{"id":4978,"date":"2017-12-29T18:02:15","date_gmt":"2017-12-29T18:02:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/?post_type=chapter&#038;p=4978"},"modified":"2018-05-17T01:06:42","modified_gmt":"2018-05-17T01:06:42","slug":"rational-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/rational-equations\/","title":{"raw":"Rational Equations","rendered":"Rational Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Methods for solving rational equations\r\n<ul>\r\n \t<li>Solve rational equations by clearing denominators<\/li>\r\n \t<li>Identify extraneous solutions in a rational equation<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n&nbsp;\r\n\r\nEquations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex]\\displaystyle \\frac{2x+1}{4}=\\frac{7}{x}[\/latex] is a rational equation.\u00a0Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of\u00a0proportional relationships.\r\n\r\nOne of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator, then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:\r\n\r\nSolve \u00a0[latex]\\frac{2}{3}x - \\frac{5}{6} = \\frac{3}{4}[\/latex] by clearing the fractions in the equation first.\r\n<p style=\"text-align: center\">[latex]\\begin{eqnarray*} \\frac{2}{3}\\,x - \\frac{5}{6}\\,= \\frac{3}{4}\\,&amp; &amp; \\text{ Multiply}\\,\\text{ each}\r\n\\text{ term}\\,\\text{ by}\\,\\text{ LCD}, 12\\\\\r\n&amp; &amp; \\\\\r\n\\frac{2 (12)}{3}\\,x - \\frac{5 (12)}{6}\\,= \\frac{3 (12)}{4}\\,&amp; &amp;\r\n\\text{ Reduce}\\,\\text{ fractions}\\\\\r\n&amp; &amp; \\\\\r\n2 (4) x - 5 (2) = 3 (3) &amp; &amp; \\text{ Multiply}\\\\\r\n8 x - 10 = 9 &amp; &amp; \\text{ Solve}\\\\\r\n\\underline{+ 10 + 10}\\,&amp; &amp; \\text{ Add}\\,10 \\text{ to}\\,\\text{ both}\r\n\\text{ sides}\\\\\r\n8 x = 19 &amp; &amp; \\text{ Divide}\\,\\text{ both}\\,\\text{ sides}\\,\\text{ by}\\,8\\\\\r\n\\overline{8}\\, \\overline{8}\\,&amp; &amp; \\\\\r\nx = \\frac{19}{8}\\,&amp; &amp; \\text{ Our}\\,\\text{ Solution}\r\n\\end{eqnarray*}[\/latex]<\/p>\r\nWe could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations. \u00a0The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve the equation: [latex]\\displaystyle 3 x - \\frac{1}{2}\\,= \\frac{1}{x}[\/latex]\r\n[reveal-answer q=\"837141\"]Show Solution[\/reveal-answer]\r\n<p style=\"text-align: center\">[hidden-answer a=\"837141\"][latex]\\begin{eqnarray*} 3 x - \\frac{1}{2}\\,= \\frac{1}{x}\\,&amp; &amp; \\text{ Multiply}\\,\\text{ each} \\text{ term}\\,\\text{ by}\\,\\text{ LCD}, (2 x)\\\\ 3 x (2 x) - \\frac{(2 x)}{2}\\,= \\frac{(2 x)}{x}\\,&amp; &amp; \\text{ Reduce} \\text{ fractions}\\\\ 6 x^2 - x = 2 &amp; &amp; \\text{ Distribute}\\\\ \\underline{- 2 - 2}\\,&amp; &amp; \\text{ Subtract}\\,2 \\text{ from}\\,\\text{ both} \\text{ sides}\\\\ 6 x^2 - x - 2 = 0 &amp; &amp; \\text{ Factor}\\\\ (3 x - 2) (2 x + 1) = 0 &amp; &amp; \\text{ Set}\\,\\text{ each}\\,\\text{ factor} \\text{ equal}\\,\\text{ to}\\,\\text{ zero}\\\\ 3 x - 2 = 0 \\text{ or}\\,2 x + 1 = 0 &amp; &amp; \\text{ Solve}\\,\\text{ each} \\text{ equation}\\\\ \\underline{+ 2 \\quad + 2}\\, \\underline{- 1 - 1}\\,&amp; &amp; \\\\ 3 x = 2 \\hspace{3em}\\,2 x = - 1 &amp; &amp; \\\\ x = \\frac{2}{3}\\,\\text{ or}\\,x = - \\frac{1}{2}\\,&amp; &amp; \\text{ Check} \\text{ solutions}, \\text{ LCD}\\,\\text{ can}' t \\text{ be}\\,\\text{ zero}\\\\ 2 \\left( \\frac{3}{2}\\,\\right) =2 \\left( - \\frac{1}{2}\\,\\right) = - 1 &amp; &amp; \\text{ Neither}\\,\\text{ make}\\,\\text{ LCD}\\,\\text{ zero}, \\text{ both}\\,\\text{ are} \\text{ solutions}\\\\ x = \\frac{2}{3}\\,\\text{ or}\\,x = - \\frac{1}{2}\\,&amp; &amp; \\text{ Our} \\text{ Solution} \\end{eqnarray*}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle x = \\frac{2}{3}\\,\\text{ or}\\,x = - \\frac{1}{2}\\,[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the next two examples, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they don't share any common factors.