{"id":177,"date":"2023-11-08T16:10:15","date_gmt":"2023-11-08T16:10:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-or-watch-complex-rational-expressions-2\/"},"modified":"2024-08-02T21:22:42","modified_gmt":"2024-08-02T21:22:42","slug":"4-4-simplifying-complex-rational-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-4-simplifying-complex-rational-expressions\/","title":{"raw":"4.4 Simplifying Complex Rational Expressions","rendered":"4.4 Simplifying Complex Rational Expressions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Simplify complex rational expressions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nFractions and rational expressions can be interpreted as quotients. When both, the dividend (numerator) and divisor (denominator) include fractions or rational expressions, you have something more <i>complex<\/i> than usual. Do not fear \u2013 you have all the tools to simplify these quotients!\r\n\r\nA <strong>complex fraction<\/strong> is a fraction in which the numerator and\/or denominator include a fraction, e.g.,\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\dfrac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,}[\/latex].<\/p>\r\nThese complex fractions are never considered to be in the simplest form, but they can always be simplified using different methods. For the simplicity of explanation, we will refer to the numerator of the complex fraction as the\u00a0\u201cbig numerator\u201d and to\u00a0the denominator of the complex fraction as the\u00a0 \u201cbig denominator\u201d.\u00a0More complicated complex fractions might involve several fractions\u00a0in the \u201cbig numerator\u201d or \u201cbig denominator\u201d (or both). We also introduce the notion of \u201csmall fractions\u201d:\r\n<p style=\"text-align: center;\">[latex]\\require{color}\\dfrac{\\quad \\fcolorbox{Green}{SpringGreen}{$\\dfrac{3}{4} + \\dfrac{1}{2}$}\\quad }{\\quad \\fcolorbox{Blue}{SkyBlue}{$\\dfrac{4}{5} - \\dfrac{1}{10}$}\\quad }\\begin{matrix}\r\n{\\color{Green}\\textsf{\u201cbig numerator\u201d}} \\\\\r\n\\\\\r\n{\\color{Blue}\\textsf{\u201cbig denominator\u201d}}\r\n\\end{matrix}\\hspace{1in}\r\n\\dfrac{\\quad \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{3}{4}$} + \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{1}{2}$}\\quad }{\\quad \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{4}{5}$} - \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{1}{10}$}\\quad }\\ {\\color{BrickRed}\\textsf{\u201csmall fractions\u201d}}[\/latex]<\/p>\r\nWe will describe two methods for simplifying complex fractions. The first method relies on the fact that to divide means to multiply by the reciprocal. Once you present the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d, respectively, as simplified fractions (the\u00a0\u201cbig numerator\u201d should be just a single fraction and\u00a0\u201cbig denominator\u201d should be just a single fraction), rewrite the complex fraction as division and then rewrite the division as multiplication by the reciprocal.\r\n\r\n<i>Before<\/i> you multiply the numbers, it is often helpful to factor the fractions. You can then divide out common factors.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\dfrac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,}[\/latex]<\/p>\r\n[reveal-answer q=\"770219\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"770219\"]\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,} &amp;= \\frac{12}{35}\\div \\frac{6}{7} &amp;&amp; {\\color{blue}\\textsf{rewrite as division}}\\\\[5pt]\r\n&amp;= \\frac{12}{35}\\cdot \\frac{7}{6} &amp;&amp; {\\color{blue}\\textsf{rewrite division as multiplication by the reciprocal}}\\\\[5pt]\r\n&amp;= \\frac{2\\cdot 6}{5\\cdot 7}\\cdot \\frac{7}{6} &amp;&amp; {\\color{blue}\\textsf{factor looking for common factors to simplify}}\\\\[5pt]\r\n&amp;= \\frac{2\\cdot {\\color{red}\\cancel{\\color{black}{6}}}}{5\\cdot {\\color{red}\\cancel{\\color{black}{7}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 1} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{2}{5} &amp;&amp; {\\color{blue}\\textsf{simplify}}\\\\[5pt]\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe second method relies on\u00a0\u201celiminating\u201d all\u00a0\u201csmall fractions\u201d to reduce the complex fraction to a single fraction. You first find the LCM of the denominators of all\u00a0\u201csmall fractions\u201d and multiply the complex fraction by the form of [latex]1[\/latex] consisting of that LCM in the numerator and denominator. In other words, you multiply that LCM into both, the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d. Then, focusing separately on the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d, you simplify each to obtain a single fraction that you simplify at the end if possible. Presenting all\u00a0denominators of \u201csmall fractions\u201d and the multiplied-in LCM in factored forms may expediate the simplification process.\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\dfrac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,}[\/latex]<\/p>\r\n[reveal-answer q=\"179488\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"179488\"]\r\n\r\nThe\u00a0\u201csmall fractions\u201d are [latex]\\dfrac{12}{35}[\/latex] and [latex]\\dfrac{6}{7}[\/latex]. The LCM of [latex]35[\/latex] and [latex]7[\/latex] is [latex]35[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,} &amp;= \\frac{\\,\\dfrac{12}{5\\cdot 7}{\\color{red}{} \\cdot 5\\cdot 7}\\,}{\\,\\dfrac{6}{7}{\\color{red}{} \\cdot 5\\cdot 7}\\,} &amp;&amp; {\\color{blue}\\textsf{multiply by the proper form of $1$}}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{12}{{\\color{red}\\cancel{\\color{black}{5\\cdot 7}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{5\\cdot 7}}}\\cdot 1}{1}\\,}{\\,\\dfrac{6}{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 1}\\cdot \\dfrac{5\\cdot {\\color{red}\\cancel{\\color{black}{7}}}}{1}\\,} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{12}{30} &amp;&amp; {\\color{blue}\\textsf{simplify the \u201cbig numerator\u201d and \u201cbig denominator\u201d}}\\\\[5pt]\r\n&amp;= \\frac{2\\cdot {\\color{red}\\cancel{\\color{black}{6}}}}{5\\cdot {\\color{red}\\cancel{\\color{black}{6}}}} &amp;&amp; {\\color{blue}\\textsf{factor and divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{2}{5} &amp;&amp; {\\color{blue}\\textsf{simplify}}\\\\[5pt]\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe second method is usually faster but we will practice both and let you decide your preference.