{"id":16344,"date":"2019-10-03T04:18:53","date_gmt":"2019-10-03T04:18:53","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/6-1-3-adding-and-subtracting-rational-expressions\/"},"modified":"2024-05-01T19:04:16","modified_gmt":"2024-05-01T19:04:16","slug":"6-1-3-adding-and-subtracting-rational-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/6-1-3-adding-and-subtracting-rational-expressions\/","title":{"raw":"Adding and Subtracting Rational Expressions Part I","rendered":"Adding and Subtracting Rational Expressions Part I"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Add and subtract rational expressions with like denominators<\/li>\r\n \t<li>Add and subtract rational expressions with unlike denominators using a greatest common denominator<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators. By working carefully and writing down the steps along the way, you can keep track of all of the numbers and variables and perform the operations accurately.\r\n\r\n[caption id=\"attachment_5017\" align=\"aligncenter\" width=\"339\"]<img class=\" wp-image-5017\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/20042339\/Screen-Shot-2016-06-19-at-9.23.12-PM.png\" alt=\"Add and Subtract\" width=\"339\" height=\"163\" \/> Add and Subtract[\/caption]\r\n<h2>Adding and Subtracting Rational Expressions with Like Denominators<\/h2>\r\nAdding rational expressions with the same denominator is the simplest place to start, so let\u2019s begin there.\r\n\r\nTo add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\frac{2}{9}+\\frac{4}{9}=\\frac{6}{9}\\\\\\\\\\frac{6}{9}=\\frac{3\\cdot 2}{3\\cdot 3}=\\frac{3}{3}\\cdot \\frac{2}{3}=1\\cdot \\frac{2}{3}=\\frac{2}{3}\\end{array}[\/latex]<\/p>\r\nFollow the same process to add rational expressions with like denominators. Let's try one.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd [latex] \\displaystyle \\frac{2{{x}^{2}}}{x+4}+\\frac{8x}{x+4}[\/latex], and define the domain.\r\n\r\nState the sum in simplest form.\r\n\r\n[reveal-answer q=\"324146\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324146\"]\r\n\r\nSince the denominators are the same, add the numerators.\r\n<p style=\"text-align: center;\">[latex]\\frac{2{{x}^{2}}+8x}{x+4}[\/latex]<\/p>\r\nFactor the numerator.\r\n<p style=\"text-align: center;\">[latex]\\frac{2x(x+4)}{x+4}[\/latex]<\/p>\r\nSimplify\u00a0common factors and.\r\n<p style=\"text-align: center;\">[latex]\\large\\begin{array}{c}\\frac{2x\\cancel{(x+4)}}{\\cancel{x+4}}\\\\\\\\=\\frac{2x}{1}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The domain is found by setting the denominators in the original sum\u00a0equal to zero.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+4=0\\\\x=-4\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The domain is [latex]x\\ne-4[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\displaystyle \\frac{2{{x}^{2}}}{x+4}+\\frac{8x}{x+4}=2x,x\\ne -4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\"><img class=\" wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"70\" height=\"62\" \/>Caution! \u00a0Remember to define\u00a0the domain of a sum or difference <em>before<\/em> simplifying. \u00a0You may lose important information when you simplify. In the example above, the domain is [latex]x\\ne-4[\/latex]. \u00a0If we were to have defined the domain after simplifying, we would find that the domain is all real numbers which is incorrect.<\/div>\r\nTo subtract rational expressions with like denominators, follow the same process you use to subtract fractions with like denominators. The process is just like the addition of rational expressions, except that you subtract instead of add.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSubtract[latex]\\frac{4x+7}{x+6}-\\frac{2x+8}{x+6}[\/latex], and define the domain.\r\n\r\nState the difference in simplest form.\r\n\r\n[reveal-answer q=\"681427\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"681427\"]\r\n\r\nSubtract the second numerator from the first and keep the denominator the same.\r\n<p style=\"text-align: center;\">[latex]\\frac{4x+7-(2x+8)}{x+6}[\/latex]<\/p>\r\nBe careful to distribute the negative to both terms of the second numerator.\r\n<p style=\"text-align: center;\">[latex]\\frac{4x+7-2x-8}{x+6}[\/latex]<\/p>\r\nCombine like terms. This rational expression cannot be simplified any further.\r\n<p style=\"text-align: center;\">[latex]\\frac{2x-1}{x+6}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The domain is found from the denominators of original\u00a0expression.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+6=0\\\\x=-6\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The domain is [latex]x\\ne-6[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\displaystyle \\frac{4x+7}{x+6}-\\frac{2x+8}{x+6}=\\frac{2x-1}{x+6},\\text{}x\\ne-6[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]40242[\/ohm_question]\r\n\r\n<\/div>\r\nIn the video that follows, we present\u00a0more examples of adding rational expressions with like denominators. Additionally, we review finding the domain of a rational expression.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=BeaHKtxB868&amp;feature=youtu.be\r\n<h2>Adding and Subtracting Rational Expressions with Unlike Denominators<\/h2>\r\n[caption id=\"attachment_4984\" align=\"aligncenter\" width=\"407\"]<img class=\" wp-image-4984\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/17210449\/Screen-Shot-2016-06-17-at-1.59.26-PM-300x207.png\" alt=\"skateboard, motorcycle, bike, and rollerskates to illustrate that they are different but have wheels in common.\" width=\"407\" height=\"281\" \/> What do they have in common?[\/caption]\r\n\r\nBefore adding and subtracting rational expressions with <i>unlike<\/i> denominators, you need to find a common denominator. Once again, this process is similar to the one used for adding and subtracting numeric fractions with unlike denominators.\u00a0Remember how to do this?\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{5}{6}+\\frac{8}{10}+\\frac{3}{4}[\/latex]<\/p>\r\nSince the denominators are [latex]6[\/latex], [latex]10[\/latex], and [latex]4[\/latex], you want to find the <strong>least common denominator<\/strong> and express each fraction with this denominator before adding. (BTW, you can add fractions by finding <i>any<\/i> common denominator; it does not have to be the least. You focus on using the <i>least<\/i> because then there is less simplifying to do. But either way works.)\r\n\r\nFinding the least common denominator is the same as finding the <strong>least common multiple<\/strong> of \u00a0[latex]4[\/latex], [latex]6[\/latex], and [latex]10[\/latex]. There are a couple of ways to do this. The first is to list the multiples of each number and determine which multiples they have in common. The least of these numbers will be the least common denominator.