{"id":2868,"date":"2022-06-20T23:58:01","date_gmt":"2022-06-20T23:58:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=2868"},"modified":"2026-01-18T03:36:45","modified_gmt":"2026-01-18T03:36:45","slug":"7-6-algebraic-analysis-on-intersection-points","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/7-6-algebraic-analysis-on-intersection-points\/","title":{"raw":"7.6: Algebraic Analysis on Intersection Points","rendered":"7.6: Algebraic Analysis on Intersection Points"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\n<ul>\r\n \t<li>Describe the meaning of solving rational equations<\/li>\r\n \t<li>Solve rational equations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Meaning of Solving Exponential Equations<\/h2>\r\n<h3>Intersection points of functions<\/h3>\r\nIn chapter 3, we learned that the meaning of solving an equation is to find the intersection point(s) between two functions.\u00a0The intersection point(s) between the graphs of any two functions [latex]f(x)[\/latex] and\u00a0[latex]g(x)[\/latex]\u00a0can be found algebraically by setting the two functions equal to each other:\r\n<p style=\"text-align: center;\">[latex]f(x)=g(x)[\/latex]<\/p>\r\nWhen the functions are equal, the value of [latex]x[\/latex]\u00a0is the same for both functions, as is the function value. In other words, [latex]f(x)=g(x)[\/latex]\u00a0means that the two functions have the same input\u00a0[latex]x[\/latex] as well as the same output (i.e\u00a0[latex]f(x)=g(x)[\/latex]).\u00a0For example, solving the equation [latex]-\\dfrac{1}{x}=\\dfrac{2}{x}+3[\/latex] is equivalent to\u00a0determining the intersection point between the two functions [latex]f(x)=-\\dfrac{1}{x}[\/latex] and [latex]g(x)=\\dfrac{2}{x}+3[\/latex]. The solution is [latex]x=-1[\/latex]\u00a0<span style=\"font-size: 1em;\">(figure 1)<\/span><span style=\"font-size: 1rem; text-align: initial;\">.<\/span>\r\n\r\n[caption id=\"attachment_3447\" align=\"aligncenter\" width=\"417\"]<img class=\"wp-image-3447\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/25224244\/7-6-Meaning1-300x300.png\" alt=\"Blue graph of f(x)=-1\/x and green graph of g(x)=(2\/x) + 3 intersecting at the point (-1,1).\" width=\"417\" height=\"417\" \/> Figure 1. The Intersection of two functions at the point [latex](-1, 1)[\/latex].[\/caption]&nbsp;\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nUse graphing to solve the rational equation [latex]\\dfrac{-1}{x-4}=\\dfrac{1}{x^2+2x}[\/latex].\r\n<h4>Solution<\/h4>\r\nLet [latex]f(x)=\\dfrac{-1}{x-4}[\/latex] and [latex]g(x)=\\dfrac{1}{x^2+2x}[\/latex]\r\n\r\nThen use <a href=\"http:\/\/desmos.com\/calculator\">Desmos<\/a> to graph the functions.\r\n\r\nLook for the intersection points of the two graphs.\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-3346 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-300x300.png\" alt=\"Blue graph of f(x)=-1\/(x-4) and green graph of g(x)=1\/(x^2+2x) intersecting at two points where the x coordinates are -4 and 1.\" width=\"300\" height=\"300\" \/><\/p>\r\n&nbsp;\r\n\r\nThe graphs intersect at [latex]x=-4[\/latex] and [latex]x=1[\/latex].\r\n\r\nTherefore, the solution of the equation is\u00a0[latex]x=-4,\\,1[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nUse graphing to solve the rational equation [latex]\\dfrac{1}{x^2}=\\dfrac{1}{2x+3}[\/latex].\r\n\r\n[reveal-answer q=\"hjm662\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm662\"]\r\n\r\n<img class=\"aligncenter wp-image-3347 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-300x300.png\" alt=\"Blue graph of g(x)=1\/(2x+3) and green graph of f(x)=1\/x^2 intersecting at two points where the x coordinates are -1 and 3.\" width=\"300\" height=\"300\" \/>\r\n\r\n[latex]x=-1,\\,3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h4>Finding the [latex]x[\/latex]-value given a function value<\/h4>\r\nGiven the equation\u00a0[latex]\\dfrac{8}{x+1}=2[\/latex], in addition to the interpretation of finding the intersection point between the two functions\u00a0[latex]f(x)=\\dfrac{8}{x+1}[\/latex] and [latex]g(x)=2[\/latex], another interpretation of the equation is finding the [latex]x[\/latex] value when the function value of [latex]f(x)=\\dfrac{8}{x+1}[\/latex] is 2 (figure 2).