\r\n\r\nWhen using the first method for more complicated complex fractions,\u00a0start with combining the fractions in\u00a0the \u201cbig numerator\u201d and \u201cbig denominator\u201d accordingly.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify using both methods.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\,\\dfrac{3}{4}+\\dfrac{1}{2}\\,}{\\,\\dfrac{4}{5}-\\dfrac{1}{10}\\,}[\/latex]<\/p>\r\n[reveal-answer q=\"96511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"96511\"]\r\n<p style=\"text-align: left;\">For the first method, we start with combining the\u00a0\u201cbig numerator\u201d and \u201cbig denominator\u201d to single fractions accordingly. For that purpose, we acknowledge that [latex]\\operatorname{LCM}(2,4) = 4[\/latex] and [latex]\\operatorname{LCM}(5,10) = 10[\/latex]. Now,<\/p>\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{3}{4}+\\dfrac{1}{2}\\,}{\\,\\dfrac{4}{5}-\\dfrac{1}{10}\\,} &amp;= \\frac{\\,\\dfrac{3}{4}+\\dfrac{1{\\color{red}{} \\cdot 2}}{2{\\color{red}{} \\cdot 2}}\\,}{\\,\\dfrac{4{\\color{red}{} \\cdot 2}}{5{\\color{red}{} \\cdot 2}}-\\dfrac{1}{10}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{3}{4}+\\dfrac{2}{4}\\,}{\\,\\dfrac{8}{10}-\\dfrac{1}{10}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{5}{4}\\,}{\\,\\dfrac{7}{10}\\,}\\\\[5pt]\r\n&amp;= \\frac{5}{4}\\div \\frac{7}{10}\\\\[5pt]\r\n&amp;= \\frac{5}{4}\\cdot \\frac{10}{7}\\\\[5pt]\r\n&amp;= \\frac{5}{2\\cdot {\\color{red}\\cancel{\\color{black}{2}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{2}}} \\cdot 5}{7}\\\\[5pt]\r\n&amp;= \\frac{25}{14}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\nFor the second method, we start with the LCM of the denominators of all\u00a0\u201csmall fractions\u201d. (Please refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-add-and-subtract-rational-expressions-part-i\/#LCMdescr\" target=\"_blank\" rel=\"noopener\">this description<\/a> to review finding the LCM.)\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\operatorname{LCM}(4,2,5,10) &amp;= \\operatorname{LCM}\\left(2^2,2^1,5^1,5^1\\cdot 2^1\\right)\\\\\r\n&amp;= 2^2\\cdot 5^1\\\\\r\n&amp;= 20\r\n\\end{align}\r\n[\/latex]<\/p>\r\nNow,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{3}{4}+\\dfrac{1}{2}\\,}{\\,\\dfrac{4}{5}-\\dfrac{1}{10}\\,} &amp;= \\frac{\\,\\left(\\dfrac{3}{4}+\\dfrac{1}{2}\\right){\\color{red}{} \\cdot 20}\\,}{\\,\\left(\\dfrac{4}{5}-\\dfrac{1}{10}\\right){\\color{red}{} \\cdot 20}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{3}{4}\\cdot 20+\\dfrac{1}{2}\\cdot 20\\,}{\\,\\dfrac{4}{5}\\cdot 20-\\dfrac{1}{10}\\cdot 20\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{3}{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 5}{1}+\\dfrac{1}{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 10}{1}\\,}{\\,\\dfrac{4}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 4}{1}-\\dfrac{1}{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 2}{1}\\,}\\\\[5pt]\r\n&amp;= \\frac{15+10}{16-2}\\\\[5pt]\r\n&amp;= \\frac{25}{14}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we will show a couple more examples of how to simplify complex fractions.\r\n\r\nhttps:\/\/youtu.be\/lQCwze2w7OU\r\n<h2>Complex Rational Expressions<\/h2>\r\nA <strong>complex rational expression<\/strong> is a quotient with rational expressions in the dividend, divisor, or in both. We simplify these in the exact same way as we would a complex fraction, remembering to identify the restricted values of the variable.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nState the restricted values of the variable and simplify using both methods.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\,\\dfrac{x+5}{x^2-16}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,}[\/latex]<\/p>\r\n[reveal-answer q=\"245262\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245262\"]\r\n\r\nFor the restricted values of [latex]x[\/latex], we need to acknowledge that we have three denominators\u00a0\u2013 the denominators of the \u201csmall fractions\u201d and the \u201cbig denominator\u201d. The\u00a0\u201cbig denominator\u201d is itself a rational expression, so it's equal to zero only if its numerator is equal to zero. Since [latex]x^2-16=(x+4)(x-4)[\/latex], this denominator is equal to zero if [latex]x=-4[\/latex] or if [latex]x=4[\/latex] by the Zero-Product Property. The denominator [latex]x-4[\/latex] is equal to zero if [latex]x=4[\/latex], and since [latex]x^2-25 = (x+5)(x-5)[\/latex], the numerator of the\u00a0\u201cbig denominator\u201d is equal to zero if [latex]x=-5[\/latex] or if [latex]x=5[\/latex] by the Zero-Product Property. Summarizing, the restricted values of [latex]x[\/latex] are [latex]-5[\/latex], [latex]-4[\/latex], [latex]4[\/latex], and [latex]5[\/latex].\r\n<p style=\"text-align: left;\">For the first method, we start with rewriting the complex rational expression as division.<\/p>\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{x+5}{x^2-16}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,} &amp;= \\frac{x+5}{x^2-16}\\div\\frac{x^2-25}{x-4}\\\\[5pt]\r\n&amp;= \\frac{x+5}{x^2-16}\\cdot \\frac{x-4}{x^2-25}\\\\[5pt]\r\n&amp;= \\frac{(x+5)}{(x+4)(x-4)}\\cdot \\frac{(x-4)}{(x+5)(x-5)}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(x+5)}}}\\cdot 1}{(x+4)\\cdot {\\color{red}\\cancel{\\color{black}{(x-4)}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{(x-4)}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{(x+5)}}}\\cdot (x-5)}\\\\[5pt]\r\n&amp;= \\frac{1}{(x+4)(x-5)}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\nFor the second method, we start with the LCM of the denominators of all\u00a0\u201csmall fractions\u201d.