\r\n<table style=\"width: 80%;\">\r\n<thead>\r\n<tr style=\"height: 43px;\">\r\n<th style=\"text-align: center; height: 43px; width: 6%;\">Number<\/th>\r\n<th style=\"height: 43px; width: 84.2962%;\" colspan=\"16\">\r\n<p style=\"text-align: center;\"><b>Multiples<\/b><\/p>\r\n<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"height: 15px; width: 6%;\">[latex]4[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]8[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]12[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]16[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]20[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 7%;\">[latex]24[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 7%;\">[latex]28[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 7%;\">[latex]32[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 3%;\">[latex]36[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 10%;\">[latex]40[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 3%;\">[latex]44[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 3%;\">[latex]48[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 3%;\">[latex]52[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 3%;\">[latex]56[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 4%;\">[latex]\\textbf{60}[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 10%;\">[latex]64[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 0.296209%;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15.2812px;\">\r\n<td style=\"height: 15.2812px; width: 6%;\">[latex]6[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 6%;\">[latex]12[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 6%;\">[latex]18[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 6%;\">[latex]24[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 6%;\">[latex]30[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 7%;\">[latex]36[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 7%;\">[latex]42[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 7%;\">[latex]48[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 3%;\">[latex]54[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 10%;\">[latex]\\textbf{60}[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 3%;\">[latex]66[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 3%;\">[latex]68[\/latex]<\/td>\r\n<td style=\"height: 15.2812px; width: 3%;\"><\/td>\r\n<td style=\"height: 15.2812px; width: 3%;\"><\/td>\r\n<td style=\"height: 15.2812px; width: 4%;\"><\/td>\r\n<td style=\"height: 15.2812px; width: 10%;\"><\/td>\r\n<td style=\"height: 15.2812px; width: 0.296209%;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"height: 15px; width: 6%;\">[latex]10[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]20[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]30[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]40[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 6%;\">[latex]50[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 7%;\">[latex]\\textbf{60}[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 7%;\">\u00a0[latex]70[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 7%;\">[latex]80[\/latex]<\/td>\r\n<td style=\"height: 15px; width: 3%;\"><\/td>\r\n<td style=\"height: 15px; width: 10%;\"><\/td>\r\n<td style=\"height: 15px; width: 3%;\"><\/td>\r\n<td style=\"height: 15px; width: 3%;\"><\/td>\r\n<td style=\"height: 15px; width: 3%;\"><\/td>\r\n<td style=\"height: 15px; width: 3%;\"><\/td>\r\n<td style=\"height: 15px; width: 4%;\"><\/td>\r\n<td style=\"height: 15px; width: 10%;\"><\/td>\r\n<td style=\"height: 15px; width: 0.296209%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe other method is to use <strong>prime factorization<\/strong>, the process of finding the prime factors of a number. This is how the method works with numbers.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse prime factorization to find the least common multiple of [latex]6[\/latex], [latex]10[\/latex], and [latex]4[\/latex].\r\n\r\n[reveal-answer q=\"371893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"371893\"]First, find the prime factorization of each denominator.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}6=3\\cdot2\\\\10=5\\cdot2\\\\4=2\\cdot2\\end{array}[\/latex]<\/p>\r\nThe LCM will contain factors of \u00a0[latex]2[\/latex],[latex]3[\/latex], and [latex]5[\/latex]. Multiply each number the maximum number of times it appears in a single factorization.\r\n\r\nIn this case, [latex]3[\/latex] appears once, [latex]5[\/latex] appears once, and [latex]2[\/latex] is used twice because it appears twice in the prime factorization of [latex]4[\/latex].\r\n\r\nTherefore, the LCM of [latex]6[\/latex], [latex]10[\/latex], and [latex]4[\/latex] is [latex]3\\cdot5\\cdot2\\cdot2[\/latex], or [latex]60[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6=3\\cdot2\\\\\\,\\,\\,\\,\\,\\,\\,\\,10=5\\cdot2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4=2\\cdot2\\\\\\text{LCM}=3\\cdot5\\cdot2\\cdot2\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe least common multiple of \u00a0[latex]6, 10[\/latex], and [latex]4[\/latex] is [latex]60[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBoth methods give the same result, but prime factorization is faster. Your choice!\r\n\r\nNow that you have found the least common multiple, you can use that number as the least common denominator of the fractions. Multiply each fraction by the fractional form of [latex]1[\/latex] that will produce a denominator of [latex]60[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\frac{5}{6}\\cdot \\frac{10}{10}=\\frac{50}{60}\\\\\\\\\\frac{8}{10}\\cdot\\frac{6}{6}=\\frac{48}{60}\\\\\\\\\\frac{3}{4}\\cdot\\frac{15}{15}=\\frac{45}{60}\\end{array}[\/latex]<\/p>\r\nNow that you have like denominators, add the fractions:\r\n<p style=\"text-align: center;\">[latex]\\frac{50}{60}+\\frac{48}{60}+\\frac{45}{60}=\\frac{143}{60}[\/latex]<\/p>\r\nIn the next example, we show how to find the least common multiple of a rational expression with a monomial in the denominator.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd[latex]\\frac{2n}{15m^{2}}+\\frac{3n}{21m}[\/latex], and give the domain.\r\n\r\nState the sum in simplest form.\r\n\r\n[reveal-answer q=\"926091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"926091\"]\r\n\r\nFind the prime factorization of each denominator.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}15m^{2}\\,=\\,3\\cdot5\\cdot{m}\\cdot{m}\\\\\\,\\,\\,21m\\,=\\,3\\cdot7\\cdot{m}\\end{array}[\/latex]<\/p>\r\nFind the least common multiple. [latex]3[\/latex] appears exactly once in both of the expressions, so it will appear once in the least common multiple. Both [latex]5[\/latex] and [latex]7[\/latex] appear at most once. For the variables, the most <i>m<\/i> appears is twice.\r\n\r\nUse the least common multiple for your new common denominator, it will be the LCD.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}15m^{2}\\,=\\,3\\cdot5\\cdot{m}\\cdot{m}\\\\21m\\,\\,\\,=\\,3\\cdot7\\cdot{m}\\\\\\text{LCM}:3\\cdot5\\cdot7\\cdot{m}\\cdot{m}\\\\\\text{LCM}:105m^{2}\\end{array}[\/latex]<\/p>\r\nCompare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]105m^{2}[\/latex]<i>. <\/i>Remember that <i>m<\/i> cannot be 0 because the denominators would be [latex]0[\/latex].