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve the equation [latex]\\displaystyle \\frac{8}{x+1}=\\frac{4}{3}[\/latex].\r\n\r\n[reveal-answer q=\"331190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"331190\"]Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3\\left(x+1\\right)[\/latex] since [latex]3\\text{ and }x+1[\/latex] don't have any common factors.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}3\\left(x+1\\right)\\left(\\frac{8}{x+1}\\right)=3\\left(x+1\\right)\\left(\\frac{4}{3}\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Simplify common factors.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}3\\cancel{\\left(x+1\\right)}\\left(\\frac{8}{\\cancel{x+1}}\\right)=\\cancel{3}\\left(x+1\\right)\\left(\\frac{4}{\\cancel{3}}\\right)\\\\24=4\\left(x+1\\right)\\\\24=4x+4\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}24=4x+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\20=4x\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\x=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck the solution in the original equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\frac{8}{\\left(x+1\\right)}=\\frac{4}{3}\\\\\\\\\\frac{8}{\\left(5+1\\right)}=\\frac{4}{3}\\\\\\\\\\frac{8}{6}=\\frac{4}{3}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Reduce the fraction [latex]\\frac{8}{6}[\/latex] by simplifying\u00a0the common factor of 2:<\/p>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{\\cancel{2}\\cdot4}{\\cancel{2}\\cdot3}\\normalsize=\\large\\frac{4}{3}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve the equation: [latex]\\displaystyle \\frac{5 x + 5}{x + 2}\\,+ 3 x = \\frac{x^2}{x + 2}[\/latex]\r\n<p style=\"text-align: left\">[reveal-answer q=\"993275\"]Show Answer[\/reveal-answer]<\/p>\r\n<p style=\"text-align: center\">[hidden-answer a=\"993275\"][latex] \\begin{eqnarray*}\\frac{5 x + 5}{x + 2}\\,+ 3 x = \\frac{x^2}{x + 2}\\,&amp; &amp; \\text{ Multiply} \\text{ each}\\,\\text{ term}\\,\\text{ by}\\,\\text{ LCD}, (x + 2)\\\\ &amp; &amp; \\\\ \\frac{(5 x + 5) (x + 2)}{x + 2}\\,+ 3 x (x + 2) = \\frac{x^2 (x + 2)}{x + 2} &amp; &amp; \\text{ Reduce}\\,\\text{ fractions}\\\\ &amp; &amp; \\\\ 5 x + 5 + 3 x (x + 2) = x^2 &amp; &amp; \\text{ Distribute}\\\\ 5 x + 5 + 3 x^2 + 6 x = x^2 &amp; &amp; \\text{ Combine}\\,\\text{ like}\\,\\text{ terms}\\\\ 3 x^2 + 11 x + 5 = x^2 &amp; &amp; \\text{ Make}\\,\\text{ equation}\\,\\text{ equal} \\text{ zero}\\\\ \\underline{- x^2 - x^2}\\,&amp; &amp; \\text{ Subtract}\\,x^2 \\text{ from}\\,\\text{ both} \\text{ sides}\\\\ 2 x^2 + 11 x + 5 = 0 &amp; &amp; \\text{ Factor}\\\\ (2 x + 1) (x + 5) = 0 &amp; &amp; \\text{ Set}\\,\\text{ each}\\,\\text{ factor} \\text{ equal}\\,\\text{ to}\\,\\text{ zero}\\\\ 2 x + 1 = 0 \\text{ or}\\,x + 5 = 0 &amp; &amp; \\text{ Solve}\\,\\text{ each} \\text{ equation}\\\\ 2 x = - 1 \\text{ or}\\,x = - 5 &amp; &amp; \\\\ x = - \\frac{1}{2}\\,\\text{ or}\\,- 5 &amp; &amp; \\text{ Check}\\,\\text{ solutions}, \\text{ LCD}\\,\\text{ can}' t \\text{ be}\\,\\text{ zero}\\\\ - \\frac{1}{2}\\,+ 2 = \\frac{3}{2}\\,- 5 + 2 = - 3 &amp; &amp; \\text{ Neither} \\text{ make}\\,\\text{ LCD}\\,\\text{ zero}, \\text{ both}\\,\\text{ are} \\text{ solutions}\\\\ x = - \\frac{1}{2}\\,\\text{ or}\\,- 5 &amp; &amp; \\text{ Our}\\,\\text{ Solution} \\end{eqnarray*}[\/latex]<\/p>\r\n<strong>Answer<\/strong>\r\n\r\n[latex]x = - \\frac{1}{2}\\,\\text{ or}\\,x = - 5[\/latex]\r\n<p style=\"text-align: left\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nIn the video that follows we present two ways to solve rational equations with both integer and variable denominators.\r\n\r\nhttps:\/\/youtu.be\/R9y2D9VFw0I\r\n<h2>Excluded Values and Extraneous Solutions<\/h2>\r\nSome rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called <strong>excluded values<\/strong>. Let\u2019s look at an example.