{"id":2077,"date":"2025-09-03T14:33:10","date_gmt":"2025-09-03T14:33:10","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=2077"},"modified":"2025-09-03T14:34:57","modified_gmt":"2025-09-03T14:34:57","slug":"advanced-integration-techniques-background-youll-need-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/advanced-integration-techniques-background-youll-need-5\/","title":{"raw":"Advanced Integration Techniques: Background You\u2019ll Need 5","rendered":"Advanced Integration Techniques: Background You\u2019ll Need 5"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Solve equations that include fractions with variables<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Solving a Rational Equation<\/h2>\r\n<p class=\"whitespace-pre-wrap break-words\">A <strong>rational equation<\/strong> is an equation that contains at least one rational expression\u2014a fraction where the numerator, denominator, or both are polynomials. Although these equations may appear more complex at first glance, many can be manipulated to reveal a underlying linear structure.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>rational equation<\/h3>\r\nA\u00a0<span id=\"term-00009\" data-type=\"term\">rational equation<\/span>\u00a0contains at least one rational expression where the variable appears in at least one of the denominators.\r\n\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-pre-wrap break-words\">Recall that a rational number is the ratio of two numbers, such as [latex]\\dfrac{3}{4}[\/latex] or [latex]\\dfrac{7}{2}[\/latex]. A rational expression is the ratio, or quotient, of two polynomials - [latex]\\dfrac{x+1}{x^2-4}, \\dfrac{1}{x-3}, or \\dfrac{4}{x^2+x-2}[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-pre-wrap break-words\">Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">Solve the rational equation:<\/p>\r\n\r\n<center>[latex]\\dfrac{7}{2x} - \\dfrac{5}{3x} = \\dfrac{22}{3}[\/latex]<\/center>\r\n<p class=\"whitespace-pre-wrap break-words\">[reveal-answer q=\"644596\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"644596\"]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">We have three denominators: [latex]2x[\/latex], [latex]3x[\/latex], and [latex]3[\/latex]. The LCD must contain [latex]2x[\/latex], [latex]3x[\/latex], and [latex]3[\/latex]. An LCD of [latex]6x[\/latex] contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD [latex]6x[\/latex].<\/p>\r\n\r\n<center>[latex]\\begin{array}{rcll}\r\n(6x)(\\frac{7}{2x} - \\frac{5}{3x}) &amp;=&amp; (\\frac{22}{3})(6x) &amp; \\\\\r\n(6x)(\\frac{7}{2x}) - (6x)(\\frac{5}{3x}) &amp;=&amp; (\\frac{22}{3})(6x) &amp; \\text{Use the distributive property.} \\\\\r\n(\\cancel{6x})(\\frac{7}{\\cancel{2x}}) - (\\cancel{6x})(\\frac{5}{\\cancel{3x}}) &amp;=&amp; (\\frac{22}{3})(\\cancel{6x}) &amp; \\text{Cancel out the common factors.} \\\\\r\n3(7) - 2(5) &amp;=&amp; 22(2x) &amp; \\text{Multiply remaining factors by each} \\\\\r\n21 - 10 &amp;=&amp; 44x &amp; \\\\\r\n11 &amp;=&amp; 44x &amp; \\\\\r\n\\frac{11}{44} &amp;=&amp; x &amp; \\\\\r\n\\frac{1}{4} &amp;=&amp; x &amp;\r\n\\end{array}[\/latex]&lt;.center&gt;<\/center>\r\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\r\n\r\n<\/section>A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial\u2014two terms added or subtracted\u2014such as [latex](x + 1)[\/latex]. Always consider a binomial as an individual factor\u2014the terms cannot be separated.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">For example, suppose a problem has three terms and the denominators are [latex]x[\/latex], [latex]1[\/latex], and [latex]3x - 3[\/latex].<\/p>\r\nFirst, factor all denominators. We then have [latex]x[\/latex], [latex]1[\/latex], and [latex]3(x - 1)[\/latex] as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of [latex](x - 1)[\/latex].