\r\n\r\n[caption id=\"attachment_2945\" align=\"aligncenter\" width=\"380\"]<img class=\"wp-image-2945\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/22153535\/7-6-Meaning2-300x300.png\" alt=\"Graph of f(x)=8\/(x+1) showing a point on the curve with an unknown x-coordinate and a y-coordinate of 2.\" width=\"380\" height=\"380\" \/> Figure 2. Find the [latex]x[\/latex] coordinate given that the [latex]y[\/latex] coordinate is 2.[\/caption]Graphically, this means finding the [latex]x[\/latex] value for a given [latex]y[\/latex]-value on a graph. Determine where the [latex]y[\/latex]-value meets the graph then move vertically down to the [latex]x[\/latex]-axis to determine the corresponding [latex]x[\/latex]-value. In figure 2, to solve [latex]\\dfrac{8}{x+1}=2[\/latex], we graph the function [latex]f(x)=\\dfrac{8}{x+1}[\/latex], look for 2 on the [latex]y[\/latex]-axis then determine which [latex]x[\/latex]-value has a function value at 2. In this case, [latex]x=3[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nGraphically solve the equation [latex]\\dfrac{1}{x+3}=-1[\/latex].\r\n<h4>Solution<\/h4>\r\nLet [latex]f(x)=\\dfrac{1}{x+3}[\/latex]. Use <a href=\"http:\/\/desmos.con\/calculator\">Desmos<\/a> to graph the function then determine where the [latex]y[\/latex]-value equals \u20131. Then determine the corresponding [latex]x[\/latex]-value.\r\n\r\n<img class=\"aligncenter size-medium wp-image-3348\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/19015344\/desmos-graph-2022-07-18T195126.423-300x300.png\" alt=\"f(x)=1\/(x+3)\" width=\"300\" height=\"300\" \/>\r\n<h4>Answer<\/h4>\r\n[latex]x=-4[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nGraphically solve the equation [latex]\\dfrac{1}{x-4}=1[\/latex].\r\n\r\n[reveal-answer q=\"hjm857\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm857\"]\r\n\r\n<img class=\"aligncenter size-medium wp-image-3349\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/19020023\/desmos-graph-2022-07-18T195942.121-300x300.png\" alt=\"f(x)=1\/(x-4)\" width=\"300\" height=\"300\" \/>\r\n\r\n[latex]x=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIt's more likely that the [latex]x[\/latex]-value given any function value will not be an integer. This can be difficult to show on a graph. Consequently, it is necessary to use algebraic methods to solve rational equations.\r\n<h3>Solving an equation in one variable<\/h3>\r\nAlgebraically, [latex]-\\dfrac{1}{x}=\\dfrac{2}{x}+4[\/latex] or [latex]\\dfrac{8}{x+1}=2[\/latex] is an equation in one variable. When solving an equation in one variable, we find the value of the variable that satisfies the equation (e.g., the [latex]x[\/latex]-value). There is no function value to report as the equation is in just one variable.\r\n\r\nFor example, when solving the equation, [latex]\\dfrac{8}{x+1}=2[\/latex], we find the value of [latex]x[\/latex] that makes the equation true. The value of\u00a0[latex]x[\/latex] is 3 because [latex]\\dfrac{8}{3+1}=\\dfrac{8}{4}=2[\/latex]. We need algebraic methods to solve equations, including the properties of equality covered in chapter 3.\r\n\r\nIn summary, when we set functions equal to each other, we are finding the intersection point between two functions (e.g., [latex]f(x)=\\dfrac{8}{x+1}[\/latex] and [latex]g(x)=2[\/latex]). In this example, the intersection point between the graphs of the two functions [latex]f(x)[\/latex] and\u00a0[latex]g(x)[\/latex]\u00a0is [latex](3, 2)[\/latex] because the [latex]y[\/latex]-value is the function value [latex]f(3)=\\dfrac{8}{3+1}=2[\/latex] and [latex]g(3)=2[\/latex]. However, when we are solving an equation algebraically, there is no function in sight so we can just report the value of the variable (e.g., [latex]x=3[\/latex]).\r\n<h2>Solving Rational Equations<\/h2>\r\nWe have already learned to add\/subtract and multiply\/divide rational expressions. We will use these skills to clear fractions (or rational expressions) in a rational equation by multiplying both sides by the lowest common denominator. After clearing the rational expressions, we will no longer have denominators to deal with and it will be easier for us to solve the resulting equation for the variable.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nFind the intersection point between the two functions [latex]f(x)=\\dfrac{1}{x+3}[\/latex] and\u00a0[latex]g(x)=\\dfrac{1}{3x-5}[\/latex].