\r\n\r\nThe first denominator is [latex]15m^{2}[\/latex]\u00a0and the LCD is [latex]105m^{2}[\/latex]. You need to multiply [latex]15m^{2}[\/latex]\u00a0by [latex]7[\/latex] to get the LCD, so multiply the entire rational expression by [latex]\\frac{7}{7}[\/latex].\r\n\r\nThe second denominator is [latex]21m[\/latex] and the LCD is [latex]105m^{2}[\/latex]. You need to multiply [latex]21m[\/latex] by [latex]5m[\/latex] to get the LCD, so multiply the entire rational expression by [latex]\\frac{5m}{5m}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2n}{15m^{2}}\\cdot\\frac{7}{7}=\\frac{14n}{105m^{2}}\\\\\\\\\\frac{3n}{21m}\\cdot\\frac{5m}{5m}=\\frac{15mn}{105m^{2}}\\end{array}[\/latex]<\/p>\r\nAdd the numerators and keep the denominator the same.\r\n<p style=\"text-align: center;\">[latex]\\frac{14n}{105{{m}^{2}}}+\\frac{15mn}{105{{m}^{2}}}=\\frac{14n+15mn}{105{{m}^{2}}}[\/latex]<\/p>\r\nIf possible, simplify by finding common factors in the numerator and denominator. This rational expression is already in simplest form because the numerator and denominator have no factors in common.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{n(14+15m)}{105{{m}^{2}}}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\displaystyle \\frac{2n}{15{{m}^{2}}}+\\frac{3n}{21m}=\\frac{n(14+15m)}{105{{m}^{2}}},m\\ne 0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThat took a while, but you got through it. Adding rational expressions can be a lengthy process, but taken one step at a time, it can be done.\u00a0 Now let's take a look at some examples where the denominator is not a monomial.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">To find the least common denominator (LCD) of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, consider the following rational expressions:<\/span>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{6}{\\left(x+3\\right)\\left(x+4\\right)},\\text{ and }\\frac{9x}{\\left(x+4\\right)\\left(x+5\\right)}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The LCD would be [latex]\\left(x+3\\right)\\left(x+4\\right)\\left(x+5\\right)[\/latex].<\/p>\r\nTo find the LCD, we count the greatest number of times a factor appears in each denominator and include it in the LCD that many times.\r\n\r\nFor example, in\u00a0[latex]\\dfrac{6}{\\left(x+3\\right)\\left(x+4\\right)}[\/latex], [latex]\\left(x+3\\right)[\/latex] is represented once and \u00a0[latex]\\left(x+4\\right)[\/latex] is represented once, so they both appear exactly once in the LCD.\r\n\r\nIn [latex]\\dfrac{9x}{\\left(x+4\\right)\\left(x+5\\right)}[\/latex], [latex]\\left(x+4\\right)[\/latex] appears once and [latex]\\left(x+5\\right)[\/latex] appears once.\r\n\r\nWe have already accounted for\u00a0[latex]\\left(x+4\\right)[\/latex], so the LCD just needs one factor of\u00a0[latex]\\left(x+5\\right)[\/latex] to be complete.\r\n\r\nOnce we find the LCD, we need to multiply each expression by the form of\u00a0[latex]1[\/latex] that will change the denominator to the LCD.\r\n\r\nWhat do we mean by \" the form of\u00a0[latex]1[\/latex]\"?\r\n\r\n[latex]\\frac{x+5}{x+5}=1[\/latex] so multiplying an expression\u00a0by it will not change its value.\r\n\r\nFor example, we would need to multiply the expression [latex]\\dfrac{6}{\\left(x+3\\right)\\left(x+4\\right)}[\/latex] by [latex]\\frac{x+5}{x+5}[\/latex] and the expression [latex]\\frac{9x}{\\left(x+4\\right)\\left(x+5\\right)}[\/latex] by [latex]\\frac{x+3}{x+3}[\/latex].\r\n\r\nHopefully this process will become clear after you practice it yourself. \u00a0As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which \"form of 1\" you will need to multiply\u00a0each expression by so that it has the LCD.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd the rational expressions [latex]\\frac{5}{x}+\\frac{6}{y}[\/latex] and define the domain.\r\n\r\nState the sum in simplest form.\r\n\r\n[reveal-answer q=\"324147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324147\"]\r\n\r\nFirst, define the domain of each expression. Since we have x and y in the denominators, we can say [latex]x\\ne0 ,\\text{ and }y\\ne0[\/latex].\r\n\r\nNow we have to find the LCD. Since x appears once and y appears once, \u00a0the LCD will be [latex]xy[\/latex]. \u00a0We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[\/latex] as the denominator for each fraction.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{5}{x}\\cdot \\frac{y}{y}+\\frac{6}{y}\\cdot \\frac{x}{x}\\\\ \\frac{5y}{xy}+\\frac{6x}{xy}\\end{array}[\/latex]<\/div>\r\nNow that the expressions have the same denominator, we simply add the numerators to find the sum.\r\n<div style=\"text-align: center;\">[latex]\\frac{6x+5y}{xy}[\/latex]<\/div>\r\n[latex]\\frac{5}{x}+\\frac{6}{y}=\\frac{(6x+5y)}{xy}[\/latex],\u00a0[latex]x\\ne0 ,\\text{ and }y\\ne0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is one more example of adding rational expressions where the denominators are multi-term polynomials. First, we will factor and then find the LCD. Note that [latex]x^2-4[\/latex] is a difference of squares and can be factored using special products.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify[latex]\\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\\frac{x}{x-2}[\/latex] and give the domain.\r\n\r\nState the result in simplest form.\r\n\r\n[reveal-answer q=\"57692\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"57692\"]\r\n\r\nFind the least common denominator by factoring each denominator. The least common denominator includes the maximum number of times it appears in a single factorization. Remember that <i>x<\/i> cannot be [latex]2[\/latex] or [latex]-2[\/latex] because the denominators would be\u00a0[latex]0[\/latex].\r\n\r\n[latex]\\left(x+2\\right)[\/latex] appears a maximum of one time and so does [latex]\\left(x\u20132\\right)[\/latex]. This means the LCD is [latex]\\left(x+2\\right)\\left(x\u20132\\right)[\/latex].\r\n\r\nMultiply each expression by the equivalent of\u00a0[latex]1[\/latex] that will give it the common denominator. Notice the first fraction already has the LCD and as a result remains unchanged.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\\\\frac{x}{x-2}\\cdot \\frac{x+2}{x+2}=\\frac{x(x+2)}{(x+2)(x-2)}\\end{array}[\/latex]<\/p>\r\nRewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{2{{x}^{2}}}{(x+2)(x-2)}+\\frac{x(x+2)}{(x+2)(x-2)}[\/latex]<\/p>\r\nCombine the numerators.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\frac{2{{x}^{2}}+x(x+2)}{(x+2)(x-2)}\\\\\\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x}{(x+2)(x-2)}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[\/latex]<\/p>\r\nCheck for simplest form. Since neither [latex]\\left(x+2\\right)[\/latex] nor [latex]\\left(x-2\\right)[\/latex] is a factor of [latex]3{{x}^{2}}+2x[\/latex], this expression is in simplest form.\r\n\r\n[latex] \\displaystyle \\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\\frac{x}{x-2}=\\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[\/latex] \u00a0\u00a0 [latex] \\displaystyle x\\ne 2,-2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe video that follows contains an example of adding rational expressions whose denominators are not alike. \u00a0The denominators of both expressions contain only monomials.