{"id":2929,"date":"2016-07-22T16:55:46","date_gmt":"2016-07-22T16:55:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2929"},"modified":"2019-08-06T19:09:15","modified_gmt":"2019-08-06T19:09:15","slug":"read-add-and-subtract-rational-expressions-part-i","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-add-and-subtract-rational-expressions-part-i\/","title":{"raw":"Add and Subtract Rational Expressions","rendered":"Add and Subtract Rational Expressions"},"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<\/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.\r\n\r\n<img class=\"aligncenter wp-image-5017\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162750\/Screen-Shot-2016-06-19-at-9.23.12-PM.png\" alt=\"Add and Subtract Sign\" width=\"339\" height=\"163\" \/>\r\n<h2>Adding Rational Expressions<\/h2>\r\nTo 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:\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=\"324146\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324146\"]\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, but in this case, the expressions have denominators with 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=\"57691\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"57691\"]\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\nIn the video that follows, we present an example of adding two rational expression whose denominators are binomials with no common factors.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=CKGpiTE5vIg&amp;feature=youtu.be\r\n<h2>Subtracting Rational Expressions<\/h2>\r\nTo subtract rational expressions, follow the same process you use to add rational expressions.\u00a0You will need to be careful with signs though.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSubtract[latex]\\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}[\/latex] and 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 LCD\u00a0of each expression. [latex]t+1[\/latex] cannot be factored any further but [latex]{{t}^{2}}-t-2[\/latex] can be. Note\u00a0that <i>t<\/i> cannot be [latex]-1[\/latex] or [latex]2[\/latex] because the denominators would be\u00a0[latex]0[\/latex].\r\n\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 LCD. In this case, it is easier to leave the common denominator in terms of the factors. You will not need to multiply it out.\r\n\r\nUse the LCD as your new common denominator.\r\n\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\nIn the first expression, 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].\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\nNow rewrite the subtraction problem using the expressions 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\u00a0[latex]t[\/latex].\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[latex] \\displaystyle \\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}=\\frac{1}{t+1}[\/latex], \u00a0 [latex]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=\"681427\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"681427\"]\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\nThe video that follows contains an example of subtracting rational expressions.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=MMlNtCrkakI&amp;feature=youtu.be","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Add and subtract rational expressions<\/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.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5017\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162750\/Screen-Shot-2016-06-19-at-9.23.12-PM.png\" alt=\"Add and Subtract Sign\" width=\"339\" height=\"163\" \/><\/p>\n<h2>Adding Rational Expressions<\/h2>\n<p>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:<\/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=\"q324146\">Show Solution<\/span><\/p>\n<div id=\"q324146\" 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, but in this case, the expressions have denominators with 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=\"q57691\">Show Solution<\/span><\/p>\n<div id=\"q57691\" 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>In the video that follows, we present an example of adding two rational expression whose denominators are binomials with no common factors.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Add Rational Expressions with Unlike Denominators\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/CKGpiTE5vIg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Subtracting Rational Expressions<\/h2>\n<p>To subtract rational expressions, follow the same process you use to add rational expressions.\u00a0You will need to be careful with signs though.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Subtract[latex]\\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}[\/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=\"q704185\">Show Solution<\/span><\/p>\n<div id=\"q704185\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the LCD\u00a0of each expression. [latex]t+1[\/latex] cannot be factored any further but [latex]{{t}^{2}}-t-2[\/latex] can be. Note\u00a0that <i>t<\/i> cannot be [latex]-1[\/latex] or [latex]2[\/latex] because the denominators would be\u00a0[latex]0[\/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 LCD. In this case, it is easier to leave the common denominator in terms of the factors. You will not need to multiply it out.<\/p>\n<p>Use the LCD as your new common denominator.<\/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>In the first expression, 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>Now rewrite the subtraction problem using the expressions 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\u00a0[latex]t[\/latex].<\/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<p>[latex]\\displaystyle \\frac{2}{t+1}-\\frac{t-2}{{{t}^{2}}-t-2}=\\frac{1}{t+1}[\/latex], \u00a0 [latex]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=\"q681427\">Show Solution<\/span><\/p>\n<div id=\"q681427\" 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].<br \/>\nThe video that follows contains an example of subtracting rational expressions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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\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-2929\">\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>Ex: Add Rational Expressions with Unlike Denominators. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=CKGpiTE5vIg&#038;feature=youtu.be\">https:\/\/www.youtube.com\/watch?v=CKGpiTE5vIg&#038;feature=youtu.be<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Subtract Rational Expressions with Unlike Denominators and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=MMlNtCrkakI&#038;feature=youtu.be\">https:\/\/www.youtube.com\/watch?v=MMlNtCrkakI&#038;feature=youtu.be<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Subtract Rational Expressions with Unlike Denominators and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/MMlNtCrkakI\">https:\/\/youtu.be\/MMlNtCrkakI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/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":21,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: Caution\",\"author\":\"\",\"organization\":\"Lumen 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