\r\n<h4>Solution<\/h4>\r\nTo find the intersection point of [latex]f(x)[\/latex] and [latex]g(x)[\/latex], we set the functions equal to one another, [latex]f(x)=g(x)[\/latex], and then solve for\u00a0[latex]x[\/latex]. The value of [latex]x[\/latex] will be the [latex]x[\/latex]-coordinate of the intersection point. We will then substitute the variable\u00a0[latex]x[\/latex] into either\u00a0[latex]f(x)[\/latex] or\u00a0[latex]g(x)[\/latex] to determine the [latex]y[\/latex] coordinate of the intersection point.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=g(x)\\\\\\\\\\dfrac{1}{x+3}&amp;=\\dfrac{1}{3x-5}\\end{aligned}[\/latex]<\/p>\r\nBefore we solve the equation for [latex]x[\/latex], we need to discuss the restrictions on the variable. Restrictions on the domain of a rational function occur when the denominator equals zero: In [latex]f(x)[\/latex], [latex]x+3=0[\/latex] solves to [latex]x=-3[\/latex]. In [latex]g(x)[\/latex], [latex]3x-5=0[\/latex] solves to [latex]x=\\dfrac{5}{3}[\/latex]. Therefore, [latex]-3, \\dfrac{5}{3}[\/latex] are restricted values of [latex]x[\/latex].\r\n\r\nNow, we will start to solve the equation by clearing out the fractions. The LCM of [latex](x+3)[\/latex] and [latex](3x-5)[\/latex] is [latex](x+3)(3x-5)[\/latex], so we will multiply both sides of the equation by\u00a0[latex](x+3)(3x-5)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{1}{x+3}&amp;=\\dfrac{1}{3x-5}\\\\\\\\ \\color{blue}{(x+3)(3x-5)}\\times\\dfrac{1}{(x+3)}&amp;=\\dfrac{1}{3x-5}\\times\\color{blue}{(x+3)(3x-5)}&amp;&amp;\\text{Multiply buy the LCM on both sides}\\\\\\\\\\cancel{(x+3)}(3x-5)\\times\\dfrac{1}{\\cancel{x+3}}&amp;=\\dfrac{1}{\\cancel{3x-5}}\\times(x+3)\\cancel{(3x-5)}&amp;&amp;\\text{Cancel common factors}\\\\\\\\3x-5&amp;=x+3&amp;&amp;\\text{This is a linear equation}\\\\\\\\3x-5\\color{blue}{-x}&amp;=x+3\\color{blue}{-x}&amp;&amp;\\text{Subtract }x\\text{ from both sides}\\\\\\\\2x-5&amp;=3\\\\\\\\2x-5\\color{blue}{+5}&amp;=3\\color{blue}{+5}&amp;&amp;\\text{Add }5\\text{ to both sides}\\\\\\\\2x&amp;=8\\\\\\\\\\dfrac{2x}{\\color{blue}{2}}&amp;=\\dfrac{2}{\\color{blue}{2}}&amp;&amp;\\text{Divide both sides by }2\\\\\\\\x&amp;=4\\end{aligned}[\/latex]<\/p>\r\n[latex]x=4[\/latex] is not a restricted value, so we have our\u00a0[latex]x[\/latex]-coordinate of the intersection point.\r\n\r\nNow, that we know the value for [latex]x[\/latex] at the intersection point, we evaluate either [latex]f(4)[\/latex] or [latex]g(4)[\/latex] to find the [latex]y[\/latex]-coordinate.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(4)&amp;=\\dfrac{1}{4+3}\\\\\\\\&amp;=\\dfrac{1}{7}\\end{aligned}[\/latex]<\/p>\r\nTherefore, the intersection point between the two functions\u00a0[latex]f(x)=\\dfrac{1}{x+3}[\/latex] and\u00a0[latex]g(x)=\\dfrac{1}{3x-5}[\/latex] is [latex]\\left(4, \\dfrac{1}{7}\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nFind the intersection point between the two functions [latex]f(x)=\\dfrac{2}{x-4}[\/latex] and\u00a0[latex]g(x)=\\dfrac{1}{2x-6}[\/latex].\r\n\r\n[reveal-answer q=\"hjm220\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm220\"]\r\n\r\n[latex]\\left(\\dfrac{8}{3},\\,-\\dfrac{3}{2}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nSolve the rational equation [latex]2+\\dfrac{4}{y-2}=\\dfrac{8}{y^2-2y}[\/latex].\r\n<h4>Solution<\/h4>\r\nBefore we solve the equation for [latex]y[\/latex], let's determine the restrictions on the variable by setting each denominator to zero: [latex]y-2=0[\/latex] solves to [latex]y=2[\/latex], and [latex]y^2-2y=0[\/latex] factors to [latex]y(y-2)=0[\/latex] then solves to [latex]y=0,\\,2[\/latex]. Therefore, [latex]y\\neq0, 2[\/latex].\r\n\r\nNow, we will start to solve the equation by clearing the fractions by multiplying by the LCM [latex]y(y-2)[\/latex].\r\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{aligned}2+\\dfrac{4}{y-2}&amp;=\\dfrac{8}{y^2-2y}\\\\\\\\2+\\dfrac{4}{y-2}&amp;=\\dfrac{8}{y(y-2)}&amp;&amp;\\text{Factor denominators}\\\\\\\\ \\color{blue}{y(y-2)} \\times \\left(2+\\dfrac{4}{y-2}\\right)&amp;=\\dfrac{8}{y(y-2)} \\times\\color{blue}{y(y-2)}&amp;&amp;\\text{Multiply both sides by the LCM}\\\\\\\\2\\cdot\\color{blue}{y(y-2)}+\\dfrac{4}{(y-2)}\\cdot\\color{blue}{y(y-2)}&amp;=\\dfrac{8}{y(y-2)}\\cdot\\color{blue}{y(y-2)}&amp;&amp;\\text{Distribute on the left side}\\\\\\\\2y(y-2) + y\\cancel{(y-2)}\\cdot\\dfrac{4}{\\cancel{(y-2)}}&amp;=\\dfrac{8}{\\cancel{y}\\cancel{(y-2)}} \\cdot \\cancel{y}\\cancel{(y-2)}&amp;&amp;\\text{Cancel common factors}\\\\\\\\2y^2-4y+4y&amp;=8&amp;&amp;\\text{Simplify. We have a quadratic equation.