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=Wk8ZZhE9ZjI&amp;feature=youtu.be\r\n<h2>Subtracting Rational Expressions<\/h2>\r\nNow let\u2019s try subtracting rational expressions. You'll use the same basic technique of finding the least common denominator and rewriting each rational expression to have that denominator.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSubtract[latex]\\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}[\/latex], define the domain.\r\n\r\nState the difference in simplest form.\r\n\r\n[reveal-answer q=\"704185\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704185\"]\r\n\r\nFind the prime factorization of each denominator. [latex]t+1[\/latex] cannot be factored any further, but [latex]{{t}^{2}}-t-2[\/latex] can be. Remember that <i>t<\/i> cannot be [latex]-1[\/latex] or [latex]2[\/latex] because the denominators would be [latex]0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}t+1=t+1\\\\t^{2}-t-2=\\left(t-2\\right)\\left(t+1\\right)\\end{array}[\/latex]<\/p>\r\nFind the least common multiple. [latex]t+1[\/latex] appears exactly once in both of the expressions, so it will appear once in the least common denominator. [latex]t\u20132[\/latex] also appears once.\r\n\r\nThis means that [latex]\\left(t-2\\right)\\left(t+1\\right)[\/latex] is the least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out.\r\n\r\nUse the least common multiple for your new common denominator, it will be the LCD.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}t+1=t+1\\\\t^{2}-t-2=\\left(t-2\\right)\\left(t+1\\right)\\\\\\text{LCM}:\\left(t+1\\right)\\left(t-1\\right)\\end{array}[\/latex]<\/p>\r\nCompare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]\\left(t+1\\right)\\left(t\u20132\\right)[\/latex]<i>.<\/i>\r\n\r\nYou need to multiply [latex]t+1[\/latex] by [latex]t\u20132[\/latex] to get the LCD, so multiply the entire rational expression by [latex] \\displaystyle \\frac{t-2}{t-2}[\/latex].\r\n\r\nThe second expression already has a denominator of [latex]\\left(t+1\\right)\\left(t\u20132\\right)[\/latex], so you do not need to multiply it by anything.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\frac{2}{t+1}\\cdot \\frac{t-2}{t-2}=\\frac{2(t-2)}{(t+1)(t-2)}\\\\\\\\\\,\\,\\,\\frac{t-2}{{{t}^{2}}-t-2}=\\frac{t-2}{(t+1)(t-2)}\\end{array}[\/latex]<\/p>\r\nThen rewrite the subtraction problem with the common denominator.\r\n<p style=\"text-align: center;\">[latex] \\frac{2\\left(t-2\\right)}{\\left(t+1\\right)\\left(t-2\\right)}-\\frac{t-2}{\\left(t+1\\right)\\left(t-2\\right)}[\/latex]<\/p>\r\nSubtract the numerators and simplify. Remember that parentheses need to be included around the second [latex]\\left(t\u20132\\right)[\/latex] in the numerator because the whole quantity is subtracted. Otherwise you would be subtracting just the <i>t<\/i>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\\\\\\frac{t-2}{(t+1)(t-2)}\\end{array}[\/latex]<\/p>\r\nThe numerator and denominator have a common factor of [latex]t\u20132[\/latex], so the rational expression can be simplified.\r\n<p style=\"text-align: center;\">[latex]\\large\\begin{array}{c}\\frac{\\cancel{t-2}}{(t+1)\\cancel{(t-2)}}\\\\\\\\=\\frac{1}{t+1}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\displaystyle \\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}=\\frac{1}{t+1},t\\ne -1,2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will give less instruction. See if you can find the LCD yourself before you look at the answer.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSubtract the rational expressions:\u00a0[latex]\\frac{6}{{x}^{2}+4x+4}-\\frac{2}{{x}^{2}-4}[\/latex], and define the domain.\r\n\r\nState the difference in simplest form.\r\n\r\n[reveal-answer q=\"681428\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"681428\"]\r\n\r\nNote that the denominator of the first expression is a perfect square trinomial, and the denominator of\u00a0the second expression is a difference of squares so they can be factored using special products.\r\n\r\n[latex]\\begin{array}{cc}\\frac{6}{{\\left(x+2\\right)}^{2}}-\\frac{2}{\\left(x+2\\right)\\left(x - 2\\right)}\\hfill &amp; \\text{Factor}.\\hfill \\\\ \\frac{6}{{\\left(x+2\\right)}^{2}}\\cdot \\frac{x - 2}{x - 2}-\\frac{2}{\\left(x+2\\right)\\left(x - 2\\right)}\\cdot \\frac{x+2}{x+2}\\hfill &amp; \\text{Multiply each fraction to get LCD as denominator}.\\hfill \\\\ \\frac{6\\left(x - 2\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}-\\frac{2\\left(x+2\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill &amp; \\text{Multiply}.\\hfill \\\\ \\frac{6x - 12-\\left(2x+4\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill &amp; \\text{Apply distributive property}.\\hfill \\\\ \\frac{4x - 16}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill &amp; \\text{Subtract}.\\hfill \\\\ \\frac{4\\left(x - 4\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill &amp; \\text{Simplify}.\\hfill \\end{array}[\/latex]\r\n<p style=\"text-align: left;\">The domain is [latex]x\\ne-2,2[\/latex]<\/p>\r\n[latex]\\frac{6}{{x}^{2}+4x+4}-\\frac{2}{{x}^{2}-4}=\\frac{4(x-4)}{(x+2)^2(x-2)}[\/latex],\u00a0\u00a0 [latex]x\\ne-2,2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the previous example, the LCD was \u00a0[latex]\\left(x+2\\right)^2\\left(x-2\\right)[\/latex]. \u00a0The reason we need to include [latex]\\left(x+2\\right)[\/latex] two times is because it appears two times in the expression [latex]\\frac{6}{{x}^{2}+4x+4}[\/latex].\r\n\r\nThe video that follows contains an example of subtracting rational expressions whose denominators are not alike. \u00a0The denominators are a trinomial and a binomial.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=MMlNtCrkakI&amp;feature=youtu.be\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question]3434[\/ohm_question]\r\n\r\n<\/div>\r\nOn the next page, we will show you how to find the greatest common denominator for a rational sum or difference that does not share any common factors. \u00a0We will also show you how to manage a sum or difference of\u00a0more than two rational expressions.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Add and subtract rational expressions with like denominators<\/li>\n<li>Add and subtract rational expressions with unlike denominators using a greatest common denominator<\/li>\n<\/ul>\n<\/div>\n<p>In beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators. By working carefully and writing down the steps along the way, you can keep track of all of the numbers and variables and perform the operations accurately.<\/p>\n<div id=\"attachment_5017\" style=\"width: 349px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5017\" class=\"wp-image-5017\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/20042339\/Screen-Shot-2016-06-19-at-9.23.12-PM.png\" alt=\"Add and Subtract\" width=\"339\" height=\"163\" \/><\/p>\n<p id=\"caption-attachment-5017\" class=\"wp-caption-text\">Add and Subtract<\/p>\n<\/div>\n<h2>Adding and Subtracting Rational Expressions with Like Denominators<\/h2>\n<p>Adding rational expressions with the same denominator is the simplest place to start, so let\u2019s begin there.