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve the equation [latex]\\displaystyle \\frac{2x-5}{x-5}=\\frac{15}{x-5}[\/latex].\r\n\r\n[reveal-answer q=\"266674\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"266674\"]Determine any values for <i>x <\/i>that would make the denominator 0.\r\n<p style=\"text-align: center\">[latex] \\frac{2x-5}{x-5}=\\frac{15}{x-5}[\/latex]<\/p>\r\n5 is an excluded value because it makes the denominator [latex]x-5[\/latex] equal to 0.\r\n\r\nSince the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for <i>x.<\/i>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2x-5=15\\\\2x=20\\\\x=10\\end{array}[\/latex]<\/p>\r\nCheck the solution in the original equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\frac{2x-5}{x-5}=\\frac{15}{x-5}\\,\\,\\\\\\\\\\frac{2(10)-5}{10-5}=\\frac{15}{10-5}\\\\\\\\\\frac{20-5}{10-5}=\\frac{15}{10-5}\\\\\\\\\\frac{15}{5}=\\frac{15}{5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=10[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we present an example of solving a rational equation with variables in the denominator.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=gGA-dF_aQQQ&amp;feature=youtu.be\r\n\r\nYou\u2019ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don\u2019t work in the original form of the equation. These types of answers are called <strong>extraneous solutions<\/strong>. That's why it is always important to check all solutions in the original equations\u2014you may find that they yield untrue statements or produce undefined expressions.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve the equation [latex]\\displaystyle \\frac{16}{m+4}=\\frac{{{m}^{2}}}{m+4}[\/latex].\r\n\r\n[reveal-answer q=\"450589\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"450589\"]Determine any values for <i>m <\/i>that would make the denominator 0. [latex]\u22124[\/latex] is an excluded value because it makes [latex]m+4[\/latex]\u00a0equal to 0.\r\n\r\nSince the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for <i>m.<\/i>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}16=m^{2}\\\\\\,\\,\\,0={{m}^{2}}-16\\\\\\,\\,\\,0=\\left( m+4 \\right)\\left( m-4 \\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0=m+4\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,0=m-4\\\\m=-4\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,m=4\\\\m=4,-4\\end{array}[\/latex]<\/p>\r\nCheck the solutions in the original equation.\r\n\r\nSince [latex]m=\u22124[\/latex] leads to division by 0, it is an extraneous solution.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\frac{16}{m+4}=\\frac{{{m}^{2}}}{m+4}\\\\\\\\\\frac{16}{-4+4}=\\frac{{{(-4)}^{2}}}{-4+4}\\\\\\\\\\frac{16}{0}=\\frac{16}{0}\\end{array}[\/latex]<\/p>\r\n[latex]-4[\/latex] is excluded because\u00a0it leads to division by 0.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\frac{16}{4+4}=\\frac{{{(4)}^{2}}}{4+4}\\\\\\\\\\frac{16}{8}=\\frac{16}{8}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]m=4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>More Rational Equations<\/h3>\r\nSometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Rational Equation Leading to a Quadratic<\/h3>\r\nSolve the following rational equation: [latex]\\displaystyle \\frac{-4x}{x - 1}+\\frac{4}{x+1}=\\frac{-8}{{x}^{2}-1}[\/latex].\r\n\r\n[reveal-answer q=\"164755\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"164755\"]\r\nWe want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Then, the LCD is [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Next, we multiply the whole equation by the LCD.