\r\n\r\nThe [latex]x[\/latex] in the first denominator is separate from the [latex]x[\/latex] in the [latex](x - 1)[\/latex] denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor.\r\n\r\nThe LCD in this instance is found by multiplying together the [latex]x[\/latex], one factor of [latex](x - 1)[\/latex], and the 3. Thus, the LCD is the following:\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]x(x - 1)3 = 3x(x - 1)[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">So, both sides of the equation would be multiplied by [latex]3x(x - 1)[\/latex]. Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-pre-wrap break-words\">Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{a}{b} = \\dfrac{c}{d}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\text{If } \\dfrac{a}{b} = \\dfrac{c}{d} \\text{, then } a \\cdot d = b \\cdot c[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Multiply [latex]a(d)[\/latex] and [latex]b(c)[\/latex], which results in [latex]ad = bc[\/latex].<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: <\/strong><strong>Given a rational equation, solve it.<\/strong>\r\n<ol id=\"fs-id2667782\" type=\"1\"><\/ol>\r\n<ol id=\"fs-id2667782\" type=\"1\">\r\n \t<li>Factor all denominators in the equation.<\/li>\r\n \t<li>Find and exclude values that set each denominator equal to zero.<\/li>\r\n \t<li>Find the LCD.<\/li>\r\n \t<li>Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.<\/li>\r\n \t<li>Solve the remaining equation.<\/li>\r\n \t<li>Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">Solve the following rational equation:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{2}{x} - \\dfrac{3}{2} = \\dfrac{7}{2x}[\/latex]<\/p>\r\n[reveal-answer q=\"717736\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"717736\"]We have three denominators: [latex]x[\/latex], [latex]2[\/latex], and [latex]2x[\/latex]. No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is [latex]2x[\/latex]. Only one value is excluded from a solution set, [latex]0[\/latex]. Next, multiply the whole equation (both sides of the equal sign) by [latex]2x[\/latex].\r\n\r\n<center>[latex]\\begin{array}{rcll} 2x(\\frac{2}{x} - \\frac{3}{2}) &amp;=&amp; (\\frac{7}{2x})2x &amp; \\\\ 2x(\\frac{2}{x}) - 2x(\\frac{3}{2}) &amp;=&amp; (\\frac{7}{2x})2x &amp; \\text{Distribute }2x. \\\\ 2(2) - 3x &amp;=&amp; 7 &amp; \\text{Denominators cancel out.} \\\\ 4 - 3x &amp;=&amp; 7 &amp; \\\\ -3x &amp;=&amp; 3 &amp; \\\\ x &amp;=&amp; -1 &amp; \\\\ \\text{or } \\{-1\\} &amp; &amp; &amp; \\end{array}[\/latex]<\/center>The proposed solution is [latex]-1[\/latex], which is not an excluded value, so the solution set contains one number,[latex] -1,[\/latex] or [latex]{-1}[\/latex] written in set notation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">Solve the following rational equation:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{1}{x} = \\dfrac{1}{10} - \\dfrac{3}{4x}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[reveal-answer q=\"846436\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"846436\"]First find the common denominator. The three denominators in factored form are [latex]x[\/latex], [latex]10 = 2 \\cdot 5[\/latex], and [latex]4x = 2 \\cdot 2 \\cdot x[\/latex]. The smallest expression that is divisible by each one of the denominators is [latex]20x[\/latex]. Only [latex]x = 0[\/latex] is an excluded value. Multiply the whole equation by [latex]20x[\/latex].