<\/p>\n<p>To add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2}{9}+\\frac{4}{9}=\\frac{6}{9}\\\\\\\\\\frac{6}{9}=\\frac{3\\cdot 2}{3\\cdot 3}=\\frac{3}{3}\\cdot \\frac{2}{3}=1\\cdot \\frac{2}{3}=\\frac{2}{3}\\end{array}[\/latex]<\/p>\n<p>Follow the same process to add rational expressions with like denominators. Let&#8217;s try one.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add [latex]\\displaystyle \\frac{2{{x}^{2}}}{x+4}+\\frac{8x}{x+4}[\/latex], and define the domain.<\/p>\n<p>State the sum in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324146\">Show Solution<\/span><\/p>\n<div id=\"q324146\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the denominators are the same, add the numerators.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2{{x}^{2}}+8x}{x+4}[\/latex]<\/p>\n<p>Factor the numerator.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2x(x+4)}{x+4}[\/latex]<\/p>\n<p>Simplify\u00a0common factors and.<\/p>\n<p style=\"text-align: center;\">[latex]\\large\\begin{array}{c}\\frac{2x\\cancel{(x+4)}}{\\cancel{x+4}}\\\\\\\\=\\frac{2x}{1}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">The domain is found by setting the denominators in the original sum\u00a0equal to zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+4=0\\\\x=-4\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">The domain is [latex]x\\ne-4[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{2{{x}^{2}}}{x+4}+\\frac{8x}{x+4}=2x,x\\ne -4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"70\" height=\"62\" \/>Caution! \u00a0Remember to define\u00a0the domain of a sum or difference <em>before<\/em> simplifying. \u00a0You may lose important information when you simplify. In the example above, the domain is [latex]x\\ne-4[\/latex]. \u00a0If we were to have defined the domain after simplifying, we would find that the domain is all real numbers which is incorrect.<\/div>\n<p>To subtract rational expressions with like denominators, follow the same process you use to subtract fractions with like denominators. The process is just like the addition of rational expressions, except that you subtract instead of add.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Subtract[latex]\\frac{4x+7}{x+6}-\\frac{2x+8}{x+6}[\/latex], and define the domain.<\/p>\n<p>State the difference in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q681427\">Show Solution<\/span><\/p>\n<div id=\"q681427\" class=\"hidden-answer\" style=\"display: none\">\n<p>Subtract the second numerator from the first and keep the denominator the same.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{4x+7-(2x+8)}{x+6}[\/latex]<\/p>\n<p>Be careful to distribute the negative to both terms of the second numerator.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{4x+7-2x-8}{x+6}[\/latex]<\/p>\n<p>Combine like terms. This rational expression cannot be simplified any further.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2x-1}{x+6}[\/latex]<\/p>\n<p style=\"text-align: left;\">The domain is found from the denominators of original\u00a0expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+6=0\\\\x=-6\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">The domain is [latex]x\\ne-6[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{4x+7}{x+6}-\\frac{2x+8}{x+6}=\\frac{2x-1}{x+6},\\text{}x\\ne-6[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm40242\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=40242&theme=oea&iframe_resize_id=ohm40242&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the video that follows, we present\u00a0more examples of adding rational expressions with like denominators. Additionally, we review finding the domain of a rational expression.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Add and Subtract Rational Expressions with Like Denominators and Give the Domain\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BeaHKtxB868?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Adding and Subtracting Rational Expressions with Unlike Denominators<\/h2>\n<div id=\"attachment_4984\" style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4984\" class=\"wp-image-4984\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/17210449\/Screen-Shot-2016-06-17-at-1.59.26-PM-300x207.png\" alt=\"skateboard, motorcycle, bike, and rollerskates to illustrate that they are different but have wheels in common.\" width=\"407\" height=\"281\" \/><\/p>\n<p id=\"caption-attachment-4984\" class=\"wp-caption-text\">What do they have in common?<\/p>\n<\/div>\n<p>Before adding and subtracting rational expressions with <i>unlike<\/i> denominators, you need to find a common denominator. Once again, this process is similar to the one used for adding and subtracting numeric fractions with unlike denominators.\u00a0Remember how to do this?<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{5}{6}+\\frac{8}{10}+\\frac{3}{4}[\/latex]<\/p>\n<p>Since the denominators are [latex]6[\/latex], [latex]10[\/latex], and [latex]4[\/latex], you want to find the <strong>least common denominator<\/strong> and express each fraction with this denominator before adding. (BTW, you can add fractions by finding <i>any<\/i> common denominator; it does not have to be the least. You focus on using the <i>least<\/i> because then there is less simplifying to do. But either way works.)<\/p>\n<p>Finding the least common denominator is the same as finding the <strong>least common multiple<\/strong> of \u00a0[latex]4[\/latex], [latex]6[\/latex], and [latex]10[\/latex]. There are a couple of ways to do this. The first is to list the multiples of each number and determine which multiples they have in common. The least of these numbers will be the least common denominator.<\/p>\n<table style=\"width: 80%;\">\n<thead>\n<tr style=\"height: 43px;\">\n<th style=\"text-align: center; height: 43px; width: 6%;\">Number<\/th>\n<th style=\"height: 43px; width: 84.2962%;\" colspan=\"16\">\n<p style=\"text-align: center;\"><b>Multiples<\/b><\/p>\n<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px; width: 6%;\">[latex]4[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]8[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]12[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]16[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]20[\/latex]<\/td>\n<td style=\"height: 15px; width: 7%;\">[latex]24[\/latex]<\/td>\n<td style=\"height: 15px; width: 7%;\">[latex]28[\/latex]<\/td>\n<td style=\"height: 15px; width: 7%;\">[latex]32[\/latex]<\/td>\n<td style=\"height: 15px; width: 3%;\">[latex]36[\/latex]<\/td>\n<td style=\"height: 15px; width: 10%;\">[latex]40[\/latex]<\/td>\n<td style=\"height: 15px; width: 3%;\">[latex]44[\/latex]<\/td>\n<td style=\"height: 15px; width: 3%;\">[latex]48[\/latex]<\/td>\n<td style=\"height: 15px; width: 3%;\">[latex]52[\/latex]<\/td>\n<td style=\"height: 15px; width: 