{"id":2851,"date":"2022-06-20T19:24:27","date_gmt":"2022-06-20T19:24:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=2851"},"modified":"2026-01-18T03:23:01","modified_gmt":"2026-01-18T03:23:01","slug":"7-5-the-addition-and-subtraction-of-rational-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/7-5-the-addition-and-subtraction-of-rational-expressions\/","title":{"raw":"7.5: Addition and Subtraction of Rational Functions","rendered":"7.5: Addition and Subtraction of Rational Functions"},"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=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-163\" class=\"standard post-163 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Perform addition and subtraction of rational expressions<\/li>\r\n \t<li>Perform addition and subtraction of rational functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"wp-nocaption aligncenter wp-image-5017\" style=\"text-align: left;\">The concept and skills for adding or subtracting rational expressions are similar to those for adding or subtracting numerical fractions. We will find a common denominator (or common multiple) among the rational expressions, build each rational expression to get all rational expressions with the same denominator, and then perform the addition or subtraction on the numerators. The common denominator is the common multiple of the denominators. Although any common multiple of the denominators can be used as the common denominator for adding or subtracting the rational expressions, the least common multiple is recommended because it could help us avoid complicated algebra or further simplification of our answer. Therefore, we use the Least Common Multiple (LCM) as the common denominator when adding or subtracting fractions.<\/div>\r\n<h2>Least Common Multiple<\/h2>\r\nTo find the least common multiple (LCM) of two or more algebraic expressions, we factor the expressions and multiply all of the distinct factors.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nDetermine the least common multiple of [latex]x^2-4,\\;x^2-4x-5[\/latex] and [latex]x^2-7x+10[\/latex].\r\n<h4>Solution<\/h4>\r\nWe factor each expression:\r\n\r\n[latex]x^2-4=\\color{blue}{(x-2)}(x+2)[\/latex]\r\n\r\n[latex]x^2-4x-5=\\color{red}{(x-5)}(x+1)[\/latex]\r\n\r\n[latex]x^2-7x+10=\\color{red}{(x-5)}\\color{blue}{(x-2)}[\/latex]\r\n\r\nThe LCM must contain the highest power of every factor:\r\n\r\nLCM [latex]=\\color{blue}{(x-2)}(x+2)\\color{red}{(x-5)}(x+1)[\/latex]\r\n\r\nThe colors indicate factors that are common to two expressions. Notice that they are written in the LCM only once.\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nDetermine the least common multiple of [latex]x^2-9,\\;x^2-x-6[\/latex] and [latex]x^2-6x+9[\/latex].\r\n\r\n[reveal-answer q=\"hjm985\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm985\"]\r\n\r\nLCM [latex]=(x-3)^2(x+2)(x+3)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Adding and Subtracting Rational Expressions<\/h2>\r\nTo add rational expressions we need a common denominator. We can use the LCM of the denominators as the common denominator.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nAdd:\u00a0[latex]\\dfrac{5x-3}{x}+\\dfrac{6x+4}{x}[\/latex]. State any restrictions on the variable.\r\n<h4>Solution<\/h4>\r\nIn this example we already have a common denominator, [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{5x-3}{x}+\\dfrac{6x+4}{x}&amp;=\\dfrac{5x-3+6x+4}{x}&amp;&amp;\\text{Combine numerators with the common denominator. }x\\neq 0\\\\\\\\&amp;=\\dfrac{11x+1}{x}&amp;&amp;\\text{Simplify numerator}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nAdd:\u00a0[latex]\\dfrac{4x-1}{x+2}+\\dfrac{-3x+4}{x+2}[\/latex]. State any restrictions on the variable.\r\n\r\n[reveal-answer q=\"hjm435\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm435\"][latex]\\dfrac{x+3}{x+2}\\;\\;\\;\\;x\\neq -2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nTo subtract rational expressions, we need to subtract the complete numerator of the second fraction.