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\operatorname{LCM}\\left(x^2-16, x-4\\right) &amp;= \\operatorname{LCM}\\left((x+4)^1\\cdot (x-4)^1, (x-4)^1\\right)\\\\\r\n&amp;= (x+4)^1\\cdot (x-4)^1\\\\\r\n&amp;= (x+4)(x-4)\r\n\\end{align}\r\n[\/latex]<\/p>\r\nNow,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{x+5}{x^2-16}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,} &amp;= \\frac{\\, \\dfrac{x+5}{x^2-16}{\\color{red}{} \\cdot (x+4)\\cdot (x-4)}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,{\\color{red}{} \\cdot (x+4)\\cdot (x-4)}}\\\\[5pt]\r\n&amp;= \\frac{\\, \\dfrac{(x+5)}{{\\color{red}\\cancel{\\color{black}{(x+4)\\cdot (x-4)}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{(x+4)\\cdot (x-4)}}}\\cdot 1}{1}\\,}{\\,\\dfrac{(x+5)(x-5)}{{\\color{red}\\cancel{\\color{black}{(x-4)}}} \\cdot 1} \\cdot \\dfrac{(x+4)\\cdot {\\color{red}\\cancel{\\color{black}{(x-4)}}}}{1}}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(x+5)}}} \\cdot 1}{{\\color{red}\\cancel{\\color{black}{(x+5)}}}\\cdot (x-5)\\cdot (x+4)}\\\\[5pt]\r\n&amp;= \\frac{1}{(x-5)(x+4)}\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThis time, in the final answer we may conveniently multiply the denominator out and simplify to obtain an equivalent form [latex]\\dfrac{1}{x^2-x-20}[\/latex].\u00a0 It's not expected or necessary though.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next video contains another example of simplifying a complex rational expression with just two \u201csmall fractions\u201d.\r\n\r\nhttps:\/\/youtu.be\/fAaqo8gGW9Y\r\n\r\nThe same ideas can be used when simplifying complex rational expressions that include more than one rational expression\u00a0in the \u201cbig numerator\u201d or \u201cbig denominator\u201d (or both).\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nState the restricted values of the variable and simplify using both methods.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle\\frac{\\,1-\\dfrac{9}{x^2}\\,}{\\,1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\,}[\/latex]<\/p>\r\n[reveal-answer q=\"344101\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"344101\"]\r\n\r\nFor the restricted values of [latex]x[\/latex], we need to acknowledge that we have four denominators\u00a0\u2013 the denominators of the \u201csmall fractions\u201d and the \u201cbig denominator\u201d. While the denominators of all \u201csmall fractions\u201d are just powers of [latex]x[\/latex], meaning only [latex]x=0[\/latex] would make any of them equal to zero, the \u201cbig denominator\u201d is not yet a single rational expression, so we first combine the \u201cbig denominator\u201d into a single rational expression to acknowledge its numerator for restricted values of the variable purposes. The LCM of [latex]x[\/latex] and [latex]x^2[\/latex] is [latex]x^2[\/latex], and [latex]1=\\dfrac{1}{1}[\/latex], thus\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{1}{1}+\\frac{5}{x}+\\frac{6}{x^2} &amp;= \\frac{1{\\color{red}{} \\cdot x^2}}{1{\\color{red}{} \\cdot x^2}}+\\frac{5{\\color{red}{} \\cdot x}}{x{\\color{red}{} \\cdot x}}+\\frac{6}{x^2}\\\\[5pt]\r\n&amp;= \\frac{x^2+5x+6}{x^2}\\\\[5pt]\r\n&amp;= \\frac{(x+2)(x+3)}{x^2}\r\n\\end{align}\r\n[\/latex]<\/p>\r\nHence, the numerator of\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression is equal to zero if [latex]x=-2[\/latex] or if [latex]x=-3[\/latex] by the Zero-Product Property. Summarizing, the restricted values of [latex]x[\/latex] are [latex]0[\/latex], [latex]-2[\/latex], and [latex]-3[\/latex].\r\n\r\nSince we already have\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression, for the first method we only need to rewrite\u00a0the \u201cbig numerator\u201d as a single rational expression and proceed accordingly.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,1-\\dfrac{9}{x^2}\\,}{\\,1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\,} &amp;= \\frac{\\,\\dfrac{1}{1}-\\dfrac{9}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{1{\\color{red}{} \\cdot x^2}}{1{\\color{red}{} \\cdot x^2}}-\\dfrac{9}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{x^2-9}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{(x+3)(x-3)}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]\r\n&amp;= \\dfrac{(x+3)(x-3)}{x^2}\\div\\dfrac{(x+2)(x+3)}{x^2}\\\\[5pt]\r\n&amp;= \\dfrac{(x+3)(x-3)}{x^2}\\cdot\\dfrac{x^2}{(x+2)(x+3)}\\\\[5pt]\r\n&amp;= \\dfrac{{\\color{red}\\cancel{\\color{black}{(x+3)}}}\\cdot (x-3)}{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}{(x+2)\\cdot {\\color{red}\\cancel{\\color{black}{(x+3)}}}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\nFor the second method, we acknowledge that the LCM of the denominators of all\u00a0\u201csmall fractions\u201d is [latex]x^2[\/latex], and proceed accordingly.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,1-\\dfrac{9}{x^2}\\,}{\\,1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\,} &amp;= \\frac{\\,\\left(1-\\dfrac{9}{x^2}\\right){\\color{red}\\cdot\\: x^2}\\,}{\\,\\left(1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\right){\\color{red} \\cdot\\: x^2}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,1\\cdot x^2-\\dfrac{9}{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}{1}\\,}{\\,1\\cdot x^2+\\dfrac{5}{{\\color{red}\\cancel{\\color{black}{x}}} \\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{x}}}\\cdot x}{1}+\\dfrac{6}{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}{1}\\,}\\\\[5pt]\r\n&amp;= \\frac{x^2-9}{x^2+5x+6}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(x+3)}}}\\cdot (x-3)}{{\\color{red}\\cancel{\\color{black}{(x+3)}}}\\cdot(x+2)}\\\\[5pt]\r\n&amp;= \\frac{x-3}{x+2}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next video contains another example of simplifying a complex rational expression similar to the one in the example above.\r\n\r\nhttps:\/\/youtu.be\/P5dfmX_FNPk\r\n\r\nWe conclude with an example in which the\u00a0LCM of the denominators of all\u00a0\u201csmall fractions\u201d will be more complicated.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nState the restricted values of the variable and simplify using both methods.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle\\frac{\\,\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\,}{\\,\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\,}[\/latex]<\/p>\r\n[reveal-answer q=\"234174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"234174\"]\r\n\r\nFor the restricted values of [latex]x[\/latex], we need to acknowledge that we have five denominators\u00a0\u2013 the denominators of the \u201csmall fractions\u201d and the \u201cbig denominator\u201d. For the denominators of all \u201csmall fractions\u201d, we solve the following equations:\r\n<ul>\r\n \t<li>[latex]x^2-1=0[\/latex], equivalent to [latex](x-1)(x+1)=0[\/latex], resulting in\u00a0[latex]x=1[\/latex] or [latex]x=-1[\/latex] by the Zero-Factor Property,<\/li>\r\n \t<li>[latex]x-3=0[\/latex], resulting in [latex]x=3[\/latex],<\/li>\r\n \t<li>[latex]x+1=0[\/latex], resulting in [latex]x=-1[\/latex], and<\/li>\r\n \t<li>[latex]x^2-2x-3=0[\/latex], equivalent to [latex](x-3)(x+1)=0[\/latex], resulting in\u00a0[latex]x=3[\/latex] or [latex]x=-1[\/latex] by the Zero-Factor Property.