}\\\\\\\\2y^2-8&amp;=0&amp;&amp;\\text{Subtract 8 from both sides}\\\\\\\\2\\left(y^2-4\\right)&amp;=0&amp;&amp;\\text{Factor}\\\\\\\\2(y-2)(y+2)&amp;=0&amp;&amp;\\text{Factor: difference of two squares}\\\\\\\\y-2=0\\text{ or }y+2&amp;=0&amp;&amp;\\text{Zero product property}\\\\\\\\y&amp;=\\pm 2&amp;&amp;\\text{Solve}\\end{aligned}[\/latex]<\/p>\r\nSince [latex]2[\/latex] is a restricted value, the solution is [latex]y=-2[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nSolve the rational equation [latex]\\dfrac{4}{x-4}-\\dfrac{3}{x-3}=1[\/latex]. Remember to check the restrictions on the variable.\r\n\r\n[reveal-answer q=\"hjm037\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm037\"]\r\n\r\n[latex]x=2,\\,6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nSolve the rational equation [latex]\\dfrac{7}{y^2+y-12}-\\dfrac{4y}{y^2+7y+12}=\\dfrac{6}{y^2-9}[\/latex].\r\n<h4>Solution<\/h4>\r\nLet's start by determining the restrictions on the variable.\u00a0 [latex]y^2+y-12=0[\/latex] factors to [latex](y+4)(y-3)=0[\/latex]. So, [latex]y=-4, 3[\/latex]. Next, [latex]y^2+7y+12=0[\/latex] factors to [latex](y+4)(y+3)=0[\/latex]. So, [latex]y=-4, -3[\/latex]. Finally, [latex]y^2-9=0[\/latex] factors to [latex](y-3)(y+3)=0[\/latex]. So, [latex]y=3, -3[\/latex]. Therefore, [latex]y\\neq\\pm3,\\,-4[\/latex].\r\n\r\nAdditionally, the LCM is [latex](y+4)(y-3)(y+3)[\/latex].\r\n\r\nNow, we will start to solve the equation by multiplying both sides by the LCM to clear the fractions.\r\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{aligned}\\dfrac{7}{y^2+y-12}-\\dfrac{4y}{y^2+7y+12}&amp;=\\dfrac{6}{y^2-9}\\\\\\\\\\dfrac{7}{(y+4)(y-3)}-\\dfrac{4y}{(y+4)(y+3)}&amp;=\\dfrac{6}{(y+3)(y-3)}&amp;&amp;\\text{Factor}\\\\\\\\\\color{blue}{(y+4)(y+3)(y-3)\\cdot}\\left(\\dfrac{7}{(y+4)(y-3)}-\\dfrac{4y}{(y+4)(y+3)}\\right)&amp;=\\dfrac{6}{(y+3)(y-3)}\\color{blue}{\\cdot(y+4)(y+3)(y-3)}&amp;&amp;\\text{Multiply} \\\\&amp;&amp;&amp;\\text{by LCM}\\\\\\\\\\color{blue}{(y+4)(y+3)(y-3)}\\cdot\\dfrac{7}{(y+4)(y-3)}&amp;-\\color{blue}{(y+4)(y+3)(y-3)}\\cdot\\dfrac{4y}{(y+4)(y+3)}\\\\\\\\&amp;=\\dfrac{6}{(y+3)(y-3)}\\cdot(y+4)(y+3)(y-3)&amp;&amp;\\text{Distribute}\\\\\\\\\\cancel{(y+4)}(y+3)\\cancel{(y-3)}\\cdot\\dfrac{7}{\\cancel{(y+4)}\\cancel{(y-3)}}&amp;-\\cancel{(y+4)}\\cancel{(y+3)}(y-3)\\cdot\\dfrac{4y}{\\cancel{(y+4)}\\cancel{(y+3)}}\\\\\\\\&amp;=\\dfrac{6}{\\cancel{(y+3)}\\cancel{(y-3)}}\\cdot(y+4)\\cancel{(y+3)}\\cancel{(y-3)}&amp;&amp;\\text{Cancel}\\\\\\\\7(y+3)-4y(y-3)&amp;=6(y+4)&amp;&amp;\\text{Simplify}\\\\\\\\7y+21-4y^2+12y&amp;=6y+24&amp;&amp;\\text{Distribute}\\\\\\\\4y^2-13y+3&amp;=0&amp;&amp;\\text{Simplify}\\\\\\\\(4y-1)(y-3)&amp;=0&amp;&amp;\\text{Factor}\\\\\\\\4y-1=0\\text{ or }y-3&amp;=0&amp;&amp;\\text{Zero product}\\\\&amp;&amp;&amp;\\text {property}\\\\\\\\y&amp;=\\dfrac{1}{4},\\,3\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">However, [latex]y=3[\/latex] is a restricted value since [latex]y\\neq\\pm3,\\,-4[\/latex].<\/p>\r\nConsequently, the only solution is\u00a0[latex]y=\\dfrac{1}{4}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nSolve the rational equation [latex]\\dfrac{x+11}{x^2-5x+4}=\\dfrac{5}{x-4}-\\dfrac{3}{x-1}[\/latex].\r\n\r\n[reveal-answer q=\"hjm166\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm166\"]The equation has no solution.[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<ul>\n<li>Describe the meaning of solving rational equations<\/li>\n<li>Solve rational equations<\/li>\n<\/ul>\n<\/div>\n<h2>The Meaning of Solving Exponential Equations<\/h2>\n<h3>Intersection points of functions<\/h3>\n<p>In chapter 3, we learned that the meaning of solving an equation is to find the intersection point(s) between two functions.\u00a0The intersection point(s) between the graphs of any two functions [latex]f(x)[\/latex] and\u00a0[latex]g(x)[\/latex]\u00a0can be found algebraically by setting the two functions equal to each other:<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=g(x)[\/latex]<\/p>\n<p>When the functions are equal, the value of [latex]x[\/latex]\u00a0is the same for both functions, as is the function value. In other words, [latex]f(x)=g(x)[\/latex]\u00a0means that the two functions have the same input\u00a0[latex]x[\/latex] as well as the same output (i.e\u00a0[latex]f(x)=g(x)[\/latex]).