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}\\left(x+1\\right)\\left(x - 1\\right)\\left[\\frac{-4x}{x - 1}+\\frac{4}{x+1}\\right]\\hfill&amp;=\\left[\\frac{-8}{\\left(x+1\\right)\\left(x - 1\\right)}\\right]\\left(x+1\\right)\\left(x - 1\\right)\\hfill \\\\ -4x\\left(x+1\\right)+4\\left(x - 1\\right)\\hfill&amp;=-8\\hfill \\\\ -4{x}^{2}-4x+4x - 4\\hfill&amp;=-8\\hfill \\\\ -4{x}^{2}+4\\hfill&amp;=0\\hfill \\\\ -4\\left({x}^{2}-1\\right)\\hfill&amp;=0\\hfill \\\\ -4\\left(x+1\\right)\\left(x - 1\\right)\\hfill&amp;=0\\hfill \\\\ x\\hfill&amp;=-1\\hfill \\\\ x\\hfill&amp;=1\\hfill \\end{array}[\/latex]<\/div>\r\nIn this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\displaystyle \\frac{3x+2}{x - 2}+\\frac{1}{x}=\\frac{-2}{{x}^{2}-2x}[\/latex].[reveal-answer q=\"586054\"]Answer[\/reveal-answer]\r\n[hidden-answer a=\"586054\"][latex]x=-1[\/latex], ([latex]x=0[\/latex] is not a solution).[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3496&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]44861[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Methods for solving rational equations\n<ul>\n<li>Solve rational equations by clearing denominators<\/li>\n<li>Identify extraneous solutions in a rational equation<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Equations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex]\\displaystyle \\frac{2x+1}{4}=\\frac{7}{x}[\/latex] is a rational equation.\u00a0Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of\u00a0proportional relationships.<\/p>\n<p>One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator, then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:<\/p>\n<p>Solve \u00a0[latex]\\frac{2}{3}x - \\frac{5}{6} = \\frac{3}{4}[\/latex] by clearing the fractions in the equation first.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{eqnarray*} \\frac{2}{3}\\,x - \\frac{5}{6}\\,= \\frac{3}{4}\\,& & \\text{ Multiply}\\,\\text{ each}  \\text{ term}\\,\\text{ by}\\,\\text{ LCD}, 12\\\\  & & \\\\  \\frac{2 (12)}{3}\\,x - \\frac{5 (12)}{6}\\,= \\frac{3 (12)}{4}\\,& &  \\text{ Reduce}\\,\\text{ fractions}\\\\  & & \\\\  2 (4) x - 5 (2) = 3 (3) & & \\text{ Multiply}\\\\  8 x - 10 = 9 & & \\text{ Solve}\\\\  \\underline{+ 10 + 10}\\,& & \\text{ Add}\\,10 \\text{ to}\\,\\text{ both}  \\text{ sides}\\\\  8 x = 19 & & \\text{ Divide}\\,\\text{ both}\\,\\text{ sides}\\,\\text{ by}\\,8\\\\  \\overline{8}\\, \\overline{8}\\,& & \\\\  x = \\frac{19}{8}\\,& & \\text{ Our}\\,\\text{ Solution}  \\end{eqnarray*}[\/latex]<\/p>\n<p>We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations. \u00a0The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation: [latex]\\displaystyle 3 x - \\frac{1}{2}\\,= \\frac{1}{x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q837141\">Show Solution<\/span><\/p>\n<p style=\"text-align: center\">\n<div id=\"q837141\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{eqnarray*} 3 x - \\frac{1}{2}\\,= \\frac{1}{x}\\,& & \\text{ Multiply}\\,\\text{ each} \\text{ term}\\,\\text{ by}\\,\\text{ LCD}, (2 x)\\\\ 3 x (2 x) - \\frac{(2 x)}{2}\\,= \\frac{(2 x)}{x}\\,& & \\text{ Reduce} \\text{ fractions}\\\\ 6 x^2 - x = 2 & & \\text{ Distribute}\\\\ \\underline{- 2 - 2}\\,& & \\text{ Subtract}\\,2 \\text{ from}\\,\\text{ both} \\text{ sides}\\\\ 6 x^2 - x - 2 = 0 & & \\text{ Factor}\\\\ (3 x - 2) (2 x + 1) = 0 & & \\text{ Set}\\,\\text{ each}\\,\\text{ factor} \\text{ equal}\\,\\text{ to}\\,\\text{ zero}\\\\ 3 x - 2 = 0 \\text{ or}\\,2 x + 1 = 0 & & \\text{ Solve}\\,\\text{ each} \\text{ equation}\\\\ \\underline{+ 2 \\quad + 2}\\, \\underline{- 1 - 1}\\,& & \\\\ 3 x = 2 \\hspace{3em}\\,2 x = - 1 & & \\\\ x = \\frac{2}{3}\\,\\text{ or}\\,x = - \\frac{1}{2}\\,& & \\text{ Check} \\text{ solutions}, \\text{ LCD}\\,\\text{ can}' t \\text{ be}\\,\\text{ zero}\\\\ 2 \\left( \\frac{3}{2}\\,\\right) =2 \\left( - \\frac{1}{2}\\,\\right) = - 1 & & \\text{ Neither}\\,\\text{ make}\\,\\text{ LCD}\\,\\text{ zero}, \\text{ both}\\,\\text{ are} \\text{ solutions}\\\\ x = \\frac{2}{3}\\,\\text{ or}\\,x = - \\frac{1}{2}\\,& & \\text{ Our} \\text{ Solution} \\end{eqnarray*}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle x = \\frac{2}{3}\\,\\text{ or}\\,x = - \\frac{1}{2}\\,[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the next two examples, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they don&#8217;t share any common factors.