<\/li>\r\n<\/ul>\r\nThe \u201cbig denominator\u201d is not yet a single rational expression, so we first combine the \u201cbig denominator\u201d into a single rational expression to acknowledge its numerator for restricted values of the variable purposes. Based on the factorization [latex](x-3)(x+1)[\/latex] of [latex]x^2-2x-3[\/latex], the LCM of [latex]x+1[\/latex] and [latex]x^2-2x-3[\/latex] is [latex](x-3)(x+1)[\/latex] and thus\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{2}{x+1}-\\frac{3x}{x^2-2x-3} &amp;= \\frac{2{\\color{red}\\: \\cdot\\: (x-3)}}{(x+1){\\color{red}\\: \\cdot\\: (x-3)}}-\\frac{3x}{(x-3)(x+1)}\\\\[5pt]\r\n&amp;= \\frac{2x-6-3x}{(x-3)(x+1)}\\\\[5pt]\r\n&amp;= \\frac{-x-6}{(x-3)(x+1)}\r\n\\end{align}\r\n[\/latex]<\/p>\r\nHence, the numerator of\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression, [latex]-x-6[\/latex] is equal to zero if [latex]x=-6[\/latex]. Summarizing, the restricted values of [latex]x[\/latex] are [latex]1[\/latex], [latex]-1[\/latex], [latex]3[\/latex], and [latex]-6[\/latex].\r\n\r\nSince we already have\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression, for the first method we only need to rewrite\u00a0the \u201cbig numerator\u201d as a single rational expression and proceed accordingly. Notice that [latex]\\operatorname{LCM}\\left(x^2-1, x-3\\right)=\\operatorname{LCM}\\left((x+1)(x-1), x-3\\right)=(x+1)(x-1)(x-3)[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\,}{\\,\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\,} &amp;= \\frac{\\,\\dfrac{x{\\color{red}\\: \\cdot\\: (x-3)}}{(x+1)\\cdot(x-1){\\color{red}\\: \\cdot\\: (x-3)}}-\\dfrac{(x+1){\\color{red}\\: \\cdot\\: (x-1)\\cdot(x+1)}}{(x-3){\\color{red}\\: \\cdot\\: (x-1)\\cdot(x+1)}}\\,}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{x^2-3x-\\left(\\left(x^2-1\\right)\\cdot(x+1)\\right)}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{x^2-3x-\\left(x^3+x^2-x-1\\right)}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{x^2-3x-x^3-x^2+x+1}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]\r\n&amp;= \\frac{\\,\\dfrac{-x^3-2x+1}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]\r\n&amp;= \\frac{-x^3-2x+1}{(x-3)\\cdot(x-1)\\cdot(x+1)}\\div\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot\\left(x^3+2x-1\\right)}{{\\color{red}\\cancel{\\color{black}{(x-3)}}}\\cdot(x-1)\\cdot{\\color{red}\\cancel{\\color{black}{(x+1)}}}}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{(x-3)}}}\\cdot{\\color{red}\\cancel{\\color{black}{(x+1)}}}}{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot(x+6)}\\\\[5pt]\r\n&amp;= \\frac{x^3+2x-1}{(x-1)(x+6)}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\nFor the second method, let's begin with the LCM of the denominators of all\u00a0\u201csmall fractions\u201d:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\operatorname{LCM}\\left(x^2-1, x-3, x+1, x^2-2x-3\\right)&amp;=\\operatorname{LCM}\\left((x+1)^1\\cdot(x-1)^1, (x-3)^1, (x+1)^1, (x-3)^1\\cdot(x+1)^1\\right)\\\\[5pt]\r\n&amp;=(x+1)(x-1)(x-3)\r\n\\end{align}\r\n[\/latex]<\/p>\r\nNow,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{\\,\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\,}{\\,\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\,} &amp;= \\frac{\\,\\left(\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\right){\\color{red}\\: \\cdot\\: (x+1)\\cdot(x-1)\\cdot(x-3)}\\,}{\\,\\left(\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\right){\\color{red}\\: \\cdot\\: (x+1)\\cdot(x-1)\\cdot(x-3)}\\,}\\\\[5pt]\r\n&amp;=\\frac{\\,\\dfrac{x}{{\\color{red}\\cancel{\\color{black}{\\left(x^2-1\\right)}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{\\left(x^2-1\\right)}}}\\cdot (x-3)}{1}-\\dfrac{(x+1)}{{\\color{red}\\cancel{\\color{black}{(x-3)}}}\\cdot 1}\\cdot\\dfrac{\\left(x^2-1\\right)\\cdot {\\color{red}\\cancel{\\color{black}{(x-3)}}}}{1}\\,}{\\,\\dfrac{2}{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot\\left(x^2-4x+3\\right)}{1}-\\dfrac{3x}{{\\color{red}\\cancel{\\color{black}{\\left(x^2-2x-3\\right)}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{\\left(x^2-2x-3\\right)}}}\\cdot(x-1)}{1}\\,}\\\\[5pt]\r\n&amp;= \\frac{x^2-3x-\\left(x^3-x+x^2-1\\right)}{2x^2-8x+6-\\left(3x^2-3x\\right)}\\\\[5pt]\r\n&amp;= \\frac{x^2-3x-x^3+x-x^2+1}{2x^2-8x+6-3x^2+3x}\\\\[5pt]\r\n&amp;= \\frac{-x^3-2x+1}{-x^2-5x+6}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot\\left(x^3+2x-1\\right)}{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot\\left(x^2+5x-6\\right)}\\\\[5pt]\r\n&amp;= \\frac{x^3+2x-1}{(x-1)(x+6)}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nComplex rational expressions are quotients with rational expressions in the divisor, dividend, or both. When written in fraction form, they appear to be fractions within a fraction. These can be simplified by first treating the quotient as a division problem, after you present the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d, respectively, as simplified fractions. Then you can rewrite the division as multiplication by the reciprocal of the divisor. Or you can simplify the complex rational expression by multiplying both the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d by\u00a0the LCM of the denominators of all\u00a0\u201csmall fractions\u201d. This can help simplify the complex expression even faster.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Simplify complex rational expressions.<\/li>\n<\/ul>\n<\/div>\n<p>Fractions and rational expressions can be interpreted as quotients. When both, the dividend (numerator) and divisor (denominator) include fractions or rational expressions, you have something more <i>complex<\/i> than usual. Do not fear \u2013 you have all the tools to simplify these quotients!