\u00a0For example, solving the equation [latex]-\\dfrac{1}{x}=\\dfrac{2}{x}+3[\/latex] is equivalent to\u00a0determining the intersection point between the two functions [latex]f(x)=-\\dfrac{1}{x}[\/latex] and [latex]g(x)=\\dfrac{2}{x}+3[\/latex]. The solution is [latex]x=-1[\/latex]\u00a0<span style=\"font-size: 1em;\">(figure 1)<\/span><span style=\"font-size: 1rem; text-align: initial;\">.<\/span><\/p>\n<div id=\"attachment_3447\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3447\" class=\"wp-image-3447\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/25224244\/7-6-Meaning1-300x300.png\" alt=\"Blue graph of f(x)=-1\/x and green graph of g(x)=(2\/x) + 3 intersecting at the point (-1,1).\" width=\"417\" height=\"417\" \/><\/p>\n<p id=\"caption-attachment-3447\" class=\"wp-caption-text\">Figure 1. The Intersection of two functions at the point [latex](-1, 1)[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Use graphing to solve the rational equation [latex]\\dfrac{-1}{x-4}=\\dfrac{1}{x^2+2x}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Let [latex]f(x)=\\dfrac{-1}{x-4}[\/latex] and [latex]g(x)=\\dfrac{1}{x^2+2x}[\/latex]<\/p>\n<p>Then use <a href=\"http:\/\/desmos.com\/calculator\">Desmos<\/a> to graph the functions.<\/p>\n<p>Look for the intersection points of the two graphs.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3346 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-300x300.png\" alt=\"Blue graph of f(x)=-1\/(x-4) and green graph of g(x)=1\/(x^2+2x) intersecting at two points where the x coordinates are -4 and 1.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T192607.467.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>The graphs intersect at [latex]x=-4[\/latex] and [latex]x=1[\/latex].<\/p>\n<p>Therefore, the solution of the equation is\u00a0[latex]x=-4,\\,1[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Use graphing to solve the rational equation [latex]\\dfrac{1}{x^2}=\\dfrac{1}{2x+3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm662\">Show Answer<\/span><\/p>\n<div id=\"qhjm662\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3347 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-300x300.png\" alt=\"Blue graph of g(x)=1\/(2x+3) and green graph of f(x)=1\/x^2 intersecting at two points where the x coordinates are -1 and 3.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/06\/desmos-graph-2022-07-18T193543.692.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]x=-1,\\,3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h4>Finding the [latex]x[\/latex]-value given a function value<\/h4>\n<p>Given the equation\u00a0[latex]\\dfrac{8}{x+1}=2[\/latex], in addition to the interpretation of finding the intersection point between the two functions\u00a0[latex]f(x)=\\dfrac{8}{x+1}[\/latex] and [latex]g(x)=2[\/latex], another interpretation of the equation is finding the [latex]x[\/latex] value when the function value of [latex]f(x)=\\dfrac{8}{x+1}[\/latex] is 2 (figure 2).<\/p>\n<div id=\"attachment_2945\" style=\"width: 390px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2945\" class=\"wp-image-2945\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/22153535\/7-6-Meaning2-300x300.png\" alt=\"Graph of f(x)=8\/(x+1) showing a point on the curve with an unknown x-coordinate and a y-coordinate of 2.\" width=\"380\" height=\"380\" \/><\/p>\n<p id=\"caption-attachment-2945\" class=\"wp-caption-text\">Figure 2. Find the [latex]x[\/latex] coordinate given that the [latex]y[\/latex] coordinate is 2.<\/p>\n<\/div>\n<p>Graphically, this means finding the [latex]x[\/latex] value for a given [latex]y[\/latex]-value on a graph. Determine where the [latex]y[\/latex]-value meets the graph then move vertically down to the [latex]x[\/latex]-axis to determine the corresponding [latex]x[\/latex]-value. In figure 2, to solve [latex]\\dfrac{8}{x+1}=2[\/latex], we graph the function [latex]f(x)=\\dfrac{8}{x+1}[\/latex], look for 2 on the [latex]y[\/latex]-axis then determine which [latex]x[\/latex]-value has a function value at 2. In this case, [latex]x=3[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Graphically solve the equation [latex]\\dfrac{1}{x+3}=-1[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Let [latex]f(x)=\\dfrac{1}{x+3}[\/latex]. Use <a href=\"http:\/\/desmos.con\/calculator\">Desmos<\/a> to graph the function then determine where the [latex]y[\/latex]-value equals \u20131. Then determine the corresponding [latex]x[\/latex]-value.