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation [latex]\\displaystyle \\frac{8}{x+1}=\\frac{4}{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q331190\">Show Solution<\/span><\/p>\n<div id=\"q331190\" class=\"hidden-answer\" style=\"display: none\">Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3\\left(x+1\\right)[\/latex] since [latex]3\\text{ and }x+1[\/latex] don&#8217;t have any common factors.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}3\\left(x+1\\right)\\left(\\frac{8}{x+1}\\right)=3\\left(x+1\\right)\\left(\\frac{4}{3}\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Simplify common factors.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}3\\cancel{\\left(x+1\\right)}\\left(\\frac{8}{\\cancel{x+1}}\\right)=\\cancel{3}\\left(x+1\\right)\\left(\\frac{4}{\\cancel{3}}\\right)\\\\24=4\\left(x+1\\right)\\\\24=4x+4\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}24=4x+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\20=4x\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\x=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Check the solution in the original equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\frac{8}{\\left(x+1\\right)}=\\frac{4}{3}\\\\\\\\\\frac{8}{\\left(5+1\\right)}=\\frac{4}{3}\\\\\\\\\\frac{8}{6}=\\frac{4}{3}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Reduce the fraction [latex]\\frac{8}{6}[\/latex] by simplifying\u00a0the common factor of 2:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{\\cancel{2}\\cdot4}{\\cancel{2}\\cdot3}\\normalsize=\\large\\frac{4}{3}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation: [latex]\\displaystyle \\frac{5 x + 5}{x + 2}\\,+ 3 x = \\frac{x^2}{x + 2}[\/latex]<\/p>\n<p style=\"text-align: left\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q993275\">Show Answer<\/span><\/p>\n<p style=\"text-align: center\">\n<div id=\"q993275\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{eqnarray*}\\frac{5 x + 5}{x + 2}\\,+ 3 x = \\frac{x^2}{x + 2}\\,& & \\text{ Multiply} \\text{ each}\\,\\text{ term}\\,\\text{ by}\\,\\text{ LCD}, (x + 2)\\\\ & & \\\\ \\frac{(5 x + 5) (x + 2)}{x + 2}\\,+ 3 x (x + 2) = \\frac{x^2 (x + 2)}{x + 2} & & \\text{ Reduce}\\,\\text{ fractions}\\\\ & & \\\\ 5 x + 5 + 3 x (x + 2) = x^2 & & \\text{ Distribute}\\\\ 5 x + 5 + 3 x^2 + 6 x = x^2 & & \\text{ Combine}\\,\\text{ like}\\,\\text{ terms}\\\\ 3 x^2 + 11 x + 5 = x^2 & & \\text{ Make}\\,\\text{ equation}\\,\\text{ equal} \\text{ zero}\\\\ \\underline{- x^2 - x^2}\\,& & \\text{ Subtract}\\,x^2 \\text{ from}\\,\\text{ both} \\text{ sides}\\\\ 2 x^2 + 11 x + 5 = 0 & & \\text{ Factor}\\\\ (2 x + 1) (x + 5) = 0 & & \\text{ Set}\\,\\text{ each}\\,\\text{ factor} \\text{ equal}\\,\\text{ to}\\,\\text{ zero}\\\\ 2 x + 1 = 0 \\text{ or}\\,x + 5 = 0 & & \\text{ Solve}\\,\\text{ each} \\text{ equation}\\\\ 2 x = - 1 \\text{ or}\\,x = - 5 & & \\\\ x = - \\frac{1}{2}\\,\\text{ or}\\,- 5 & & \\text{ Check}\\,\\text{ solutions}, \\text{ LCD}\\,\\text{ can}' t \\text{ be}\\,\\text{ zero}\\\\ - \\frac{1}{2}\\,+ 2 = \\frac{3}{2}\\,- 5 + 2 = - 3 & & \\text{ Neither} \\text{ make}\\,\\text{ LCD}\\,\\text{ zero}, \\text{ both}\\,\\text{ are} \\text{ solutions}\\\\ x = - \\frac{1}{2}\\,\\text{ or}\\,- 5 & & \\text{ Our}\\,\\text{ Solution} \\end{eqnarray*}[\/latex]<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>[latex]x = - \\frac{1}{2}\\,\\text{ or}\\,x = - 5[\/latex]<\/p>\n<p style=\"text-align: left\"><\/div>\n<\/div>\n<\/div>\n<p>In the video that follows we present two ways to solve rational equations with both integer and variable denominators.