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nSubtract: [latex]\\dfrac{4x}{x-1}-\\dfrac{x-5}{x-1}[\/latex]\r\n<h4>Solution<\/h4>\r\nIn this example we have a common denominator, [latex]x-1[\/latex], with a restriction that [latex]x\\neq 1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{4x}{x-1}-\\dfrac{x-5}{x-1}&amp;=\\dfrac{4x-(x-5)}{x-1}&amp;&amp;\\text{Combine the numerators}\\\\\\\\&amp;=\\dfrac{4x-x+5}{x-1}&amp;&amp;\\text{Subtract using the distributive property}\\\\\\\\&amp;=\\dfrac{3x+5}{x-1}&amp;&amp;\\text{Simplify}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nSubtract: [latex]\\dfrac{4x-3}{x+5}-\\dfrac{2x-5}{x+5}[\/latex]\r\n\r\n[reveal-answer q=\"hjm548\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm548\"][latex]\\dfrac{2x+2}{x+5}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nUsually, we need to start by finding a common denominator and building equivalent fractions with that new denominator. To build a fraction, we multiply the numerator and denominator by the same factors.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nAdd: [latex]\\dfrac{2x^2}{x^2-4}+\\dfrac{x}{x-2}[\/latex]. State any restrictions on the variable.\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{2x^2}{x^2-4}+\\dfrac{x}{x-2}&amp;=\\dfrac{2x^2}{(x-2)(x+2)}+\\dfrac{x}{(x-2)}&amp;&amp;\\text{Factor. LCM }=(x-2)(x+2)\\;\\;x\\neq\\pm2\\\\\\\\&amp;=\\dfrac{2x^2}{(x-2)(x+2)}+\\dfrac{x\\color{blue}{(x+2)}}{(x-2)\\color{blue}{(x+2)}}&amp;&amp;\\text{Build fractions with a common denominator}\\\\\\\\&amp;=\\dfrac{2x^2+x(x+2)}{(x-2)(x+2)}&amp;&amp;\\text{Combine numerators; use common denominator}\\\\\\\\&amp;=\\dfrac{2x^2+x^2+2x}{(x-2)(x+2)}&amp;&amp;\\text{Distribute }x\\text{ into }(x+2)\\\\\\\\&amp;=\\dfrac{3x^2+2x}{(x-2)(x+2)}&amp;&amp;\\text{Simplify}\\\\\\\\&amp;=\\dfrac{x(3x+2)}{(x-2)(x+2)}&amp;&amp;\\text{Factor numerator to see if fraction will simplify}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nAdd: [latex]\\dfrac{4x+3}{x^2-25}+\\dfrac{5x-1}{x+5}[\/latex]. State any restrictions on the variable.\r\n\r\n[reveal-answer q=\"hjm727\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm727\"]\r\n\r\n[latex]\\dfrac{5x^2-22x+8}{(x-5)(x+5)}=\\dfrac{(x-4)(5x-2)}{(x-5)(x+5)}\\;\\;x\\neq\\pm5[\/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\nSubtract: [latex]\\dfrac{2x^2}{x^2-4x-5}-\\dfrac{3x}{x-5}[\/latex]. State any restrictions on the variable.\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{2x^2}{x^2-4x-5}-\\dfrac{3x}{x-5}&amp;=\\dfrac{2x^2}{(x-5)(x+1)}-\\dfrac{3x}{(x-5)}&amp;&amp;\\text{Factor. LCM }=(x-5)(x+1)\\;\\;x\\neq5,\\,-1\\\\\\\\&amp;=\\dfrac{2x^2}{(x-5)(x+1)}-\\dfrac{3x\\color{blue}{(x+1)}}{(x-5)\\color{blue}{(x+1)}}&amp;&amp;\\text{Build fractions using LCM as common denominator}\\\\\\\\&amp;=\\dfrac{2x^2-3x(x+1)}{(x-5)(x+1)}&amp;&amp;\\text{Combine fractions}\\\\\\\\&amp;=\\dfrac{2x^2-3x^2-3x}{(x-5)(x+1)}&amp;&amp;\\text{Distribute }-3x\\text{ into }(x+1)\\\\\\\\&amp;=\\dfrac{-x^2-3x}{(x-5)(x+1)}&amp;&amp;\\text{Simplify}\\\\\\\\&amp;=\\dfrac{-x(x+3)}{(x-5)(x+1)}&amp;&amp;\\text{Factor to look for common factors}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nSubtract: [latex]\\dfrac{4x+3}{x^2+9x+8}-\\dfrac{2x-3}{x+1}[\/latex]. State any restrictions on the variable.\r\n\r\n[reveal-answer q=\"hjm782\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm782\"]\r\n\r\n[latex]\\dfrac{-2x^2-9x+27}{(x+1)(x+8)}\\;\\;x\\neq-8,\\,-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe video presents an example of adding two rational expressions whose denominators are binomials with no common factors.\r\n\r\n<iframe title=\"Ex: Add Rational Expressions with Unlike Denominators\" src=\"https:\/\/www.youtube.com\/embed\/CKGpiTE5vIg?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<div id=\"post-163\" class=\"standard post-163 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n\r\nThe next video contains an example of subtracting rational expressions.\r\n\r\n<iframe title=\"Subtract Rational Expressions with Unlike Denominators and Give the Domain\" src=\"https:\/\/www.youtube.com\/embed\/MMlNtCrkakI?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<!-- .entry-content -->\r\n\r\n<\/div>\r\n<!-- #post-## -->\r\n\r\n<!-- CITATIONS AND ATTRIBUTIONS -->\r\n\r\n<section class=\"citations-section\" role=\"contentinfo\"><\/section>\r\n<h2>Adding and Subtracting Rational Functions<\/h2>\r\nNow, we may use the methods we learned above to add two rational functions. The domain of the sum function will equal all real numbers with the combined restrictions of each function.