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-3348\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/19015344\/desmos-graph-2022-07-18T195126.423-300x300.png\" alt=\"f(x)=1\/(x+3)\" width=\"300\" height=\"300\" \/><\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-4[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Graphically solve the equation [latex]\\dfrac{1}{x-4}=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm857\">Show Answer<\/span><\/p>\n<div id=\"qhjm857\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-3349\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/06\/19020023\/desmos-graph-2022-07-18T195942.121-300x300.png\" alt=\"f(x)=1\/(x-4)\" width=\"300\" height=\"300\" \/><\/p>\n<p>[latex]x=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>It&#8217;s more likely that the [latex]x[\/latex]-value given any function value will not be an integer. This can be difficult to show on a graph. Consequently, it is necessary to use algebraic methods to solve rational equations.<\/p>\n<h3>Solving an equation in one variable<\/h3>\n<p>Algebraically, [latex]-\\dfrac{1}{x}=\\dfrac{2}{x}+4[\/latex] or [latex]\\dfrac{8}{x+1}=2[\/latex] is an equation in one variable. When solving an equation in one variable, we find the value of the variable that satisfies the equation (e.g., the [latex]x[\/latex]-value). There is no function value to report as the equation is in just one variable.<\/p>\n<p>For example, when solving the equation, [latex]\\dfrac{8}{x+1}=2[\/latex], we find the value of [latex]x[\/latex] that makes the equation true. The value of\u00a0[latex]x[\/latex] is 3 because [latex]\\dfrac{8}{3+1}=\\dfrac{8}{4}=2[\/latex]. We need algebraic methods to solve equations, including the properties of equality covered in chapter 3.<\/p>\n<p>In summary, when we set functions equal to each other, we are finding the intersection point between two functions (e.g., [latex]f(x)=\\dfrac{8}{x+1}[\/latex] and [latex]g(x)=2[\/latex]). In this example, the intersection point between the graphs of the two functions [latex]f(x)[\/latex] and\u00a0[latex]g(x)[\/latex]\u00a0is [latex](3, 2)[\/latex] because the [latex]y[\/latex]-value is the function value [latex]f(3)=\\dfrac{8}{3+1}=2[\/latex] and [latex]g(3)=2[\/latex]. However, when we are solving an equation algebraically, there is no function in sight so we can just report the value of the variable (e.g., [latex]x=3[\/latex]).<\/p>\n<h2>Solving Rational Equations<\/h2>\n<p>We have already learned to add\/subtract and multiply\/divide rational expressions. We will use these skills to clear fractions (or rational expressions) in a rational equation by multiplying both sides by the lowest common denominator. After clearing the rational expressions, we will no longer have denominators to deal with and it will be easier for us to solve the resulting equation for the variable.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Find the intersection point between the two functions [latex]f(x)=\\dfrac{1}{x+3}[\/latex] and\u00a0[latex]g(x)=\\dfrac{1}{3x-5}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>To find the intersection point of [latex]f(x)[\/latex] and [latex]g(x)[\/latex], we set the functions equal to one another, [latex]f(x)=g(x)[\/latex], and then solve for\u00a0[latex]x[\/latex]. The value of [latex]x[\/latex] will be the [latex]x[\/latex]-coordinate of the intersection point. We will then substitute the variable\u00a0[latex]x[\/latex] into either\u00a0[latex]f(x)[\/latex] or\u00a0[latex]g(x)[\/latex] to determine the [latex]y[\/latex] coordinate of the intersection point.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=g(x)\\\\\\\\\\dfrac{1}{x+3}&=\\dfrac{1}{3x-5}\\end{aligned}[\/latex]<\/p>\n<p>Before we solve the equation for [latex]x[\/latex], we need to discuss the restrictions on the variable. Restrictions on the domain of a rational function occur when the denominator equals zero: In [latex]f(x)[\/latex], [latex]x+3=0[\/latex] solves to [latex]x=-3[\/latex]. In [latex]g(x)[\/latex], [latex]3x-5=0[\/latex] solves to [latex]x=\\dfrac{5}{3}[\/latex]. Therefore, [latex]-3, \\dfrac{5}{3}[\/latex] are restricted values of [latex]x[\/latex].