3%;\">[latex]56[\/latex]<\/td>\n<td style=\"height: 15px; width: 4%;\">[latex]\\textbf{60}[\/latex]<\/td>\n<td style=\"height: 15px; width: 10%;\">[latex]64[\/latex]<\/td>\n<td style=\"height: 15px; width: 0.296209%;\"><\/td>\n<\/tr>\n<tr style=\"height: 15.2812px;\">\n<td style=\"height: 15.2812px; width: 6%;\">[latex]6[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 6%;\">[latex]12[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 6%;\">[latex]18[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 6%;\">[latex]24[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 6%;\">[latex]30[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 7%;\">[latex]36[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 7%;\">[latex]42[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 7%;\">[latex]48[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 3%;\">[latex]54[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 10%;\">[latex]\\textbf{60}[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 3%;\">[latex]66[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 3%;\">[latex]68[\/latex]<\/td>\n<td style=\"height: 15.2812px; width: 3%;\"><\/td>\n<td style=\"height: 15.2812px; width: 3%;\"><\/td>\n<td style=\"height: 15.2812px; width: 4%;\"><\/td>\n<td style=\"height: 15.2812px; width: 10%;\"><\/td>\n<td style=\"height: 15.2812px; width: 0.296209%;\"><\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px; width: 6%;\">[latex]10[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]20[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]30[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]40[\/latex]<\/td>\n<td style=\"height: 15px; width: 6%;\">[latex]50[\/latex]<\/td>\n<td style=\"height: 15px; width: 7%;\">[latex]\\textbf{60}[\/latex]<\/td>\n<td style=\"height: 15px; width: 7%;\">\u00a0[latex]70[\/latex]<\/td>\n<td style=\"height: 15px; width: 7%;\">[latex]80[\/latex]<\/td>\n<td style=\"height: 15px; width: 3%;\"><\/td>\n<td style=\"height: 15px; width: 10%;\"><\/td>\n<td style=\"height: 15px; width: 3%;\"><\/td>\n<td style=\"height: 15px; width: 3%;\"><\/td>\n<td style=\"height: 15px; width: 3%;\"><\/td>\n<td style=\"height: 15px; width: 3%;\"><\/td>\n<td style=\"height: 15px; width: 4%;\"><\/td>\n<td style=\"height: 15px; width: 10%;\"><\/td>\n<td style=\"height: 15px; width: 0.296209%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The other method is to use <strong>prime factorization<\/strong>, the process of finding the prime factors of a number. This is how the method works with numbers.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use prime factorization to find the least common multiple of [latex]6[\/latex], [latex]10[\/latex], and [latex]4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371893\">Show Solution<\/span><\/p>\n<div id=\"q371893\" class=\"hidden-answer\" style=\"display: none\">First, find the prime factorization of each denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}6=3\\cdot2\\\\10=5\\cdot2\\\\4=2\\cdot2\\end{array}[\/latex]<\/p>\n<p>The LCM will contain factors of \u00a0[latex]2[\/latex],[latex]3[\/latex], and [latex]5[\/latex]. Multiply each number the maximum number of times it appears in a single factorization.<\/p>\n<p>In this case, [latex]3[\/latex] appears once, [latex]5[\/latex] appears once, and [latex]2[\/latex] is used twice because it appears twice in the prime factorization of [latex]4[\/latex].<\/p>\n<p>Therefore, the LCM of [latex]6[\/latex], [latex]10[\/latex], and [latex]4[\/latex] is [latex]3\\cdot5\\cdot2\\cdot2[\/latex], or [latex]60[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6=3\\cdot2\\\\\\,\\,\\,\\,\\,\\,\\,\\,10=5\\cdot2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4=2\\cdot2\\\\\\text{LCM}=3\\cdot5\\cdot2\\cdot2\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The least common multiple of \u00a0[latex]6, 10[\/latex], and [latex]4[\/latex] is [latex]60[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Both methods give the same result, but prime factorization is faster. Your choice!<\/p>\n<p>Now that you have found the least common multiple, you can use that number as the least common denominator of the fractions. Multiply each fraction by the fractional form of [latex]1[\/latex] that will produce a denominator of [latex]60[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\frac{5}{6}\\cdot \\frac{10}{10}=\\frac{50}{60}\\\\\\\\\\frac{8}{10}\\cdot\\frac{6}{6}=\\frac{48}{60}\\\\\\\\\\frac{3}{4}\\cdot\\frac{15}{15}=\\frac{45}{60}\\end{array}[\/latex]<\/p>\n<p>Now that you have like denominators, add the fractions:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{50}{60}+\\frac{48}{60}+\\frac{45}{60}=\\frac{143}{60}[\/latex]<\/p>\n<p>In the next example, we show how to find the least common multiple of a rational expression with a monomial in the denominator.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add[latex]\\frac{2n}{15m^{2}}+\\frac{3n}{21m}[\/latex], and give the domain.<\/p>\n<p>State the sum in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q926091\">Show Solution<\/span><\/p>\n<div id=\"q926091\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the prime factorization of each denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}15m^{2}\\,=\\,3\\cdot5\\cdot{m}\\cdot{m}\\\\\\,\\,\\,21m\\,=\\,3\\cdot7\\cdot{m}\\end{array}[\/latex]<\/p>\n<p>Find the least common multiple. [latex]3[\/latex] appears exactly once in both of the expressions, so it will appear once in the least common multiple. Both [latex]5[\/latex] and [latex]7[\/latex] appear at most once. For the variables, the most <i>m<\/i> appears is twice.<\/p>\n<p>Use the least common multiple for your new common denominator, it will be the LCD.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}15m^{2}\\,=\\,3\\cdot5\\cdot{m}\\cdot{m}\\\\21m\\,\\,\\,=\\,3\\cdot7\\cdot{m}\\\\\\text{LCM}:3\\cdot5\\cdot7\\cdot{m}\\cdot{m}\\\\\\text{LCM}:105m^{2}\\end{array}[\/latex]<\/p>\n<p>Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]105m^{2}[\/latex]<i>. <\/i>Remember that <i>m<\/i> cannot be 0 because the denominators would be [latex]0[\/latex].<\/p>\n<p>The first denominator is [latex]15m^{2}[\/latex]\u00a0and the LCD is [latex]105m^{2}[\/latex]. You need to multiply [latex]15m^{2}[\/latex]\u00a0by [latex]7[\/latex] to get the LCD, so multiply the entire rational expression by [latex]\\frac{7}{7}[\/latex].<\/p>\n<p>The second denominator is [latex]21m[\/latex] and the LCD is [latex]105m^{2}[\/latex]. You need to multiply [latex]21m[\/latex] by [latex]5m[\/latex] to get the LCD, so multiply the entire rational expression by [latex]\\frac{5m}{5m}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2n}{15m^{2}}\\cdot\\frac{7}{7}=\\frac{14n}{105m^{2}}\\\\\\\\\\frac{3n}{21m}\\cdot\\frac{5m}{5m}=\\frac{15mn}{105m^{2}}\\end{array}[\/latex]<\/p>\n<p>Add the numerators and keep the denominator the same.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{14n}{105{{m}^{2}}}+\\frac{15mn}{105{{m}^{2}}}=\\frac{14n+15mn}{105{{m}^{2}}}[\/latex]<\/p>\n<p>If possible, simplify by finding common factors in the numerator and denominator. This rational expression is already in simplest form because the numerator and denominator have no factors in common.