<\/p>\n<p>A <strong>complex fraction<\/strong> is a fraction in which the numerator and\/or denominator include a fraction, e.g.,<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\dfrac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,}[\/latex].<\/p>\n<p>These complex fractions are never considered to be in the simplest form, but they can always be simplified using different methods. For the simplicity of explanation, we will refer to the numerator of the complex fraction as the\u00a0\u201cbig numerator\u201d and to\u00a0the denominator of the complex fraction as the\u00a0 \u201cbig denominator\u201d.\u00a0More complicated complex fractions might involve several fractions\u00a0in the \u201cbig numerator\u201d or \u201cbig denominator\u201d (or both). We also introduce the notion of \u201csmall fractions\u201d:<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color}\\dfrac{\\quad \\fcolorbox{Green}{SpringGreen}{$\\dfrac{3}{4} + \\dfrac{1}{2}$}\\quad }{\\quad \\fcolorbox{Blue}{SkyBlue}{$\\dfrac{4}{5} - \\dfrac{1}{10}$}\\quad }\\begin{matrix}  {\\color{Green}\\textsf{\u201cbig numerator\u201d}} \\\\  \\\\  {\\color{Blue}\\textsf{\u201cbig denominator\u201d}}  \\end{matrix}\\hspace{1in}  \\dfrac{\\quad \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{3}{4}$} + \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{1}{2}$}\\quad }{\\quad \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{4}{5}$} - \\fcolorbox{BrickRed}{Lavender}{$\\dfrac{1}{10}$}\\quad }\\ {\\color{BrickRed}\\textsf{\u201csmall fractions\u201d}}[\/latex]<\/p>\n<p>We will describe two methods for simplifying complex fractions. The first method relies on the fact that to divide means to multiply by the reciprocal. Once you present the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d, respectively, as simplified fractions (the\u00a0\u201cbig numerator\u201d should be just a single fraction and\u00a0\u201cbig denominator\u201d should be just a single fraction), rewrite the complex fraction as division and then rewrite the division as multiplication by the reciprocal.<\/p>\n<p><i>Before<\/i> you multiply the numbers, it is often helpful to factor the fractions. You can then divide out common factors.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\dfrac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q770219\">Show Solution<\/span><\/p>\n<div id=\"q770219\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,} &= \\frac{12}{35}\\div \\frac{6}{7} && {\\color{blue}\\textsf{rewrite as division}}\\\\[5pt]  &= \\frac{12}{35}\\cdot \\frac{7}{6} && {\\color{blue}\\textsf{rewrite division as multiplication by the reciprocal}}\\\\[5pt]  &= \\frac{2\\cdot 6}{5\\cdot 7}\\cdot \\frac{7}{6} && {\\color{blue}\\textsf{factor looking for common factors to simplify}}\\\\[5pt]  &= \\frac{2\\cdot {\\color{red}\\cancel{\\color{black}{6}}}}{5\\cdot {\\color{red}\\cancel{\\color{black}{7}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 1} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{2}{5} && {\\color{blue}\\textsf{simplify}}\\\\[5pt]  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The second method relies on\u00a0\u201celiminating\u201d all\u00a0\u201csmall fractions\u201d to reduce the complex fraction to a single fraction. You first find the LCM of the denominators of all\u00a0\u201csmall fractions\u201d and multiply the complex fraction by the form of [latex]1[\/latex] consisting of that LCM in the numerator and denominator. In other words, you multiply that LCM into both, the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d. Then, focusing separately on the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d, you simplify each to obtain a single fraction that you simplify at the end if possible. Presenting all\u00a0denominators of \u201csmall fractions\u201d and the multiplied-in LCM in factored forms may expediate the simplification process.<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\dfrac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q179488\">Show Solution<\/span><\/p>\n<div id=\"q179488\" class=\"hidden-answer\" style=\"display: none\">\n<p>The\u00a0\u201csmall fractions\u201d are [latex]\\dfrac{12}{35}[\/latex] and [latex]\\dfrac{6}{7}[\/latex]. The LCM of [latex]35[\/latex] and [latex]7[\/latex] is [latex]35[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{12}{35}\\,}{\\,\\dfrac{6}{7}\\,} &= \\frac{\\,\\dfrac{12}{5\\cdot 7}{\\color{red}{} \\cdot 5\\cdot 7}\\,}{\\,\\dfrac{6}{7}{\\color{red}{} \\cdot 5\\cdot 7}\\,} && {\\color{blue}\\textsf{multiply by the proper form of $1$}}\\\\[5pt]  &= \\frac{\\,\\dfrac{12}{{\\color{red}\\cancel{\\color{black}{5\\cdot 7}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{5\\cdot 7}}}\\cdot 1}{1}\\,}{\\,\\dfrac{6}{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 1}\\cdot \\dfrac{5\\cdot {\\color{red}\\cancel{\\color{black}{7}}}}{1}\\,} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{12}{30} && {\\color{blue}\\textsf{simplify the \u201cbig numerator\u201d and \u201cbig denominator\u201d}}\\\\[5pt]  &= \\frac{2\\cdot {\\color{red}\\cancel{\\color{black}{6}}}}{5\\cdot {\\color{red}\\cancel{\\color{black}{6}}}} && {\\color{blue}\\textsf{factor and divide out common factors}}\\\\[5pt]  &= \\frac{2}{5} && {\\color{blue}\\textsf{simplify}}\\\\[5pt]  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The second method is usually faster but we will practice both and let you decide your preference.<\/p>\n<p>When using the first method for more complicated complex fractions,\u00a0start with combining the fractions in\u00a0the \u201cbig numerator\u201d and \u201cbig denominator\u201d accordingly.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify using both methods.