\r\n<div class=\"textbox shaded\">\r\n<h3>adding and subtracting functions<\/h3>\r\n<p style=\"text-align: center;\">[latex]\\left(f+g\\right)(x)=f(x)+g(x)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(f-g\\right)(x)=f(x)-g(x)[\/latex]<\/p>\r\nThe domain of the sum or difference function will be all real numbers except the combined restricted values of [latex]f(x)[\/latex] and [latex]g(x)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nAdd the two functions [latex]f(x)=\\dfrac{x+2}{x^2-1}[\/latex] and\u00a0[latex]g(x)=\\dfrac{x-3}{x^2+x-2}[\/latex]. State the domain of the function. State the restrictions and the domain.\r\n<h4>Solution<\/h4>\r\nBefore we perform the addition, we need to discuss the domain. The denominators of the two functions cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator [latex]x^2-1[\/latex] can be factored as\u00a0[latex](x-1)(x+1)[\/latex]. Setting [latex](x-1)(x+1)=0[\/latex] results in [latex]x=1, -1[\/latex]. Therefore, [latex]x \\neq 1, -1[\/latex]. The denominator [latex]x^2+x-2[\/latex] can be factored as [latex](x+2)(x-1)[\/latex]. Setting [latex](x+2)(x-1)=0[\/latex] results in [latex]x=-2, 1[\/latex]. Therefore, [latex]x \\neq -2, 1[\/latex]. In summary, the domain of [latex](f+g)(x)[\/latex] is all real numbers except for the numbers [latex]1, -1, -2[\/latex].\r\n\r\nDomain = [latex]\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq\\pm1,\\,-2\\}[\/latex]\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(f+g)(x)&amp;=\\dfrac{x+2}{x^2-1}+\\dfrac{x-3}{x^2+x-2}&amp;&amp;\\text{Factor}\\\\\\\\&amp;=\\dfrac{x+2}{(x-1)(x+1)}+\\dfrac{x-3}{(x+2)(x-1)}&amp;&amp;\\text{LCD }=(x-1)(x+1)(x+2)\\\\\\\\&amp;=\\dfrac{x+2}{(x-1)(x+1)}\\color{blue}{\\times\\frac{(x+2)}{(x+2)}}+\\dfrac{x-3}{(x+2)(x-1)}\\color{blue}{\\times \\frac{(x+1)}{(x+1)}}&amp;&amp;\\text{Build equivalent fractions}\\\\\\\\&amp;=\\dfrac{x^2+4x+4}{(x-1)(x+1)(x+2)}+\\dfrac{x^2-2x-3}{(x-1)(x+1)(x+2)}&amp;&amp;\\text{Combine numerators}\\\\\\\\&amp;=\\dfrac{(x^2+4x+4)+(x^2-2x-3)}{(x+1)(x-1)(x+2)}&amp;&amp;\\text{Add like terms}\\\\\\\\&amp;=\\dfrac{2x^2+2x+1}{(x+1)(x-1)(x+2)}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 6<\/h3>\r\nAdd the two functions [latex]f(x)=\\dfrac{x}{(x-3)^2}[\/latex] and\u00a0[latex]g(x)=\\dfrac{x-1}{x^2-2x-3}[\/latex]. State the domain of the function.\r\n\r\n[reveal-answer q=\"hjm824\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm824\"][latex]f(x)+g(x)=\\dfrac{2x^2-3x+3}{(x-3)^2(x+1)}\\;\\;\\;\\;x\\neq3,\\,-1[\/latex]\r\n\r\nDomain of [latex]\\left(f+g\\right)(x)=\\{x\\;|\\;\\in\\mathbb{R},\\;x\\neq3,\\,-1\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 7<\/h3>\r\nSubtract the function [latex]g(x)=\\dfrac{x+3}{3x^2+9x+6}[\/latex] from the function [latex]f(x)=\\dfrac{x+2}{x^2-2x-8}[\/latex]. State the restrictions and the domain.\r\n<h4>Solution<\/h4>\r\nBefore we perform the addition, we need to discuss the domain first. The denominators of the two functions (or rational expressions) cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator [latex]x^2-2x-8[\/latex] can be factored as\u00a0[latex](x-4)(x+2)[\/latex]. Let\u00a0[latex](x-4)(x+2)=0[\/latex]. According to the zero product rule, [latex]x=4, -2[\/latex]. Therefore, [latex]x \\neq 4, -2[\/latex]. The denominator [latex]3x^2+9x+6[\/latex] can be factored as [latex]3(x+1)(x+2)[\/latex]. Let [latex]3(x+1)(x+2)=0[\/latex]. According to the zero product rule, [latex]x=-1, -2[\/latex]. Therefore, [latex]x \\neq -1, -2[\/latex]. In summary, the domain of [latex](f-g)(x)[\/latex] is all real numbers except for the numbers [latex]4, -1, -2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(f-g)(x)&amp;=\\dfrac{x+2}{x^2-2x-8}-\\dfrac{x+3}{3x^2+9x+6}\\\\\\\\&amp;=\\dfrac{x+2}{(x-4)(x+2)}-\\dfrac{x+3}{3(x^2+3x+2)}&amp;&amp;\\text{Factor}\\\\\\\\&amp;=\\dfrac{x+2}{(x-4)(x+2)}-\\dfrac{x+3}{3(x+1)(x+2)}&amp;&amp;\\text{LCD }=3(x+1)(x+2)(x-4)\\\\\\\\&amp;=\\dfrac{x+2}{(x-4)(x+2)} \\times \\color{blue}{\\frac{3(x+1)}{3(x+1)}}-\\dfrac{x+3}{3(x+1)(x+2)} \\times \\color{blue}{\\frac{(x-4)}{(x-4)}}&amp;&amp;\\text{Build equivalent fractions}\\\\\\\\&amp;=\\dfrac{3x^2+9x+6}{3(x-4)(x+2)(x+1)}-\\dfrac{x^2-x-12}{3(x-4)(x+2)(x+1)}&amp;&amp;\\text{Combine numerators}\\\\\\\\&amp;=\\dfrac{3x^2+9x+6-(x^2-x-12)}{3(x-4)(x+2)(x+1)}&amp;&amp;\\text{Distribute the }-\\text{ sign}\\\\\\\\&amp;=\\dfrac{3x^2+9x+6-x^2+x+12}{3(x-4)(x+2)(x+1)}&amp;&amp;\\text{Combine like terms}\\\\&amp;=\\dfrac{2x^2+10x+18}{3(x-4)(x+2)(x+1)}&amp;&amp;\\text{Factor}\\\\\\\\&amp;=\\dfrac{2(x^2+5x+9)}{3(x-4)(x+2)(x+1)}\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 7<\/h3>\r\nSubtract the function [latex]g(x)=\\dfrac{2}{x^2-4}[\/latex] from the function [latex]f(x)=\\dfrac{6}{x^2+4x+4}[\/latex].