<\/p>\r\n\r\n<center>[latex]\\begin{array}{rcl} 20x(\\frac{1}{x}) &amp;=&amp; (\\frac{1}{10} - \\frac{3}{4x})20x \\\\ 20 &amp;=&amp; 2x - 15 \\\\ 35 &amp;=&amp; 2x \\\\ \\frac{35}{2} &amp;=&amp; x \\end{array}[\/latex]<\/center>\r\n<div class=\"font-claude-message pr-4 md:pr-9 relative leading-[1.65rem] [&amp;_pre&gt;div]:bg-bg-300 [&amp;_pre]:-mr-4 md:[&amp;_pre]:-mr-9\">\r\n<div>\r\n<div class=\"grid-col-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-pre-wrap break-words\">The solution is [latex]\\frac{35}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the following rational equations and state the excluded values:\r\n<ol type=\"a\">\r\n \t<li>[latex]\\dfrac{3}{x-6} = \\dfrac{5}{x}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{x}{x-3} = \\dfrac{5}{x-3} - \\dfrac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{x}{x-2} = \\dfrac{5}{x-2} - \\dfrac{1}{2}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"211203\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"211203\"]\r\n<ol type=\"a\">\r\n \t<li>The denominators [latex]x[\/latex] and [latex]x - 6[\/latex] have nothing in common. Therefore, the LCD is the product [latex]x(x - 6)[\/latex]. However, for this problem, we can cross-multiply.\r\n<center>[latex]\\begin{array}{rcll}\r\n\\frac{3}{x-6} &amp;=&amp; \\frac{5}{x} &amp; \\\\\r\n3x &amp;=&amp; 5(x-6) &amp; \\text{Distribute.} \\\\\r\n3x &amp;=&amp; 5x-30 &amp; \\\\\r\n-2x &amp;=&amp; -30 &amp; \\\\\r\nx &amp;=&amp; 15 &amp;\r\n\\end{array}[\/latex]<\/center>\r\nThe solution is [latex]15[\/latex]. The excluded values are [latex]6[\/latex] and [latex]0[\/latex].<\/li>\r\n \t<li>The LCD is [latex]2(x - 3)[\/latex]. Multiply both sides of the equation by [latex]2(x - 3)[\/latex].\r\n<center>[latex]\\begin{array}{rcl}\r\n2(x-3)(\\frac{x}{x-3}) &amp;=&amp; (\\frac{5}{x-3} - \\frac{1}{2})2(x-3) \\\\\r\n\\frac{2(x-3)x}{x-3} &amp;=&amp; \\frac{2(x-3)5}{x-3} - \\frac{2(x-3)}{2} \\\\\r\n2x &amp;=&amp; 10 - (x - 3) \\\\\r\n2x &amp;=&amp; 10 - x + 3 \\\\\r\n2x &amp;=&amp; 13 - x \\\\\r\n3x &amp;=&amp; 13 \\\\\r\nx &amp;=&amp; \\frac{13}{3}\r\n\\end{array}[\/latex]<\/center>\r\nThe solution is [latex]\\dfrac{13}{3}[\/latex]. The excluded value is [latex]3[\/latex].<\/li>\r\n \t<li>The least common denominator is [latex]2(x - 2)[\/latex]. Multiply both sides of the equation by [latex]x(x - 2)[\/latex].\r\n<center>[latex]\\begin{array}{rcl}\r\n2(x-2)(\\frac{x}{x-2}) &amp;=&amp; (\\frac{5}{x-2} - \\frac{1}{2})2(x-2) \\\\\r\n2x &amp;=&amp; 10 - (x - 2) \\\\\r\n2x &amp;=&amp; 12 - x \\\\\r\n3x &amp;=&amp; 12 \\\\\r\nx &amp;=&amp; 4\r\n\\end{array}[\/latex]<\/center>\r\nThe solution is [latex]4[\/latex]. The excluded value is [latex]2[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]290047[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Solve equations that include fractions with variables<\/span><\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Solving a Rational Equation<\/h2>\n<p class=\"whitespace-pre-wrap break-words\">A <strong>rational equation<\/strong> is an equation that contains at least one rational expression\u2014a fraction where the numerator, denominator, or both are polynomials. Although these equations may appear more complex at first glance, many can be manipulated to reveal a underlying linear structure.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>rational equation<\/h3>\n<p>A\u00a0<span id=\"term-00009\" data-type=\"term\">rational equation<\/span>\u00a0contains at least one rational expression where the variable appears in at least one of the denominators.<\/p>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-pre-wrap break-words\">Recall that a rational number is the ratio of two numbers, such as [latex]\\dfrac{3}{4}[\/latex] or [latex]\\dfrac{7}{2}[\/latex]. A rational expression is the ratio, or quotient, of two polynomials &#8211; [latex]\\dfrac{x+1}{x^2-4}, \\dfrac{1}{x-3}, or \\dfrac{4}{x^2+x-2}[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-pre-wrap break-words\">Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Solve the rational equation:<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{7}{2x} - \\dfrac{5}{3x} = \\dfrac{22}{3}[\/latex]<\/div>\n<p class=\"whitespace-pre-wrap break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q644596\">Show