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\,\\dfrac{3}{4}+\\dfrac{1}{2}\\,}{\\,\\dfrac{4}{5}-\\dfrac{1}{10}\\,}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q96511\">Show Solution<\/span><\/p>\n<div id=\"q96511\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">For the first method, we start with combining the\u00a0\u201cbig numerator\u201d and \u201cbig denominator\u201d to single fractions accordingly. For that purpose, we acknowledge that [latex]\\operatorname{LCM}(2,4) = 4[\/latex] and [latex]\\operatorname{LCM}(5,10) = 10[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{3}{4}+\\dfrac{1}{2}\\,}{\\,\\dfrac{4}{5}-\\dfrac{1}{10}\\,} &= \\frac{\\,\\dfrac{3}{4}+\\dfrac{1{\\color{red}{} \\cdot 2}}{2{\\color{red}{} \\cdot 2}}\\,}{\\,\\dfrac{4{\\color{red}{} \\cdot 2}}{5{\\color{red}{} \\cdot 2}}-\\dfrac{1}{10}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{3}{4}+\\dfrac{2}{4}\\,}{\\,\\dfrac{8}{10}-\\dfrac{1}{10}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{5}{4}\\,}{\\,\\dfrac{7}{10}\\,}\\\\[5pt]  &= \\frac{5}{4}\\div \\frac{7}{10}\\\\[5pt]  &= \\frac{5}{4}\\cdot \\frac{10}{7}\\\\[5pt]  &= \\frac{5}{2\\cdot {\\color{red}\\cancel{\\color{black}{2}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{2}}} \\cdot 5}{7}\\\\[5pt]  &= \\frac{25}{14}  \\end{align}[\/latex]<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>For the second method, we start with the LCM of the denominators of all\u00a0\u201csmall fractions\u201d. (Please refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-add-and-subtract-rational-expressions-part-i\/#LCMdescr\" target=\"_blank\" rel=\"noopener\">this description<\/a> to review finding the LCM.)<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\operatorname{LCM}(4,2,5,10) &= \\operatorname{LCM}\\left(2^2,2^1,5^1,5^1\\cdot 2^1\\right)\\\\  &= 2^2\\cdot 5^1\\\\  &= 20  \\end{align}[\/latex]<\/p>\n<p>Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{3}{4}+\\dfrac{1}{2}\\,}{\\,\\dfrac{4}{5}-\\dfrac{1}{10}\\,} &= \\frac{\\,\\left(\\dfrac{3}{4}+\\dfrac{1}{2}\\right){\\color{red}{} \\cdot 20}\\,}{\\,\\left(\\dfrac{4}{5}-\\dfrac{1}{10}\\right){\\color{red}{} \\cdot 20}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{3}{4}\\cdot 20+\\dfrac{1}{2}\\cdot 20\\,}{\\,\\dfrac{4}{5}\\cdot 20-\\dfrac{1}{10}\\cdot 20\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{3}{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 5}{1}+\\dfrac{1}{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 10}{1}\\,}{\\,\\dfrac{4}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 4}{1}-\\dfrac{1}{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 2}{1}\\,}\\\\[5pt]  &= \\frac{15+10}{16-2}\\\\[5pt]  &= \\frac{25}{14}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we will show a couple more examples of how to simplify complex fractions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Simplify a Complex Fraction (No Variables)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/lQCwze2w7OU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Complex Rational Expressions<\/h2>\n<p>A <strong>complex rational expression<\/strong> is a quotient with rational expressions in the dividend, divisor, or in both. We simplify these in the exact same way as we would a complex fraction, remembering to identify the restricted values of the variable.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>State the restricted values of the variable and simplify using both methods.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\,\\dfrac{x+5}{x^2-16}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q245262\">Show Solution<\/span><\/p>\n<div id=\"q245262\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the restricted values of [latex]x[\/latex], we need to acknowledge that we have three denominators\u00a0\u2013 the denominators of the \u201csmall fractions\u201d and the \u201cbig denominator\u201d. The\u00a0\u201cbig denominator\u201d is itself a rational expression, so it&#8217;s equal to zero only if its numerator is equal to zero. Since [latex]x^2-16=(x+4)(x-4)[\/latex], this denominator is equal to zero if [latex]x=-4[\/latex] or if [latex]x=4[\/latex] by the Zero-Product Property. The denominator [latex]x-4[\/latex] is equal to zero if [latex]x=4[\/latex], and since [latex]x^2-25 = (x+5)(x-5)[\/latex], the numerator of the\u00a0\u201cbig denominator\u201d is equal to zero if [latex]x=-5[\/latex] or if [latex]x=5[\/latex] by the Zero-Product Property. Summarizing, the restricted values of [latex]x[\/latex] are [latex]-5[\/latex], [latex]-4[\/latex], [latex]4[\/latex], and [latex]5[\/latex].<\/p>\n<p style=\"text-align: left;\">For the first method, we start with rewriting the complex rational expression as division.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{x+5}{x^2-16}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,} &= \\frac{x+5}{x^2-16}\\div\\frac{x^2-25}{x-4}\\\\[5pt]  &= \\frac{x+5}{x^2-16}\\cdot \\frac{x-4}{x^2-25}\\\\[5pt]  &= \\frac{(x+5)}{(x+4)(x-4)}\\cdot \\frac{(x-4)}{(x+5)(x-5)}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{(x+5)}}}\\cdot 1}{(x+4)\\cdot {\\color{red}\\cancel{\\color{black}{(x-4)}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{(x-4)}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{(x+5)}}}\\cdot (x-5)}\\\\[5pt]  &= \\frac{1}{(x+4)(x-5)}  \\end{align}[\/latex]<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>For the second method, we start with the LCM of the denominators of all\u00a0\u201csmall fractions\u201d.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\operatorname{LCM}\\left(x^2-16, x-4\\right) &= \\operatorname{LCM}\\left((x+4)^1\\cdot (x-4)^1, (x-4)^1\\right)\\\\  &= (x+4)^1\\cdot (x-4)^1\\\\  &= (x+4)(x-4)  \\end{align}[\/latex]<\/p>\n<p>Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{x+5}{x^2-16}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,} &= \\frac{\\, \\dfrac{x+5}{x^2-16}{\\color{red}{} \\cdot (x+4)\\cdot (x-4)}\\,}{\\,\\dfrac{x^2-25}{x-4}\\,{\\color{red}{} \\cdot (x+4)\\cdot (x-4)}}\\\\[5pt]  &= \\frac{\\, \\dfrac{(x+5)}{{\\color{red}\\cancel{\\color{black}{(x+4)\\cdot (x-4)}}}\\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{(x+4)\\cdot (x-4)}}}\\cdot 1}{1}\\,}{\\,\\dfrac{(x+5)(x-5)}{{\\color{red}\\cancel{\\color{black}{(x-4)}}} \\cdot 1} \\cdot \\dfrac{(x+4)\\cdot {\\color{red}\\cancel{\\color{black}{(x-4)}}}}{1}}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{(x+5)}}} \\cdot 1}{{\\color{red}\\cancel{\\color{black}{(x+5)}}}\\cdot (x-5)\\cdot (x+4)}\\\\[5pt]  &= \\frac{1}{(x-5)(x+4)}  \\end{align}[\/latex]<\/p>\n<p>This time, in the final answer we may conveniently multiply the denominator out and simplify to obtain an equivalent form [latex]\\dfrac{1}{x^2-x-20}[\/latex].\u00a0 It&#8217;s not expected or necessary though.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The next video contains another example of simplifying a complex rational expression with just two \u201csmall fractions\u201d.