\r\n\r\n[reveal-answer q=\"hjm382\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm382\"]\r\n\r\n[latex]f(x)-g(x)=\\dfrac{4(x-4)}{(x+2)^2(x-2)}[\/latex]\r\n\r\nDomain of [latex]\\left(f-g\\right)(x)=\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq\\pm2\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAdding and subtracting rational functions can be extended to include more than two functions.\r\n<div class=\"textbox examples\">\r\n<h3>Example 8<\/h3>\r\nCombine [latex]f(x)+g(x)-h(x)[\/latex], when [latex]f(x)=\\dfrac{2x}{x-1},\\;g(x)=\\dfrac{1}{x},\\;h(x)=\\dfrac{2x-1}{x^2-x}[\/latex]. State the domain of [latex]\\left(f+g-h\\right)(x)[\/latex].\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)+g(x)-h(x)&amp;=\\dfrac{2x}{x-1}+\\dfrac{1}{x}-\\dfrac{2x-1}{x^2-x}\\\\\\\\&amp;=\\dfrac{2x}{(x-1)}+\\dfrac{1}{x}-\\dfrac{(2x-1)}{x(x-1)}&amp;&amp;\\text{Factor. LCM}=x(x-1),\\;x\\neq0,\\,1\\\\\\\\&amp;=\\color{blue}{\\dfrac{x}{x}}\\cdot\\dfrac{2x}{(x-1)}+\\color{blue}{\\dfrac{(x-1)}{(x-1)}}\\cdot\\dfrac{1}{x}-\\dfrac{(2x-1)}{x(x-1)}&amp;&amp;\\text{Build fractions with common denominator}\\\\\\\\&amp;=\\dfrac{2x^2+(x-1)-(2x-1)}{x(x-1)}&amp;&amp;\\text{Combine numerators}\\\\\\\\&amp;=\\dfrac{2x^2+x-1-2x+1}{x(x-1)}&amp;&amp;\\text{distribute + and \u2013 signs}\\\\\\\\&amp;=\\dfrac{2x^2-x}{x(x-1)}&amp;&amp;\\text{Combine like terms}\\\\\\\\&amp;=\\dfrac{x(2x-1)}{x(x-1)}&amp;&amp;\\text{Factor numerator}\\\\\\\\&amp;=\\dfrac{2x-1}{x-1}&amp;&amp;\\text{Cancel: }\\dfrac{x}{x}=1\\end{aligned}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nDomain of\u00a0[latex]\\left(f+g-h\\right)(x)=\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq0,\\,1\\}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 8<\/h3>\r\nCombine [latex]f(x)+g(x)-h(x)[\/latex], when [latex]f(x)=\\dfrac{x}{x+1},\\;g(x)=\\dfrac{3}{x-1},\\;h(x)=\\dfrac{6}{x^2-1}[\/latex].\u00a0State the domain of [latex]\\left(f+g-h\\right)(x)[\/latex].\r\n\r\n[reveal-answer q=\"hjm186\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm186\"]\r\n\r\n[latex]\\left(f+g-h\\right)(x)=\\dfrac{x+3}{x+1}[\/latex]\r\n\r\nDomain of [latex]\\left(f+g-h\\right)(x)=\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq\\pm1\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 9<\/h3>\r\nCombine [latex]f(x)+g(x)-h(x)[\/latex], when [latex]f(x)=\\dfrac{2x}{x+5},\\;g(x)=\\dfrac{3}{x-3},\\;h(x)=\\dfrac{13x+15}{x^2+2x-15}[\/latex].\u00a0State the domain of [latex]\\left(f+g-h\\right)(x)[\/latex].\r\n\r\n[reveal-answer q=\"hjm188\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm188\"]\r\n\r\n[latex]\\left(f+g-h\\right)(x)=\\dfrac{2x(x-8)}{(x-3)(x+5)}[\/latex]\r\n\r\nDomain of [latex]\\left(f+g-h\\right)(x)=\\{x\\;|\\;\\mathbb{R},\\;x\\neq -5,\\;3\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\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=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-163\" class=\"standard post-163 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Perform addition and subtraction of rational expressions<\/li>\n<li>Perform addition and subtraction of rational functions<\/li>\n<\/ul>\n<\/div>\n<div class=\"wp-nocaption aligncenter wp-image-5017\" style=\"text-align: left;\">The concept and skills for adding or subtracting rational expressions are similar to those for adding or subtracting numerical fractions. We will find a common denominator (or common multiple) among the rational expressions, build each rational expression to get all rational expressions with the same denominator, and then perform the addition or subtraction on the numerators. The common denominator is the common multiple of the denominators. Although any common multiple of the denominators can be used as the common denominator for adding or subtracting the rational expressions, the least common multiple is recommended because it could help us avoid complicated algebra or further simplification of our answer. Therefore, we use the Least Common Multiple (LCM) as the common denominator when adding or subtracting fractions.