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{n(14+15m)}{105{{m}^{2}}}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{2n}{15{{m}^{2}}}+\\frac{3n}{21m}=\\frac{n(14+15m)}{105{{m}^{2}}},m\\ne 0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>That took a while, but you got through it. Adding rational expressions can be a lengthy process, but taken one step at a time, it can be done.\u00a0 Now let&#8217;s take a look at some examples where the denominator is not a monomial.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">To find the least common denominator (LCD) of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, consider the following rational expressions:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{6}{\\left(x+3\\right)\\left(x+4\\right)},\\text{ and }\\frac{9x}{\\left(x+4\\right)\\left(x+5\\right)}[\/latex]<\/p>\n<p style=\"text-align: left;\">The LCD would be [latex]\\left(x+3\\right)\\left(x+4\\right)\\left(x+5\\right)[\/latex].<\/p>\n<p>To find the LCD, we count the greatest number of times a factor appears in each denominator and include it in the LCD that many times.<\/p>\n<p>For example, in\u00a0[latex]\\dfrac{6}{\\left(x+3\\right)\\left(x+4\\right)}[\/latex], [latex]\\left(x+3\\right)[\/latex] is represented once and \u00a0[latex]\\left(x+4\\right)[\/latex] is represented once, so they both appear exactly once in the LCD.<\/p>\n<p>In [latex]\\dfrac{9x}{\\left(x+4\\right)\\left(x+5\\right)}[\/latex], [latex]\\left(x+4\\right)[\/latex] appears once and [latex]\\left(x+5\\right)[\/latex] appears once.<\/p>\n<p>We have already accounted for\u00a0[latex]\\left(x+4\\right)[\/latex], so the LCD just needs one factor of\u00a0[latex]\\left(x+5\\right)[\/latex] to be complete.<\/p>\n<p>Once we find the LCD, we need to multiply each expression by the form of\u00a0[latex]1[\/latex] that will change the denominator to the LCD.<\/p>\n<p>What do we mean by &#8221; the form of\u00a0[latex]1[\/latex]&#8220;?<\/p>\n<p>[latex]\\frac{x+5}{x+5}=1[\/latex] so multiplying an expression\u00a0by it will not change its value.<\/p>\n<p>For example, we would need to multiply the expression [latex]\\dfrac{6}{\\left(x+3\\right)\\left(x+4\\right)}[\/latex] by [latex]\\frac{x+5}{x+5}[\/latex] and the expression [latex]\\frac{9x}{\\left(x+4\\right)\\left(x+5\\right)}[\/latex] by [latex]\\frac{x+3}{x+3}[\/latex].<\/p>\n<p>Hopefully this process will become clear after you practice it yourself. \u00a0As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which &#8220;form of 1&#8221; you will need to multiply\u00a0each expression by so that it has the LCD.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add the rational expressions [latex]\\frac{5}{x}+\\frac{6}{y}[\/latex] and define the domain.<\/p>\n<p>State the sum in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324147\">Show Solution<\/span><\/p>\n<div id=\"q324147\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, define the domain of each expression. Since we have x and y in the denominators, we can say [latex]x\\ne0 ,\\text{ and }y\\ne0[\/latex].<\/p>\n<p>Now we have to find the LCD. Since x appears once and y appears once, \u00a0the LCD will be [latex]xy[\/latex]. \u00a0We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[\/latex] as the denominator for each fraction.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{5}{x}\\cdot \\frac{y}{y}+\\frac{6}{y}\\cdot \\frac{x}{x}\\\\ \\frac{5y}{xy}+\\frac{6x}{xy}\\end{array}[\/latex]<\/div>\n<p>Now that the expressions have the same denominator, we simply add the numerators to find the sum.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{6x+5y}{xy}[\/latex]<\/div>\n<p>[latex]\\frac{5}{x}+\\frac{6}{y}=\\frac{(6x+5y)}{xy}[\/latex],\u00a0[latex]x\\ne0 ,\\text{ and }y\\ne0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is one more example of adding rational expressions where the denominators are multi-term polynomials. First, we will factor and then find the LCD. Note that [latex]x^2-4[\/latex] is a difference of squares and can be factored using special products.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify[latex]\\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\\frac{x}{x-2}[\/latex] and give the domain.<\/p>\n<p>State the result in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q57692\">Show Solution<\/span><\/p>\n<div id=\"q57692\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the least common denominator by factoring each denominator. The least common denominator includes the maximum number of times it appears in a single factorization. Remember that <i>x<\/i> cannot be [latex]2[\/latex] or [latex]-2[\/latex] because the denominators would be\u00a0[latex]0[\/latex].<\/p>\n<p>[latex]\\left(x+2\\right)[\/latex] appears a maximum of one time and so does [latex]\\left(x\u20132\\right)[\/latex]. This means the LCD is [latex]\\left(x+2\\right)\\left(x\u20132\\right)[\/latex].<\/p>\n<p>Multiply each expression by the equivalent of\u00a0[latex]1[\/latex] that will give it the common denominator. Notice the first fraction already has the LCD and as a result remains unchanged.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\\\\frac{x}{x-2}\\cdot \\frac{x+2}{x+2}=\\frac{x(x+2)}{(x+2)(x-2)}\\end{array}[\/latex]<\/p>\n<p>Rewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{2{{x}^{2}}}{(x+2)(x-2)}+\\frac{x(x+2)}{(x+2)(x-2)}[\/latex]<\/p>\n<p>Combine the numerators.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2{{x}^{2}}+x(x+2)}{(x+2)(x-2)}\\\\\\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x}{(x+2)(x-2)}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[\/latex]<\/p>\n<p>Check for simplest form. Since neither [latex]\\left(x+2\\right)[\/latex] nor [latex]\\left(x-2\\right)[\/latex] is a factor of [latex]3{{x}^{2}}+2x[\/latex], this expression is in simplest form.<\/p>\n<p>[latex]\\displaystyle \\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\\frac{x}{x-2}=\\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[\/latex] \u00a0\u00a0 [latex]\\displaystyle x\\ne 2,-2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The video that follows contains an example of adding rational expressions whose denominators are not alike. \u00a0The denominators of both expressions contain only monomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Add Rational Expressions with Unlike Denominators and Give the Domain (Mono Denom)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Wk8ZZhE9ZjI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Subtracting Rational Expressions<\/h2>\n<p>Now let\u2019s try subtracting rational expressions. You&#8217;ll use the same basic technique of finding the least common denominator and rewriting each rational expression to have that denominator.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Subtract[latex]\\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}[\/latex], define the domain.