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 2:  Simplify a Complex Fraction (Variables)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fAaqo8gGW9Y?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The same ideas can be used when simplifying complex rational expressions that include more than one rational expression\u00a0in the \u201cbig numerator\u201d or \u201cbig denominator\u201d (or both).<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>State the restricted values of the variable and simplify using both methods.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\,1-\\dfrac{9}{x^2}\\,}{\\,1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\,}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q344101\">Show Solution<\/span><\/p>\n<div id=\"q344101\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the restricted values of [latex]x[\/latex], we need to acknowledge that we have four denominators\u00a0\u2013 the denominators of the \u201csmall fractions\u201d and the \u201cbig denominator\u201d. While the denominators of all \u201csmall fractions\u201d are just powers of [latex]x[\/latex], meaning only [latex]x=0[\/latex] would make any of them equal to zero, the \u201cbig denominator\u201d is not yet a single rational expression, so we first combine the \u201cbig denominator\u201d into a single rational expression to acknowledge its numerator for restricted values of the variable purposes. The LCM of [latex]x[\/latex] and [latex]x^2[\/latex] is [latex]x^2[\/latex], and [latex]1=\\dfrac{1}{1}[\/latex], thus<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{1}{1}+\\frac{5}{x}+\\frac{6}{x^2} &= \\frac{1{\\color{red}{} \\cdot x^2}}{1{\\color{red}{} \\cdot x^2}}+\\frac{5{\\color{red}{} \\cdot x}}{x{\\color{red}{} \\cdot x}}+\\frac{6}{x^2}\\\\[5pt]  &= \\frac{x^2+5x+6}{x^2}\\\\[5pt]  &= \\frac{(x+2)(x+3)}{x^2}  \\end{align}[\/latex]<\/p>\n<p>Hence, the numerator of\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression is equal to zero if [latex]x=-2[\/latex] or if [latex]x=-3[\/latex] by the Zero-Product Property. Summarizing, the restricted values of [latex]x[\/latex] are [latex]0[\/latex], [latex]-2[\/latex], and [latex]-3[\/latex].<\/p>\n<p>Since we already have\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression, for the first method we only need to rewrite\u00a0the \u201cbig numerator\u201d as a single rational expression and proceed accordingly.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,1-\\dfrac{9}{x^2}\\,}{\\,1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\,} &= \\frac{\\,\\dfrac{1}{1}-\\dfrac{9}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{1{\\color{red}{} \\cdot x^2}}{1{\\color{red}{} \\cdot x^2}}-\\dfrac{9}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{x^2-9}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{(x+3)(x-3)}{x^2}\\,}{\\,\\dfrac{(x+2)(x+3)}{x^2}\\,}\\\\[5pt]  &= \\dfrac{(x+3)(x-3)}{x^2}\\div\\dfrac{(x+2)(x+3)}{x^2}\\\\[5pt]  &= \\dfrac{(x+3)(x-3)}{x^2}\\cdot\\dfrac{x^2}{(x+2)(x+3)}\\\\[5pt]  &= \\dfrac{{\\color{red}\\cancel{\\color{black}{(x+3)}}}\\cdot (x-3)}{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}{(x+2)\\cdot {\\color{red}\\cancel{\\color{black}{(x+3)}}}}  \\end{align}[\/latex]<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>For the second method, we acknowledge that the LCM of the denominators of all\u00a0\u201csmall fractions\u201d is [latex]x^2[\/latex], and proceed accordingly.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,1-\\dfrac{9}{x^2}\\,}{\\,1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\,} &= \\frac{\\,\\left(1-\\dfrac{9}{x^2}\\right){\\color{red}\\cdot\\: x^2}\\,}{\\,\\left(1+\\dfrac{5}{x}+\\dfrac{6}{x^2}\\right){\\color{red} \\cdot\\: x^2}\\,}\\\\[5pt]  &= \\frac{\\,1\\cdot x^2-\\dfrac{9}{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}{1}\\,}{\\,1\\cdot x^2+\\dfrac{5}{{\\color{red}\\cancel{\\color{black}{x}}} \\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{x}}}\\cdot x}{1}+\\dfrac{6}{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}\\cdot \\dfrac{{\\color{red}\\cancel{\\color{black}{x^2}}} \\cdot 1}{1}\\,}\\\\[5pt]  &= \\frac{x^2-9}{x^2+5x+6}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{(x+3)}}}\\cdot (x-3)}{{\\color{red}\\cancel{\\color{black}{(x+3)}}}\\cdot(x+2)}\\\\[5pt]  &= \\frac{x-3}{x+2}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The next video contains another example of simplifying a complex rational expression similar to the one in the example above.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 3:  Simplify a Complex Fraction (Variables)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/P5dfmX_FNPk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We conclude with an example in which the\u00a0LCM of the denominators of all\u00a0\u201csmall fractions\u201d will be more complicated.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>State the restricted values of the variable and simplify using both methods.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{\\,\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\,}{\\,\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\,}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q234174\">Show Solution<\/span><\/p>\n<div id=\"q234174\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the restricted values of [latex]x[\/latex], we need to acknowledge that we have five denominators\u00a0\u2013 the denominators of the \u201csmall fractions\u201d and the \u201cbig denominator\u201d. For the denominators of all \u201csmall fractions\u201d, we solve the following equations:<\/p>\n<ul>\n<li>[latex]x^2-1=0[\/latex], equivalent to [latex](x-1)(x+1)=0[\/latex], resulting in\u00a0[latex]x=1[\/latex] or [latex]x=-1[\/latex] by the Zero-Factor Property,<\/li>\n<li>[latex]x-3=0[\/latex], resulting in [latex]x=3[\/latex],<\/li>\n<li>[latex]x+1=0[\/latex], resulting in [latex]x=-1[\/latex], and<\/li>\n<li>[latex]x^2-2x-3=0[\/latex], equivalent to [latex](x-3)(x+1)=0[\/latex], resulting in\u00a0[latex]x=3[\/latex] or [latex]x=-1[\/latex] by the Zero-Factor Property.