<\/div>\n<h2>Least Common Multiple<\/h2>\n<p>To find the least common multiple (LCM) of two or more algebraic expressions, we factor the expressions and multiply all of the distinct factors.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Determine the least common multiple of [latex]x^2-4,\\;x^2-4x-5[\/latex] and [latex]x^2-7x+10[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We factor each expression:<\/p>\n<p>[latex]x^2-4=\\color{blue}{(x-2)}(x+2)[\/latex]<\/p>\n<p>[latex]x^2-4x-5=\\color{red}{(x-5)}(x+1)[\/latex]<\/p>\n<p>[latex]x^2-7x+10=\\color{red}{(x-5)}\\color{blue}{(x-2)}[\/latex]<\/p>\n<p>The LCM must contain the highest power of every factor:<\/p>\n<p>LCM [latex]=\\color{blue}{(x-2)}(x+2)\\color{red}{(x-5)}(x+1)[\/latex]<\/p>\n<p>The colors indicate factors that are common to two expressions. Notice that they are written in the LCM only once.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Determine the least common multiple of [latex]x^2-9,\\;x^2-x-6[\/latex] and [latex]x^2-6x+9[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm985\">Show Answer<\/span><\/p>\n<div id=\"qhjm985\" class=\"hidden-answer\" style=\"display: none\">\n<p>LCM [latex]=(x-3)^2(x+2)(x+3)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Adding and Subtracting Rational Expressions<\/h2>\n<p>To add rational expressions we need a common denominator. We can use the LCM of the denominators as the common denominator.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Add:\u00a0[latex]\\dfrac{5x-3}{x}+\\dfrac{6x+4}{x}[\/latex]. State any restrictions on the variable.<\/p>\n<h4>Solution<\/h4>\n<p>In this example we already have a common denominator, [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{5x-3}{x}+\\dfrac{6x+4}{x}&=\\dfrac{5x-3+6x+4}{x}&&\\text{Combine numerators with the common denominator. }x\\neq 0\\\\\\\\&=\\dfrac{11x+1}{x}&&\\text{Simplify numerator}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Add:\u00a0[latex]\\dfrac{4x-1}{x+2}+\\dfrac{-3x+4}{x+2}[\/latex]. State any restrictions on the variable.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm435\">Show Answer<\/span><\/p>\n<div id=\"qhjm435\" class=\"hidden-answer\" style=\"display: none\">[latex]\\dfrac{x+3}{x+2}\\;\\;\\;\\;x\\neq -2[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>To subtract rational expressions, we need to subtract the complete numerator of the second fraction.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Subtract: [latex]\\dfrac{4x}{x-1}-\\dfrac{x-5}{x-1}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>In this example we have a common denominator, [latex]x-1[\/latex], with a restriction that [latex]x\\neq 1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{4x}{x-1}-\\dfrac{x-5}{x-1}&=\\dfrac{4x-(x-5)}{x-1}&&\\text{Combine the numerators}\\\\\\\\&=\\dfrac{4x-x+5}{x-1}&&\\text{Subtract using the distributive property}\\\\\\\\&=\\dfrac{3x+5}{x-1}&&\\text{Simplify}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Subtract: [latex]\\dfrac{4x-3}{x+5}-\\dfrac{2x-5}{x+5}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm548\">Show Answer<\/span><\/p>\n<div id=\"qhjm548\" class=\"hidden-answer\" style=\"display: none\">[latex]\\dfrac{2x+2}{x+5}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Usually, we need to start by finding a common denominator and building equivalent fractions with that new denominator. To build a fraction, we multiply the numerator and denominator by the same factors.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Add: [latex]\\dfrac{2x^2}{x^2-4}+\\dfrac{x}{x-2}[\/latex]. State any restrictions on the variable.<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{2x^2}{x^2-4}+\\dfrac{x}{x-2}&=\\dfrac{2x^2}{(x-2)(x+2)}+\\dfrac{x}{(x-2)}&&\\text{Factor. LCM }=(x-2)(x+2)\\;\\;x\\neq\\pm2\\\\\\\\&=\\dfrac{2x^2}{(x-2)(x+2)}+\\dfrac{x\\color{blue}{(x+2)}}{(x-2)\\color{blue}{(x+2)}}&&\\text{Build fractions with a common denominator}\\\\\\\\&=\\dfrac{2x^2+x(x+2)}{(x-2)(x+2)}&&\\text{Combine numerators; use common denominator}\\\\\\\\&=\\dfrac{2x^2+x^2+2x}{(x-2)(x+2)}&&\\text{Distribute }x\\text{ into }(x+2)\\\\\\\\&=\\dfrac{3x^2+2x}{(x-2)(x+2)}&&\\text{Simplify}\\\\\\\\&=\\dfrac{x(3x+2)}{(x-2)(x+2)}&&\\text{Factor numerator to see if fraction will simplify}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Add: [latex]\\dfrac{4x+3}{x^2-25}+\\dfrac{5x-1}{x+5}[\/latex]. State any restrictions on the variable.