<\/p>\n<p>Now, we will start to solve the equation by clearing out the fractions. The LCM of [latex](x+3)[\/latex] and [latex](3x-5)[\/latex] is [latex](x+3)(3x-5)[\/latex], so we will multiply both sides of the equation by\u00a0[latex](x+3)(3x-5)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{1}{x+3}&=\\dfrac{1}{3x-5}\\\\\\\\ \\color{blue}{(x+3)(3x-5)}\\times\\dfrac{1}{(x+3)}&=\\dfrac{1}{3x-5}\\times\\color{blue}{(x+3)(3x-5)}&&\\text{Multiply buy the LCM on both sides}\\\\\\\\\\cancel{(x+3)}(3x-5)\\times\\dfrac{1}{\\cancel{x+3}}&=\\dfrac{1}{\\cancel{3x-5}}\\times(x+3)\\cancel{(3x-5)}&&\\text{Cancel common factors}\\\\\\\\3x-5&=x+3&&\\text{This is a linear equation}\\\\\\\\3x-5\\color{blue}{-x}&=x+3\\color{blue}{-x}&&\\text{Subtract }x\\text{ from both sides}\\\\\\\\2x-5&=3\\\\\\\\2x-5\\color{blue}{+5}&=3\\color{blue}{+5}&&\\text{Add }5\\text{ to both sides}\\\\\\\\2x&=8\\\\\\\\\\dfrac{2x}{\\color{blue}{2}}&=\\dfrac{2}{\\color{blue}{2}}&&\\text{Divide both sides by }2\\\\\\\\x&=4\\end{aligned}[\/latex]<\/p>\n<p>[latex]x=4[\/latex] is not a restricted value, so we have our\u00a0[latex]x[\/latex]-coordinate of the intersection point.<\/p>\n<p>Now, that we know the value for [latex]x[\/latex] at the intersection point, we evaluate either [latex]f(4)[\/latex] or [latex]g(4)[\/latex] to find the [latex]y[\/latex]-coordinate.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(4)&=\\dfrac{1}{4+3}\\\\\\\\&=\\dfrac{1}{7}\\end{aligned}[\/latex]<\/p>\n<p>Therefore, the intersection point between the two functions\u00a0[latex]f(x)=\\dfrac{1}{x+3}[\/latex] and\u00a0[latex]g(x)=\\dfrac{1}{3x-5}[\/latex] is [latex]\\left(4, \\dfrac{1}{7}\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Find the intersection point between the two functions [latex]f(x)=\\dfrac{2}{x-4}[\/latex] and\u00a0[latex]g(x)=\\dfrac{1}{2x-6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm220\">Show Answer<\/span><\/p>\n<div id=\"qhjm220\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(\\dfrac{8}{3},\\,-\\dfrac{3}{2}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Solve the rational equation [latex]2+\\dfrac{4}{y-2}=\\dfrac{8}{y^2-2y}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Before we solve the equation for [latex]y[\/latex], let&#8217;s determine the restrictions on the variable by setting each denominator to zero: [latex]y-2=0[\/latex] solves to [latex]y=2[\/latex], and [latex]y^2-2y=0[\/latex] factors to [latex]y(y-2)=0[\/latex] then solves to [latex]y=0,\\,2[\/latex]. Therefore, [latex]y\\neq0, 2[\/latex].<\/p>\n<p>Now, we will start to solve the equation by clearing the fractions by multiplying by the LCM [latex]y(y-2)[\/latex].<\/p>\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{aligned}2+\\dfrac{4}{y-2}&=\\dfrac{8}{y^2-2y}\\\\\\\\2+\\dfrac{4}{y-2}&=\\dfrac{8}{y(y-2)}&&\\text{Factor denominators}\\\\\\\\ \\color{blue}{y(y-2)} \\times \\left(2+\\dfrac{4}{y-2}\\right)&=\\dfrac{8}{y(y-2)} \\times\\color{blue}{y(y-2)}&&\\text{Multiply both sides by the LCM}\\\\\\\\2\\cdot\\color{blue}{y(y-2)}+\\dfrac{4}{(y-2)}\\cdot\\color{blue}{y(y-2)}&=\\dfrac{8}{y(y-2)}\\cdot\\color{blue}{y(y-2)}&&\\text{Distribute on the left side}\\\\\\\\2y(y-2) + y\\cancel{(y-2)}\\cdot\\dfrac{4}{\\cancel{(y-2)}}&=\\dfrac{8}{\\cancel{y}\\cancel{(y-2)}} \\cdot \\cancel{y}\\cancel{(y-2)}&&\\text{Cancel common factors}\\\\\\\\2y^2-4y+4y&=8&&\\text{Simplify. We have a quadratic equation.}\\\\\\\\2y^2-8&=0&&\\text{Subtract 8 from both sides}\\\\\\\\2\\left(y^2-4\\right)&=0&&\\text{Factor}\\\\\\\\2(y-2)(y+2)&=0&&\\text{Factor: difference of two squares}\\\\\\\\y-2=0\\text{ or }y+2&=0&&\\text{Zero product property}\\\\\\\\y&=\\pm 2&&\\text{Solve}\\end{aligned}[\/latex]<\/p>\n<p>Since [latex]2[\/latex] is a restricted value, the solution is [latex]y=-2[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Solve the rational equation [latex]\\dfrac{4}{x-4}-\\dfrac{3}{x-3}=1[\/latex]. Remember to check the restrictions on the variable.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm037\">Show Answer<\/span><\/p>\n<div id=\"qhjm037\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=2,\\,6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Solve the rational equation [latex]\\dfrac{7}{y^2+y-12}-\\dfrac{4y}{y^2+7y+12}=\\dfrac{6}{y^2-9}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Let&#8217;s start by determining the restrictions on the variable.