Answer<\/button><\/p>\n<div id=\"q644596\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"whitespace-pre-wrap break-words\">We have three denominators: [latex]2x[\/latex], [latex]3x[\/latex], and [latex]3[\/latex]. The LCD must contain [latex]2x[\/latex], [latex]3x[\/latex], and [latex]3[\/latex]. An LCD of [latex]6x[\/latex] contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD [latex]6x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcll}  (6x)(\\frac{7}{2x} - \\frac{5}{3x}) &=& (\\frac{22}{3})(6x) & \\\\  (6x)(\\frac{7}{2x}) - (6x)(\\frac{5}{3x}) &=& (\\frac{22}{3})(6x) & \\text{Use the distributive property.} \\\\  (\\cancel{6x})(\\frac{7}{\\cancel{2x}}) - (\\cancel{6x})(\\frac{5}{\\cancel{3x}}) &=& (\\frac{22}{3})(\\cancel{6x}) & \\text{Cancel out the common factors.} \\\\  3(7) - 2(5) &=& 22(2x) & \\text{Multiply remaining factors by each} \\\\  21 - 10 &=& 44x & \\\\  11 &=& 44x & \\\\  \\frac{11}{44} &=& x & \\\\  \\frac{1}{4} &=& x &  \\end{array}[\/latex]&lt;.center&gt;<\/div>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<p>A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial\u2014two terms added or subtracted\u2014such as [latex](x + 1)[\/latex]. Always consider a binomial as an individual factor\u2014the terms cannot be separated.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">For example, suppose a problem has three terms and the denominators are [latex]x[\/latex], [latex]1[\/latex], and [latex]3x - 3[\/latex].<\/p>\n<p>First, factor all denominators. We then have [latex]x[\/latex], [latex]1[\/latex], and [latex]3(x - 1)[\/latex] as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of [latex](x - 1)[\/latex].<\/p>\n<p>The [latex]x[\/latex] in the first denominator is separate from the [latex]x[\/latex] in the [latex](x - 1)[\/latex] denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor.<\/p>\n<p>The LCD in this instance is found by multiplying together the [latex]x[\/latex], one factor of [latex](x - 1)[\/latex], and the 3. Thus, the LCD is the following:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]x(x - 1)3 = 3x(x - 1)[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">So, both sides of the equation would be multiplied by [latex]3x(x - 1)[\/latex]. Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.<\/p>\n<\/section>\n<p class=\"whitespace-pre-wrap break-words\">Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{a}{b} = \\dfrac{c}{d}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\text{If } \\dfrac{a}{b} = \\dfrac{c}{d} \\text{, then } a \\cdot d = b \\cdot c[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Multiply [latex]a(d)[\/latex] and [latex]b(c)[\/latex], which results in [latex]ad = bc[\/latex].<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: <\/strong><strong>Given a rational equation, solve it.<\/strong><\/p>\n<ol id=\"fs-id2667782\" type=\"1\"><\/ol>\n<ol type=\"1\">\n<li>Factor all denominators in the equation.<\/li>\n<li>Find and exclude values that set each denominator equal to zero.<\/li>\n<li>Find the LCD.<\/li>\n<li>Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.<\/li>\n<li>Solve the remaining equation.<\/li>\n<li>Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Solve the following rational equation:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{2}{x} - \\dfrac{3}{2} = \\dfrac{7}{2x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q717736\">Show Answer<\/button><\/p>\n<div id=\"q717736\" class=\"hidden-answer\" style=\"display: none\">We have three denominators: [latex]x[\/latex], [latex]2[\/latex], and [latex]2x[\/latex]. No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is [latex]2x[\/latex]. Only one value is excluded from a solution set, [latex]0[\/latex]. Next, multiply the whole equation (both sides of the equal sign) by [latex]2x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcll} 2x(\\frac{2}{x} - \\frac{3}{2}) &=& (\\frac{7}{2x})2x & \\\\ 2x(\\frac{2}{x}) - 2x(\\frac{3}{2}) &=& (\\frac{7}{2x})2x & \\text{Distribute }2x. \\\\ 2(2) - 3x &=& 7 & \\text{Denominators cancel out.