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm727\">Show Answer<\/span><\/p>\n<div id=\"qhjm727\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{5x^2-22x+8}{(x-5)(x+5)}=\\dfrac{(x-4)(5x-2)}{(x-5)(x+5)}\\;\\;x\\neq\\pm5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Subtract: [latex]\\dfrac{2x^2}{x^2-4x-5}-\\dfrac{3x}{x-5}[\/latex]. State any restrictions on the variable.<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\dfrac{2x^2}{x^2-4x-5}-\\dfrac{3x}{x-5}&=\\dfrac{2x^2}{(x-5)(x+1)}-\\dfrac{3x}{(x-5)}&&\\text{Factor. LCM }=(x-5)(x+1)\\;\\;x\\neq5,\\,-1\\\\\\\\&=\\dfrac{2x^2}{(x-5)(x+1)}-\\dfrac{3x\\color{blue}{(x+1)}}{(x-5)\\color{blue}{(x+1)}}&&\\text{Build fractions using LCM as common denominator}\\\\\\\\&=\\dfrac{2x^2-3x(x+1)}{(x-5)(x+1)}&&\\text{Combine fractions}\\\\\\\\&=\\dfrac{2x^2-3x^2-3x}{(x-5)(x+1)}&&\\text{Distribute }-3x\\text{ into }(x+1)\\\\\\\\&=\\dfrac{-x^2-3x}{(x-5)(x+1)}&&\\text{Simplify}\\\\\\\\&=\\dfrac{-x(x+3)}{(x-5)(x+1)}&&\\text{Factor to look for common factors}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Subtract: [latex]\\dfrac{4x+3}{x^2+9x+8}-\\dfrac{2x-3}{x+1}[\/latex]. State any restrictions on the variable.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm782\">Show Answer<\/span><\/p>\n<div id=\"qhjm782\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{-2x^2-9x+27}{(x+1)(x+8)}\\;\\;x\\neq-8,\\,-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The video presents an example of adding two rational expressions whose denominators are binomials with no common factors.<\/p>\n<p><iframe loading=\"lazy\" title=\"Ex: Add Rational Expressions with Unlike Denominators\" src=\"https:\/\/www.youtube.com\/embed\/CKGpiTE5vIg?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"post-163\" class=\"standard post-163 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<p>The next video contains an example of subtracting rational expressions.<\/p>\n<p><iframe loading=\"lazy\" title=\"Subtract Rational Expressions with Unlike Denominators and Give the Domain\" src=\"https:\/\/www.youtube.com\/embed\/MMlNtCrkakI?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p><!-- .entry-content --><\/p>\n<\/div>\n<p><!-- #post-## --><\/p>\n<p><!-- CITATIONS AND ATTRIBUTIONS --><\/p>\n<section class=\"citations-section\" role=\"contentinfo\"><\/section>\n<h2>Adding and Subtracting Rational Functions<\/h2>\n<p>Now, we may use the methods we learned above to add two rational functions. The domain of the sum function will equal all real numbers with the combined restrictions of each function.<\/p>\n<div class=\"textbox shaded\">\n<h3>adding and subtracting functions<\/h3>\n<p style=\"text-align: center;\">[latex]\\left(f+g\\right)(x)=f(x)+g(x)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f-g\\right)(x)=f(x)-g(x)[\/latex]<\/p>\n<p>The domain of the sum or difference function will be all real numbers except the combined restricted values of [latex]f(x)[\/latex] and [latex]g(x)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Add the two functions [latex]f(x)=\\dfrac{x+2}{x^2-1}[\/latex] and\u00a0[latex]g(x)=\\dfrac{x-3}{x^2+x-2}[\/latex]. State the domain of the function. State the restrictions and the domain.<\/p>\n<h4>Solution<\/h4>\n<p>Before we perform the addition, we need to discuss the domain. The denominators of the two functions cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator [latex]x^2-1[\/latex] can be factored as\u00a0[latex](x-1)(x+1)[\/latex]. Setting [latex](x-1)(x+1)=0[\/latex] results in [latex]x=1, -1[\/latex]. Therefore, [latex]x \\neq 1, -1[\/latex]. The denominator [latex]x^2+x-2[\/latex] can be factored as [latex](x+2)(x-1)[\/latex]. Setting [latex](x+2)(x-1)=0[\/latex] results in [latex]x=-2, 1[\/latex]. Therefore, [latex]x \\neq -2, 1[\/latex]. In summary, the domain of [latex](f+g)(x)[\/latex] is all real numbers except for the numbers [latex]1, -1, -2[\/latex].