}{(-3)-3} \\leq 2 \\rightarrow \\frac{-15}{-6} \\leq 2 \\rightarrow 2.5 \\leq 2[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=0[\/latex], [latex]\\frac{5(0)}{(0)-3} \\leq 2 \\rightarrow \\frac{0}{-3} \\leq 2 \\rightarrow 0 \\leq 2[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=4[\/latex], [latex]\\frac{5(4)}{(4)-3} \\leq 2 \\rightarrow \\frac{20}{1} \\leq 2 \\rightarrow 20 \\leq 2[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1176\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3.png\" alt=\"Number line plotted 3 (open) as restriction, -2 (closed) as solution, marked -3, 0, 4 as test values, middle intervals are marked in red\" width=\"360\" height=\"79\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3.png 589w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3-300x66.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3-65x14.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3-225x49.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-3-350x77.png 350w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/p>\n<p style=\"padding-left: 30px;\">Actually, we can check the results of test values from the graph of [latex]f(x)=\\frac{5x}{x-3}-2[\/latex] below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1175\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2.png\" alt=\"Graph of f(x)=5x\/(x-3)-2, marked below x-axis in red\" width=\"282\" height=\"238\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2.png 360w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2-300x253.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2-65x55.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2-225x190.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-6-2-350x296.png 350w\" sizes=\"auto, (max-width: 282px) 100vw, 282px\" \/><\/p>\n<p>Therefore, the solution of the rational inequality\u00a0[latex]\\frac{5x}{x-3} \\leq 2[\/latex] is [latex][-2, 3)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{4x-5}{x+1} \\geq 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q607532\">Show Answer<\/span><\/p>\n<div id=\"q607532\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find restriction(s):<\/p>\n<p style=\"padding-left: 30px;\">[latex]x+1 \\ne 0[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]x \\ne -1[\/latex]<\/p>\n<p>Second, let&#8217;s rewrite the inequality as an equation and then solve it:<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{4x-5}{x+1}=0[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac{4x-5}{x+1} = \\frac{0}{1}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]4x-5=0[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]4x=5[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]x=\\frac{5}{4}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">Third, plot the restriction and the solution on the number line. All restrictions should be an\u00a0<strong>open<\/strong> circle. Since the inequality has &#8220;[latex]\\geq[\/latex]&#8221; (greater than or equal to), which has &#8220;[latex]=[\/latex]&#8221; (equal sign), the solution should be a\u00a0<strong>closed<\/strong> circle:<\/p>\n<p style=\"padding-left: 30px;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1177\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7.png\" alt=\"Number line plotted -1 (open) as restriction, 5\/4 (closed) as solution, marked -2, 0, 2 as test values\" width=\"381\" height=\"90\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7.png 596w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-300x71.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-65x15.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-225x53.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-350x83.png 350w\" sizes=\"auto, (max-width: 381px) 100vw, 381px\" \/><\/p>\n<p style=\"padding-left: 30px;\">Note that now there are three intervals on the number line: [latex](-\\infty, -1), (-1, \\frac{5}{4}],[\/latex] and [latex][\\frac{5}{4}, \\infty)[\/latex].<\/p>\n<p>Fourth, choose your choice of test value from each interval:<\/p>\n<p style=\"padding-left: 30px;\">I chose [latex]-2, 0,[\/latex] and [latex]2[\/latex] from those three intervals.<\/p>\n<p>fifth, test each interval using the test value:<\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=-2[\/latex], [latex]\\frac{4(-2)-5}{(-2)+1} \\geq 0 \\rightarrow \\frac{-13}{-1} \\geq 0 \\rightarrow 13 \\geq 0[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=0[\/latex], [latex]\\frac{4(0)-5}{(0)+1} \\geq 0 \\rightarrow \\frac{-5}{1} \\geq 0 \\rightarrow -5 \\geq 0[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(False)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>not a solution<\/strong><\/p>\n<p style=\"padding-left: 30px;\">When [latex]x=2[\/latex], [latex]\\frac{4(2)-5}{(2)+1} \\geq 0 \\rightarrow \\frac{3}{3} \\geq 0 \\rightarrow 1 \\geq 0[\/latex]\u00a0<span style=\"color: #ff0000;\"><strong>(True)<\/strong><\/span>\u00a0 [latex]\\therefore[\/latex] <strong>solution<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1178\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1.png\" alt=\"Number line plotted -1 (open) as restriction, 5\/4 (closed) as solution, marked -2, 0, 2 as test values, first and third intervals are marked in red\" width=\"382\" height=\"89\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1.png 587w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1-300x70.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1-65x15.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1-225x53.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-1-350x82.png 350w\" sizes=\"auto, (max-width: 382px) 100vw, 382px\" \/><\/p>\n<p style=\"padding-left: 30px;\">Actually, we can check the results of test values from the graph of [latex]f(x)=\\frac{4x-5}{x+1}[\/latex] below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1179\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2.png\" alt=\"Graph of f(x)=(4x-5)\/(x+1), marked above x-axis in red\" width=\"282\" height=\"245\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2.png 389w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2-300x260.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2-65x56.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2-225x195.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.6-Poly-Ineq-7-2-350x303.png 350w\" sizes=\"auto, (max-width: 282px) 100vw, 282px\" \/><\/p>\n<p>Therefore, the solution of the rational inequality\u00a0[latex]\\frac{4x-5}{x+1} \\geq 0[\/latex] is [latex](-\\infty, -1) \\cup [\\frac{5}{4}, \\infty)[\/latex].<\/p>\n<\/div>\n<\/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-1150\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li> Solving Rational Inequalities. <strong>Authored by<\/strong>: Michelle Eunhee Chung. <strong>Provided by<\/strong>: Georgia State University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":705214,"menu_order":37,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\" Solving Rational Inequalities\",\"author\":\"Michelle Eunhee Chung\",\"organization\":\"Georgia State University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["mchung12"],"pb_section_license":""},"chapter-type":[],"contributor":[62],"license":[],"class_list":["post-1150","chapter","type-chapter","status-publish","hentry","contributor-mchung12"],"part":91,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1150","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/705214"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1150\/revisions"}],"predecessor-version":[{"id":1348,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1150\/revisions\/1348"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/91"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1150\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=1150"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1150"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1150"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=1150"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}