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solve Basic Rational Equations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/R9y2D9VFw0I?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Excluded Values and Extraneous Solutions<\/h2>\n<p>Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called <strong>excluded values<\/strong>. Let\u2019s look at an example.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation [latex]\\displaystyle \\frac{2x-5}{x-5}=\\frac{15}{x-5}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266674\">Show Solution<\/span><\/p>\n<div id=\"q266674\" class=\"hidden-answer\" style=\"display: none\">Determine any values for <i>x <\/i>that would make the denominator 0.<\/p>\n<p style=\"text-align: center\">[latex]\\frac{2x-5}{x-5}=\\frac{15}{x-5}[\/latex]<\/p>\n<p>5 is an excluded value because it makes the denominator [latex]x-5[\/latex] equal to 0.<\/p>\n<p>Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for <i>x.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2x-5=15\\\\2x=20\\\\x=10\\end{array}[\/latex]<\/p>\n<p>Check the solution in the original equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\frac{2x-5}{x-5}=\\frac{15}{x-5}\\,\\,\\\\\\\\\\frac{2(10)-5}{10-5}=\\frac{15}{10-5}\\\\\\\\\\frac{20-5}{10-5}=\\frac{15}{10-5}\\\\\\\\\\frac{15}{5}=\\frac{15}{5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we present an example of solving a rational equation with variables in the denominator.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Solve  Rational Equations with Like Denominators\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/gGA-dF_aQQQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You\u2019ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don\u2019t work in the original form of the equation. These types of answers are called <strong>extraneous solutions<\/strong>. That&#8217;s why it is always important to check all solutions in the original equations\u2014you may find that they yield untrue statements or produce undefined expressions.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation [latex]\\displaystyle \\frac{16}{m+4}=\\frac{{{m}^{2}}}{m+4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q450589\">Show Solution<\/span><\/p>\n<div id=\"q450589\" class=\"hidden-answer\" style=\"display: none\">Determine any values for <i>m <\/i>that would make the denominator 0. [latex]\u22124[\/latex] is an excluded value because it makes [latex]m+4[\/latex]\u00a0equal to 0.<\/p>\n<p>Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for <i>m.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}16=m^{2}\\\\\\,\\,\\,0={{m}^{2}}-16\\\\\\,\\,\\,0=\\left( m+4 \\right)\\left( m-4 \\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0=m+4\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,0=m-4\\\\m=-4\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,m=4\\\\m=4,-4\\end{array}[\/latex]<\/p>\n<p>Check the solutions in the original equation.<\/p>\n<p>Since [latex]m=\u22124[\/latex] leads to division by 0, it is an extraneous solution.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\frac{16}{m+4}=\\frac{{{m}^{2}}}{m+4}\\\\\\\\\\frac{16}{-4+4}=\\frac{{{(-4)}^{2}}}{-4+4}\\\\\\\\\\frac{16}{0}=\\frac{16}{0}\\end{array}[\/latex]<\/p>\n<p>[latex]-4[\/latex] is excluded because\u00a0it leads to division by 0.