<\/p>\n<p>Domain = [latex]\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq\\pm1,\\,-2\\}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(f+g)(x)&=\\dfrac{x+2}{x^2-1}+\\dfrac{x-3}{x^2+x-2}&&\\text{Factor}\\\\\\\\&=\\dfrac{x+2}{(x-1)(x+1)}+\\dfrac{x-3}{(x+2)(x-1)}&&\\text{LCD }=(x-1)(x+1)(x+2)\\\\\\\\&=\\dfrac{x+2}{(x-1)(x+1)}\\color{blue}{\\times\\frac{(x+2)}{(x+2)}}+\\dfrac{x-3}{(x+2)(x-1)}\\color{blue}{\\times \\frac{(x+1)}{(x+1)}}&&\\text{Build equivalent fractions}\\\\\\\\&=\\dfrac{x^2+4x+4}{(x-1)(x+1)(x+2)}+\\dfrac{x^2-2x-3}{(x-1)(x+1)(x+2)}&&\\text{Combine numerators}\\\\\\\\&=\\dfrac{(x^2+4x+4)+(x^2-2x-3)}{(x+1)(x-1)(x+2)}&&\\text{Add like terms}\\\\\\\\&=\\dfrac{2x^2+2x+1}{(x+1)(x-1)(x+2)}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 6<\/h3>\n<p>Add the two functions [latex]f(x)=\\dfrac{x}{(x-3)^2}[\/latex] and\u00a0[latex]g(x)=\\dfrac{x-1}{x^2-2x-3}[\/latex]. State the domain of the function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm824\">Show Answer<\/span><\/p>\n<div id=\"qhjm824\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x)+g(x)=\\dfrac{2x^2-3x+3}{(x-3)^2(x+1)}\\;\\;\\;\\;x\\neq3,\\,-1[\/latex]<\/p>\n<p>Domain of [latex]\\left(f+g\\right)(x)=\\{x\\;|\\;\\in\\mathbb{R},\\;x\\neq3,\\,-1\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 7<\/h3>\n<p>Subtract the function [latex]g(x)=\\dfrac{x+3}{3x^2+9x+6}[\/latex] from the function [latex]f(x)=\\dfrac{x+2}{x^2-2x-8}[\/latex]. State the restrictions and the domain.<\/p>\n<h4>Solution<\/h4>\n<p>Before we perform the addition, we need to discuss the domain first. The denominators of the two functions (or rational expressions) cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator [latex]x^2-2x-8[\/latex] can be factored as\u00a0[latex](x-4)(x+2)[\/latex]. Let\u00a0[latex](x-4)(x+2)=0[\/latex]. According to the zero product rule, [latex]x=4, -2[\/latex]. Therefore, [latex]x \\neq 4, -2[\/latex]. The denominator [latex]3x^2+9x+6[\/latex] can be factored as [latex]3(x+1)(x+2)[\/latex]. Let [latex]3(x+1)(x+2)=0[\/latex]. According to the zero product rule, [latex]x=-1, -2[\/latex]. Therefore, [latex]x \\neq -1, -2[\/latex]. In summary, the domain of [latex](f-g)(x)[\/latex] is all real numbers except for the numbers [latex]4, -1, -2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(f-g)(x)&=\\dfrac{x+2}{x^2-2x-8}-\\dfrac{x+3}{3x^2+9x+6}\\\\\\\\&=\\dfrac{x+2}{(x-4)(x+2)}-\\dfrac{x+3}{3(x^2+3x+2)}&&\\text{Factor}\\\\\\\\&=\\dfrac{x+2}{(x-4)(x+2)}-\\dfrac{x+3}{3(x+1)(x+2)}&&\\text{LCD }=3(x+1)(x+2)(x-4)\\\\\\\\&=\\dfrac{x+2}{(x-4)(x+2)} \\times \\color{blue}{\\frac{3(x+1)}{3(x+1)}}-\\dfrac{x+3}{3(x+1)(x+2)} \\times \\color{blue}{\\frac{(x-4)}{(x-4)}}&&\\text{Build equivalent fractions}\\\\\\\\&=\\dfrac{3x^2+9x+6}{3(x-4)(x+2)(x+1)}-\\dfrac{x^2-x-12}{3(x-4)(x+2)(x+1)}&&\\text{Combine numerators}\\\\\\\\&=\\dfrac{3x^2+9x+6-(x^2-x-12)}{3(x-4)(x+2)(x+1)}&&\\text{Distribute the }-\\text{ sign}\\\\\\\\&=\\dfrac{3x^2+9x+6-x^2+x+12}{3(x-4)(x+2)(x+1)}&&\\text{Combine like terms}\\\\&=\\dfrac{2x^2+10x+18}{3(x-4)(x+2)(x+1)}&&\\text{Factor}\\\\\\\\&=\\dfrac{2(x^2+5x+9)}{3(x-4)(x+2)(x+1)}\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 7<\/h3>\n<p>Subtract the function [latex]g(x)=\\dfrac{2}{x^2-4}[\/latex] from the function [latex]f(x)=\\dfrac{6}{x^2+4x+4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm382\">Show Answer<\/span><\/p>\n<div id=\"qhjm382\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x)-g(x)=\\dfrac{4(x-4)}{(x+2)^2(x-2)}[\/latex]<\/p>\n<p>Domain of [latex]\\left(f-g\\right)(x)=\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq\\pm2\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Adding and subtracting rational functions can be extended to include more than two functions.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 8<\/h3>\n<p>Combine [latex]f(x)+g(x)-h(x)[\/latex], when [latex]f(x)=\\dfrac{2x}{x-1},\\;g(x)=\\dfrac{1}{x},\\;h(x)=\\dfrac{2x-1}{x^2-x}[\/latex]. State the domain of [latex]\\left(f+g-h\\right)(x)[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)+g(x)-h(x)&=\\dfrac{2x}{x-1}+\\dfrac{1}{x}-\\dfrac{2x-1}{x^2-x}\\\\\\\\&=\\dfrac{2x}{(x-1)}+\\dfrac{1}{x}-\\dfrac{(2x-1)}{x(x-1)}&&\\text{Factor. LCM}=x(x-1),\\;x\\neq0,\\,1\\\\\\\\&=\\color{blue}{\\dfrac{x}{x}}\\cdot\\dfrac{2x}{(x-1)}+\\color{blue}{\\dfrac{(x-1)}{(x-1)}}\\cdot\\dfrac{1}{x}-\\dfrac{(2x-1)}{x(x-1)}&&\\text{Build fractions with common denominator}\\\\\\\\&=\\dfrac{2x^2+(x-1)-(2x-1)}{x(x-1)}&&\\text{Combine numerators}\\\\\\\\&=\\dfrac{2x^2+x-1-2x+1}{x(x-1)}&&\\text{distribute + and \u2013 signs}\\\\\\\\&=\\dfrac{2x^2-x}{x(x-1)}&&\\text{Combine like terms}\\\\\\\\&=\\dfrac{x(2x-1)}{x(x-1)}&&\\text{Factor numerator}\\\\\\\\&=\\dfrac{2x-1}{x-1}&&\\text{Cancel: }\\dfrac{x}{x}=1\\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Domain of\u00a0[latex]\\left(f+g-h\\right)(x)=\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq0,\\,1\\}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 8<\/h3>\n<p>Combine [latex]f(x)+g(x)-h(x)[\/latex], when [latex]f(x)=\\dfrac{x}{x+1},\\;g(x)=\\dfrac{3}{x-1},\\;h(x)=\\dfrac{6}{x^2-1}[\/latex].