\u00a0 [latex]y^2+y-12=0[\/latex] factors to [latex](y+4)(y-3)=0[\/latex]. So, [latex]y=-4, 3[\/latex]. Next, [latex]y^2+7y+12=0[\/latex] factors to [latex](y+4)(y+3)=0[\/latex]. So, [latex]y=-4, -3[\/latex]. Finally, [latex]y^2-9=0[\/latex] factors to [latex](y-3)(y+3)=0[\/latex]. So, [latex]y=3, -3[\/latex]. Therefore, [latex]y\\neq\\pm3,\\,-4[\/latex].<\/p>\n<p>Additionally, the LCM is [latex](y+4)(y-3)(y+3)[\/latex].<\/p>\n<p>Now, we will start to solve the equation by multiplying both sides by the LCM to clear the fractions.<\/p>\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{aligned}\\dfrac{7}{y^2+y-12}-\\dfrac{4y}{y^2+7y+12}&=\\dfrac{6}{y^2-9}\\\\\\\\\\dfrac{7}{(y+4)(y-3)}-\\dfrac{4y}{(y+4)(y+3)}&=\\dfrac{6}{(y+3)(y-3)}&&\\text{Factor}\\\\\\\\\\color{blue}{(y+4)(y+3)(y-3)\\cdot}\\left(\\dfrac{7}{(y+4)(y-3)}-\\dfrac{4y}{(y+4)(y+3)}\\right)&=\\dfrac{6}{(y+3)(y-3)}\\color{blue}{\\cdot(y+4)(y+3)(y-3)}&&\\text{Multiply} \\\\&&&\\text{by LCM}\\\\\\\\\\color{blue}{(y+4)(y+3)(y-3)}\\cdot\\dfrac{7}{(y+4)(y-3)}&-\\color{blue}{(y+4)(y+3)(y-3)}\\cdot\\dfrac{4y}{(y+4)(y+3)}\\\\\\\\&=\\dfrac{6}{(y+3)(y-3)}\\cdot(y+4)(y+3)(y-3)&&\\text{Distribute}\\\\\\\\\\cancel{(y+4)}(y+3)\\cancel{(y-3)}\\cdot\\dfrac{7}{\\cancel{(y+4)}\\cancel{(y-3)}}&-\\cancel{(y+4)}\\cancel{(y+3)}(y-3)\\cdot\\dfrac{4y}{\\cancel{(y+4)}\\cancel{(y+3)}}\\\\\\\\&=\\dfrac{6}{\\cancel{(y+3)}\\cancel{(y-3)}}\\cdot(y+4)\\cancel{(y+3)}\\cancel{(y-3)}&&\\text{Cancel}\\\\\\\\7(y+3)-4y(y-3)&=6(y+4)&&\\text{Simplify}\\\\\\\\7y+21-4y^2+12y&=6y+24&&\\text{Distribute}\\\\\\\\4y^2-13y+3&=0&&\\text{Simplify}\\\\\\\\(4y-1)(y-3)&=0&&\\text{Factor}\\\\\\\\4y-1=0\\text{ or }y-3&=0&&\\text{Zero product}\\\\&&&\\text {property}\\\\\\\\y&=\\dfrac{1}{4},\\,3\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">However, [latex]y=3[\/latex] is a restricted value since [latex]y\\neq\\pm3,\\,-4[\/latex].<\/p>\n<p>Consequently, the only solution is\u00a0[latex]y=\\dfrac{1}{4}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Solve the rational equation [latex]\\dfrac{x+11}{x^2-5x+4}=\\dfrac{5}{x-4}-\\dfrac{3}{x-1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm166\">Show Answer<\/span><\/p>\n<div id=\"qhjm166\" class=\"hidden-answer\" style=\"display: none\">The equation has no solution.<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/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-2868\">\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>The Meaning of Solving Rational Equations. <strong>Authored by<\/strong>: Leo Chang and Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Examples 3, 4, 5. <strong>Authored by<\/strong>: Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All graphs created by Desmos graphing calculator. <strong>Authored by<\/strong>: Leo Chang and Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Examples 1 and 2 &amp; Try It: hjm662, hjm857, hjm220, hjm037, hjm166. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley 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":422608,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"The Meaning of Solving Rational Equations\",\"author\":\"Leo Chang and Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Examples 3, 4, 5\",\"author\":\"Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All graphs created by Desmos graphing calculator\",\"author\":\"Leo Chang and Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Examples 1 and 2 & Try It: hjm662, hjm857, hjm220, hjm037, hjm166\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2868","chapter","type-chapter","status-publish","hentry"],"part":2581,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2868","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":42,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2868\/revisions"}],"predecessor-version":[{"id":4832,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2868\/revisions\/4832"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/2581"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2868\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=2868"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2868"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=2868"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=2868"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}