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\frac{16}{4+4}=\\frac{{{(4)}^{2}}}{4+4}\\\\\\\\\\frac{16}{8}=\\frac{16}{8}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]m=4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>More Rational Equations<\/h3>\n<p>Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Rational Equation Leading to a Quadratic<\/h3>\n<p>Solve the following rational equation: [latex]\\displaystyle \\frac{-4x}{x - 1}+\\frac{4}{x+1}=\\frac{-8}{{x}^{2}-1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164755\">Solution<\/span><\/p>\n<div id=\"q164755\" class=\"hidden-answer\" style=\"display: none\">\nWe want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Then, the LCD is [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Next, we multiply the whole equation by the LCD.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}\\left(x+1\\right)\\left(x - 1\\right)\\left[\\frac{-4x}{x - 1}+\\frac{4}{x+1}\\right]\\hfill&=\\left[\\frac{-8}{\\left(x+1\\right)\\left(x - 1\\right)}\\right]\\left(x+1\\right)\\left(x - 1\\right)\\hfill \\\\ -4x\\left(x+1\\right)+4\\left(x - 1\\right)\\hfill&=-8\\hfill \\\\ -4{x}^{2}-4x+4x - 4\\hfill&=-8\\hfill \\\\ -4{x}^{2}+4\\hfill&=0\\hfill \\\\ -4\\left({x}^{2}-1\\right)\\hfill&=0\\hfill \\\\ -4\\left(x+1\\right)\\left(x - 1\\right)\\hfill&=0\\hfill \\\\ x\\hfill&=-1\\hfill \\\\ x\\hfill&=1\\hfill \\end{array}[\/latex]<\/div>\n<p>In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\displaystyle \\frac{3x+2}{x - 2}+\\frac{1}{x}=\\frac{-2}{{x}^{2}-2x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q586054\">Answer<\/span><\/p>\n<div id=\"q586054\" class=\"hidden-answer\" style=\"display: none\">[latex]x=-1[\/latex], ([latex]x=0[\/latex] is not a solution).<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3496&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm44861\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=44861&theme=oea&iframe_resize_id=ohm44861&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/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-4978\">\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>Revision and Adaptation. <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 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>: Monterey Institute of Technology and Education. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Solve Basic Rational Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/R9y2D9VFw0I\">https:\/\/youtu.be\/R9y2D9VFw0I<\/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>Solve Rational Equations with Like Denominators. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/gGA-dF_aQQQ\">https:\/\/youtu.be\/gGA-dF_aQQQ<\/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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Question ID#44861. <strong>Authored by<\/strong>: Kulinsky,Carla. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 3496. <strong>Authored by<\/strong>: Shawn Triplett. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Rational Expressions - Equations examples. <strong>Authored by<\/strong>: Tyler Wallace. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\">http:\/\/www.wallace.ccfaculty.org\/book\/book.html<\/a>. <strong>Project<\/strong>: Beginning and Intermediate Algebra. <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":60342,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"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 and Education\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Solve Basic Rational Equations\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/R9y2D9VFw0I\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Solve Rational Equations with Like Denominators\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/gGA-dF_aQQQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"cc\",\"description\":\"Question ID#44861\",\"author\":\"Kulinsky,Carla\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 3496\",\"author\":\"Shawn Triplett\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Rational Expressions - 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