\u00a0State the domain of [latex]\\left(f+g-h\\right)(x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm186\">Show Answer<\/span><\/p>\n<div id=\"qhjm186\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(f+g-h\\right)(x)=\\dfrac{x+3}{x+1}[\/latex]<\/p>\n<p>Domain of [latex]\\left(f+g-h\\right)(x)=\\{x\\;|\\;x\\in\\mathbb{R},\\;x\\neq\\pm1\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 9<\/h3>\n<p>Combine [latex]f(x)+g(x)-h(x)[\/latex], when [latex]f(x)=\\dfrac{2x}{x+5},\\;g(x)=\\dfrac{3}{x-3},\\;h(x)=\\dfrac{13x+15}{x^2+2x-15}[\/latex].\u00a0State the domain of [latex]\\left(f+g-h\\right)(x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm188\">Show Answer<\/span><\/p>\n<div id=\"qhjm188\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(f+g-h\\right)(x)=\\dfrac{2x(x-8)}{(x-3)(x+5)}[\/latex]<\/p>\n<p>Domain of [latex]\\left(f+g-h\\right)(x)=\\{x\\;|\\;\\mathbb{R},\\;x\\neq -5,\\;3\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\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-2851\">\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>Examples 1 and 8, Try It: hjm985; hjm435; hjm548; hjm727; hjm782; hjm824; hjm382; hjm186; hjm188. <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><li>Least Common Multiple. <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><li>Revision and Adaptation. <strong>Authored by<\/strong>: Hazel McKenna and 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>Ex: Add Rational Expressions with Unlike Denominators. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=CKGpiTE5vIg&#038;feature=youtu.be\">https:\/\/www.youtube.com\/watch?v=CKGpiTE5vIg&#038;feature=youtu.be<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Subtract Rational Expressions with Unlike Denominators and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/MMlNtCrkakI\">https:\/\/youtu.be\/MMlNtCrkakI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex6: Adding Rational Functions, Ex7: Subtracting Rational Functions. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 15: Rational Expressions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Lumen Learning. <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":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Unit 15: Rational Expressions, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Examples 1 and 8, Try It: hjm985; hjm435; hjm548; hjm727; hjm782; hjm824; hjm382; hjm186; hjm188\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Least Common Multiple\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex: Add Rational Expressions with Unlike Denominators\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=CKGpiTE5vIg&feature=youtu.be\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Subtract Rational Expressions with Unlike Denominators and Give the Domain\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/MMlNtCrkakI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex6: Adding Rational Functions, Ex7: Subtracting Rational Functions\",\"author\":\"Leo Chang\",\"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-2851","chapter","type-chapter","status-publish","hentry"],"part":2581,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2851","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":49,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2851\/revisions"}],"predecessor-version":[{"id":4831,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2851\/revisions\/4831"}],"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\/2851\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=2851"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2851"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=2851"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=2851"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}