<\/p>\n<p>State the difference in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q704185\">Show Solution<\/span><\/p>\n<div id=\"q704185\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the prime factorization of each denominator. [latex]t+1[\/latex] cannot be factored any further, but [latex]{{t}^{2}}-t-2[\/latex] can be. Remember that <i>t<\/i> cannot be [latex]-1[\/latex] or [latex]2[\/latex] because the denominators would be [latex]0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}t+1=t+1\\\\t^{2}-t-2=\\left(t-2\\right)\\left(t+1\\right)\\end{array}[\/latex]<\/p>\n<p>Find the least common multiple. [latex]t+1[\/latex] appears exactly once in both of the expressions, so it will appear once in the least common denominator. [latex]t\u20132[\/latex] also appears once.<\/p>\n<p>This means that [latex]\\left(t-2\\right)\\left(t+1\\right)[\/latex] is the least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out.<\/p>\n<p>Use the least common multiple for your new common denominator, it will be the LCD.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}t+1=t+1\\\\t^{2}-t-2=\\left(t-2\\right)\\left(t+1\\right)\\\\\\text{LCM}:\\left(t+1\\right)\\left(t-1\\right)\\end{array}[\/latex]<\/p>\n<p>Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]\\left(t+1\\right)\\left(t\u20132\\right)[\/latex]<i>.<\/i><\/p>\n<p>You need to multiply [latex]t+1[\/latex] by [latex]t\u20132[\/latex] to get the LCD, so multiply the entire rational expression by [latex]\\displaystyle \\frac{t-2}{t-2}[\/latex].<\/p>\n<p>The second expression already has a denominator of [latex]\\left(t+1\\right)\\left(t\u20132\\right)[\/latex], so you do not need to multiply it by anything.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2}{t+1}\\cdot \\frac{t-2}{t-2}=\\frac{2(t-2)}{(t+1)(t-2)}\\\\\\\\\\,\\,\\,\\frac{t-2}{{{t}^{2}}-t-2}=\\frac{t-2}{(t+1)(t-2)}\\end{array}[\/latex]<\/p>\n<p>Then rewrite the subtraction problem with the common denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2\\left(t-2\\right)}{\\left(t+1\\right)\\left(t-2\\right)}-\\frac{t-2}{\\left(t+1\\right)\\left(t-2\\right)}[\/latex]<\/p>\n<p>Subtract the numerators and simplify. Remember that parentheses need to be included around the second [latex]\\left(t\u20132\\right)[\/latex] in the numerator because the whole quantity is subtracted. Otherwise you would be subtracting just the <i>t<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\\\\\\frac{t-2}{(t+1)(t-2)}\\end{array}[\/latex]<\/p>\n<p>The numerator and denominator have a common factor of [latex]t\u20132[\/latex], so the rational expression can be simplified.<\/p>\n<p style=\"text-align: center;\">[latex]\\large\\begin{array}{c}\\frac{\\cancel{t-2}}{(t+1)\\cancel{(t-2)}}\\\\\\\\=\\frac{1}{t+1}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}=\\frac{1}{t+1},t\\ne -1,2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will give less instruction. See if you can find the LCD yourself before you look at the answer.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Subtract the rational expressions:\u00a0[latex]\\frac{6}{{x}^{2}+4x+4}-\\frac{2}{{x}^{2}-4}[\/latex], and define the domain.<\/p>\n<p>State the difference in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q681428\">Show Solution<\/span><\/p>\n<div id=\"q681428\" class=\"hidden-answer\" style=\"display: none\">\n<p>Note that the denominator of the first expression is a perfect square trinomial, and the denominator of\u00a0the second expression is a difference of squares so they can be factored using special products.<\/p>\n<p>[latex]\\begin{array}{cc}\\frac{6}{{\\left(x+2\\right)}^{2}}-\\frac{2}{\\left(x+2\\right)\\left(x - 2\\right)}\\hfill & \\text{Factor}.\\hfill \\\\ \\frac{6}{{\\left(x+2\\right)}^{2}}\\cdot \\frac{x - 2}{x - 2}-\\frac{2}{\\left(x+2\\right)\\left(x - 2\\right)}\\cdot \\frac{x+2}{x+2}\\hfill & \\text{Multiply each fraction to get LCD as denominator}.\\hfill \\\\ \\frac{6\\left(x - 2\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}-\\frac{2\\left(x+2\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill & \\text{Multiply}.\\hfill \\\\ \\frac{6x - 12-\\left(2x+4\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill & \\text{Apply distributive property}.\\hfill \\\\ \\frac{4x - 16}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill & \\text{Subtract}.\\hfill \\\\ \\frac{4\\left(x - 4\\right)}{{\\left(x+2\\right)}^{2}\\left(x - 2\\right)}\\hfill & \\text{Simplify}.\\hfill \\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">The domain is [latex]x\\ne-2,2[\/latex]<\/p>\n<p>[latex]\\frac{6}{{x}^{2}+4x+4}-\\frac{2}{{x}^{2}-4}=\\frac{4(x-4)}{(x+2)^2(x-2)}[\/latex],\u00a0\u00a0 [latex]x\\ne-2,2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the previous example, the LCD was \u00a0[latex]\\left(x+2\\right)^2\\left(x-2\\right)[\/latex]. \u00a0The reason we need to include [latex]\\left(x+2\\right)[\/latex] two times is because it appears two times in the expression [latex]\\frac{6}{{x}^{2}+4x+4}[\/latex].<\/p>\n<p>The video that follows contains an example of subtracting rational expressions whose denominators are not alike. \u00a0The denominators are a trinomial and a binomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Subtract Rational Expressions with Unlike Denominators and Give the Domain\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/MMlNtCrkakI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm3434\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3434&theme=oea&iframe_resize_id=ohm3434&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>On the next page, we will show you how to find the greatest common denominator for a rational sum or difference that does not share any common factors. \u00a0We will also show you how to manage a sum or difference of\u00a0more than two rational expressions.<\/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-16344\">\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>Screenshot: Caution. <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><li>Image: What do they have in common?. <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><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><li>Add and Subtract Rational Expressions with Like Denominators and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BeaHKtxB868\">https:\/\/youtu.be\/BeaHKtxB868<\/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>Add Rational Expressions with Unlike Denominators and Give the Domain (Mono Denom). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Wk8ZZhE9ZjI\">https:\/\/youtu.be\/Wk8ZZhE9ZjI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Subtract Rational Expressions with Unlike Denominators and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/MMlNtCrkakI\">https:\/\/youtu.be\/MMlNtCrkakI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 15: Rational Expressions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: Caution\",\"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\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Image: What do they have in common?\",\"author\":\"\",\"organization\":\"Lumen 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