<\/li>\n<\/ul>\n<p>The \u201cbig denominator\u201d is not yet a single rational expression, so we first combine the \u201cbig denominator\u201d into a single rational expression to acknowledge its numerator for restricted values of the variable purposes. Based on the factorization [latex](x-3)(x+1)[\/latex] of [latex]x^2-2x-3[\/latex], the LCM of [latex]x+1[\/latex] and [latex]x^2-2x-3[\/latex] is [latex](x-3)(x+1)[\/latex] and thus<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{2}{x+1}-\\frac{3x}{x^2-2x-3} &= \\frac{2{\\color{red}\\: \\cdot\\: (x-3)}}{(x+1){\\color{red}\\: \\cdot\\: (x-3)}}-\\frac{3x}{(x-3)(x+1)}\\\\[5pt]  &= \\frac{2x-6-3x}{(x-3)(x+1)}\\\\[5pt]  &= \\frac{-x-6}{(x-3)(x+1)}  \\end{align}[\/latex]<\/p>\n<p>Hence, the numerator of\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression, [latex]-x-6[\/latex] is equal to zero if [latex]x=-6[\/latex]. Summarizing, the restricted values of [latex]x[\/latex] are [latex]1[\/latex], [latex]-1[\/latex], [latex]3[\/latex], and [latex]-6[\/latex].<\/p>\n<p>Since we already have\u00a0the \u201cbig denominator\u201d rewritten as a single rational expression, for the first method we only need to rewrite\u00a0the \u201cbig numerator\u201d as a single rational expression and proceed accordingly. Notice that [latex]\\operatorname{LCM}\\left(x^2-1, x-3\\right)=\\operatorname{LCM}\\left((x+1)(x-1), x-3\\right)=(x+1)(x-1)(x-3)[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\,}{\\,\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\,} &= \\frac{\\,\\dfrac{x{\\color{red}\\: \\cdot\\: (x-3)}}{(x+1)\\cdot(x-1){\\color{red}\\: \\cdot\\: (x-3)}}-\\dfrac{(x+1){\\color{red}\\: \\cdot\\: (x-1)\\cdot(x+1)}}{(x-3){\\color{red}\\: \\cdot\\: (x-1)\\cdot(x+1)}}\\,}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{x^2-3x-\\left(\\left(x^2-1\\right)\\cdot(x+1)\\right)}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{x^2-3x-\\left(x^3+x^2-x-1\\right)}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{x^2-3x-x^3-x^2+x+1}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]  &= \\frac{\\,\\dfrac{-x^3-2x+1}{(x-3)\\cdot(x-1)\\cdot(x+1)}}{\\,\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\,}\\\\[5pt]  &= \\frac{-x^3-2x+1}{(x-3)\\cdot(x-1)\\cdot(x+1)}\\div\\dfrac{-x-6}{(x-3)\\cdot(x+1)}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot\\left(x^3+2x-1\\right)}{{\\color{red}\\cancel{\\color{black}{(x-3)}}}\\cdot(x-1)\\cdot{\\color{red}\\cancel{\\color{black}{(x+1)}}}}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{(x-3)}}}\\cdot{\\color{red}\\cancel{\\color{black}{(x+1)}}}}{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot(x+6)}\\\\[5pt]  &= \\frac{x^3+2x-1}{(x-1)(x+6)}  \\end{align}[\/latex]<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>For the second method, let&#8217;s begin with the LCM of the denominators of all\u00a0\u201csmall fractions\u201d:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\operatorname{LCM}\\left(x^2-1, x-3, x+1, x^2-2x-3\\right)&=\\operatorname{LCM}\\left((x+1)^1\\cdot(x-1)^1, (x-3)^1, (x+1)^1, (x-3)^1\\cdot(x+1)^1\\right)\\\\[5pt]  &=(x+1)(x-1)(x-3)  \\end{align}[\/latex]<\/p>\n<p>Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{\\,\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\,}{\\,\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\,} &= \\frac{\\,\\left(\\dfrac{x}{x^2-1}-\\dfrac{x+1}{x-3}\\right){\\color{red}\\: \\cdot\\: (x+1)\\cdot(x-1)\\cdot(x-3)}\\,}{\\,\\left(\\dfrac{2}{x+1}-\\dfrac{3x}{x^2-2x-3}\\right){\\color{red}\\: \\cdot\\: (x+1)\\cdot(x-1)\\cdot(x-3)}\\,}\\\\[5pt]  &=\\frac{\\,\\dfrac{x}{{\\color{red}\\cancel{\\color{black}{\\left(x^2-1\\right)}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{\\left(x^2-1\\right)}}}\\cdot (x-3)}{1}-\\dfrac{(x+1)}{{\\color{red}\\cancel{\\color{black}{(x-3)}}}\\cdot 1}\\cdot\\dfrac{\\left(x^2-1\\right)\\cdot {\\color{red}\\cancel{\\color{black}{(x-3)}}}}{1}\\,}{\\,\\dfrac{2}{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{(x+1)}}}\\cdot\\left(x^2-4x+3\\right)}{1}-\\dfrac{3x}{{\\color{red}\\cancel{\\color{black}{\\left(x^2-2x-3\\right)}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{\\left(x^2-2x-3\\right)}}}\\cdot(x-1)}{1}\\,}\\\\[5pt]  &= \\frac{x^2-3x-\\left(x^3-x+x^2-1\\right)}{2x^2-8x+6-\\left(3x^2-3x\\right)}\\\\[5pt]  &= \\frac{x^2-3x-x^3+x-x^2+1}{2x^2-8x+6-3x^2+3x}\\\\[5pt]  &= \\frac{-x^3-2x+1}{-x^2-5x+6}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot\\left(x^3+2x-1\\right)}{{\\color{red}\\cancel{\\color{black}{(-1)}}}\\cdot\\left(x^2+5x-6\\right)}\\\\[5pt]  &= \\frac{x^3+2x-1}{(x-1)(x+6)}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>Complex rational expressions are quotients with rational expressions in the divisor, dividend, or both. When written in fraction form, they appear to be fractions within a fraction. These can be simplified by first treating the quotient as a division problem, after you present the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d, respectively, as simplified fractions. Then you can rewrite the division as multiplication by the reciprocal of the divisor. Or you can simplify the complex rational expression by multiplying both the\u00a0\u201cbig numerator\u201d and\u00a0\u201cbig denominator\u201d by\u00a0the LCM of the denominators of all\u00a0\u201csmall fractions\u201d. This can help simplify the complex expression even faster.<\/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-177\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Simplify a Complex Fraction (No Variables). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/lQCwze2w7OU\">https:\/\/youtu.be\/lQCwze2w7OU<\/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>Unit 15: Rational Expressions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Simplify a Complex Fraction (Variables). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fAaqo8gGW9Y\">https:\/\/youtu.be\/fAaqo8gGW9Y<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 3: Simplify a Complex Fraction (Variables). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/P5dfmX_FNPk\">https:\/\/youtu.be\/P5dfmX_FNPk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex 1: Simplify a Complex Fraction (No Variables)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/lQCwze2w7OU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 15: Rational Expressions, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Simplify a Complex Fraction (Variables)\",\"author\":\"James Sousa (Mathispower4u.com) 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