} \\\\ 4 - 3x &=& 7 & \\\\ -3x &=& 3 & \\\\ x &=& -1 & \\\\ \\text{or } \\{-1\\} & & & \\end{array}[\/latex]<\/div>\n<p>The proposed solution is [latex]-1[\/latex], which is not an excluded value, so the solution set contains one number,[latex]-1,[\/latex] or [latex]{-1}[\/latex] written in set notation.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Solve the following rational equation:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\dfrac{1}{x} = \\dfrac{1}{10} - \\dfrac{3}{4x}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q846436\">Show Answer<\/button><\/p>\n<div id=\"q846436\" class=\"hidden-answer\" style=\"display: none\">First find the common denominator. The three denominators in factored form are [latex]x[\/latex], [latex]10 = 2 \\cdot 5[\/latex], and [latex]4x = 2 \\cdot 2 \\cdot x[\/latex]. The smallest expression that is divisible by each one of the denominators is [latex]20x[\/latex]. Only [latex]x = 0[\/latex] is an excluded value. Multiply the whole equation by [latex]20x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcl} 20x(\\frac{1}{x}) &=& (\\frac{1}{10} - \\frac{3}{4x})20x \\\\ 20 &=& 2x - 15 \\\\ 35 &=& 2x \\\\ \\frac{35}{2} &=& x \\end{array}[\/latex]<\/div>\n<div class=\"font-claude-message pr-4 md:pr-9 relative leading-&#091;1.65rem&#093; &#091;&amp;_pre&gt;div&#093;:bg-bg-300 &#091;&amp;_pre&#093;:-mr-4 md:&#091;&amp;_pre&#093;:-mr-9\">\n<div>\n<div class=\"grid-col-1 grid gap-2.5 &#091;&amp;_&gt;_*&#093;:min-w-0\">\n<p class=\"whitespace-pre-wrap break-words\">The solution is [latex]\\frac{35}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the following rational equations and state the excluded values:<\/p>\n<ol type=\"a\">\n<li>[latex]\\dfrac{3}{x-6} = \\dfrac{5}{x}[\/latex]<\/li>\n<li>[latex]\\dfrac{x}{x-3} = \\dfrac{5}{x-3} - \\dfrac{1}{2}[\/latex]<\/li>\n<li>[latex]\\dfrac{x}{x-2} = \\dfrac{5}{x-2} - \\dfrac{1}{2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q211203\">Show Answer<\/button><\/p>\n<div id=\"q211203\" class=\"hidden-answer\" style=\"display: none\">\n<ol type=\"a\">\n<li>The denominators [latex]x[\/latex] and [latex]x - 6[\/latex] have nothing in common. Therefore, the LCD is the product [latex]x(x - 6)[\/latex]. However, for this problem, we can cross-multiply.\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcll}  \\frac{3}{x-6} &=& \\frac{5}{x} & \\\\  3x &=& 5(x-6) & \\text{Distribute.} \\\\  3x &=& 5x-30 & \\\\  -2x &=& -30 & \\\\  x &=& 15 &  \\end{array}[\/latex]<\/div>\n<p>The solution is [latex]15[\/latex]. The excluded values are [latex]6[\/latex] and [latex]0[\/latex].<\/li>\n<li>The LCD is [latex]2(x - 3)[\/latex]. Multiply both sides of the equation by [latex]2(x - 3)[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcl}  2(x-3)(\\frac{x}{x-3}) &=& (\\frac{5}{x-3} - \\frac{1}{2})2(x-3) \\\\  \\frac{2(x-3)x}{x-3} &=& \\frac{2(x-3)5}{x-3} - \\frac{2(x-3)}{2} \\\\  2x &=& 10 - (x - 3) \\\\  2x &=& 10 - x + 3 \\\\  2x &=& 13 - x \\\\  3x &=& 13 \\\\  x &=& \\frac{13}{3}  \\end{array}[\/latex]<\/div>\n<p>The solution is [latex]\\dfrac{13}{3}[\/latex]. The excluded value is [latex]3[\/latex].<\/li>\n<li>The least common denominator is [latex]2(x - 2)[\/latex]. Multiply both sides of the equation by [latex]x(x - 2)[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcl}  2(x-2)(\\frac{x}{x-2}) &=& (\\frac{5}{x-2} - \\frac{1}{2})2(x-2) \\\\  2x &=& 10 - (x - 2) \\\\  2x &=& 12 - x \\\\  3x &=& 12 \\\\  x &=& 4  \\end{array}[\/latex]<\/div>\n<p>The solution is [latex]4[\/latex]. 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