{"id":4132,"date":"2020-04-06T02:48:42","date_gmt":"2020-04-06T02:48:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/mathforlibscoreq\/?post_type=chapter&#038;p=4132"},"modified":"2021-02-06T00:03:08","modified_gmt":"2021-02-06T00:03:08","slug":"adding-and-subtracting-fractions-with-different-denominators","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/adding-and-subtracting-fractions-with-different-denominators\/","title":{"raw":"Adding and Subtracting Fractions With Different Denominators","rendered":"Adding and Subtracting Fractions With Different Denominators"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Add or subtract fractions with different denominators<\/li>\r\n \t<li>Identify and use fraction operations<\/li>\r\n<\/ul>\r\n<\/div>\r\nTo add or subtract fractions with different denominators, we first must write them as equivalent fractions having the same denominator. We'll use the techniques from the previous section to find the LCM of the denominators of the fractions. Recall that we call this the LCD (the least common denominator).\u00a0We only use the denominators of the fractions, not the numerators, when finding the LCD.\r\n\r\nThen we can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The steps for finding the LCD and the Equivalent Fractions Property are repeated below for reference.\r\n<div class=\"textbox shaded\">\r\n<h3>Least Common Denominator<\/h3>\r\nThe least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox shaded\">\r\n<h3>Equivalent Fractions Property<\/h3>\r\nIf [latex]a,b,c[\/latex] are whole numbers where [latex]b\\ne 0,c\\ne 0,\\text{then}[\/latex]\r\n<p style=\"text-align: center\">[latex]\\Large\\frac{a}{b}=\\Large\\frac{a\\cdot c}{b\\cdot c}\\normalsize\\text{ and }\\Large\\frac{a\\cdot c}{b\\cdot c}=\\Large\\frac{a}{b}[\/latex]<\/p>\r\n\r\n<\/div>\r\nOnce we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators. Try the examples and practice problems below to refresh these skills.\r\n<div class=\"textbox shaded\">\r\n<h3>Add or subtract fractions with different denominators<\/h3>\r\n<ol id=\"eip-id1168468303196\" class=\"stepwise\">\r\n \t<li>Find the LCD.<\/li>\r\n \t<li>Convert each fraction to an equivalent form with the LCD as the denominator.<\/li>\r\n \t<li>Add or subtract the fractions.<\/li>\r\n \t<li>Write the result in simplified form.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAdd: [latex]\\Large\\frac{1}{2}+\\Large\\frac{1}{3}[\/latex]\r\n\r\nSolution:\r\n<table id=\"eip-id1168466144855\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 1 half plus 1 third. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\Large\\frac{1}{2}+\\Large\\frac{1}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the LCD of [latex]2[\/latex], [latex]3[\/latex].<\/td>\r\n<td><img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221040\/CNX_BMath_Figure_04_05_029_img-01.png\" alt=\".\" width=\"81\" height=\"110\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Change into equivalent fractions with the LCD [latex]6[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{1\\cdot\\color{red}{3}}{2\\cdot\\color{red}{3}} +\\Large\\frac{1\\cdot\\color{red}{2}}{3\\cdot\\color{red}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerators and denominators.<\/td>\r\n<td>[latex]\\Large\\frac{3}{6}+\\Large\\frac{2}{6}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]\\Large\\frac{5}{6}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nRemember, always check to see if the answer can be simplified. Since [latex]5[\/latex] and [latex]6[\/latex] have no common factors, the fraction [latex]\\Large\\frac{5}{6}[\/latex] cannot be reduced.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question height=\"230\"]146262[\/ohm_question]\r\n\r\n<\/div>\r\nWatch the following video to see more examples and explanation about how to add two fractions with unlike denominators.\r\n\r\nhttps:\/\/youtu.be\/zV4q7j1-89I\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSubtract: [latex]\\Large\\frac{1}{2}-\\left(-\\Large\\frac{1}{4}\\right)[\/latex]\r\n[reveal-answer q=\"841760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"841760\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168466426391\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 1 half minus negative 1 fourth. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\Large\\frac{1}{2}-\\left(-\\Large\\frac{1}{4}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the LCD of [latex]2[\/latex] and [latex]4[\/latex].<\/td>\r\n<td><img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221041\/CNX_BMath_Figure_04_05_030_img-01.png\" alt=\".\" width=\"100\" height=\"76\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite as equivalent fractions using the LCD [latex]4[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{1\\cdot\\color{red}{2}}{2\\cdot\\color{red}{2}} - (--\\Large\\frac{1}{4})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the first fraction.<\/td>\r\n<td>[latex]\\Large\\frac{2}{4}-\\left(-\\Large\\frac{1}{4}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]\\Large\\frac{2-\\left(-1\\right)}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]\\Large\\frac{3}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOne of the fractions already had the least common denominator, so we only had to convert the other fraction.\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 height=\"230\"]146264[\/ohm_question]\r\n\r\n<\/div>\r\nThe following video provides two more examples of how to subtract two fractions with unlike denominators.\r\n\r\nhttps:\/\/youtu.be\/aXlkygPPzQ8\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAdd: [latex]\\Large\\frac{7}{12}+\\Large\\frac{5}{18}[\/latex]\r\n[reveal-answer q=\"826911\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"826911\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168468285390\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 7 over 12 plus 5 over 18. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\Large\\frac{7}{12}+\\Large\\frac{5}{18}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the LCD of [latex]12[\/latex] and [latex]18[\/latex].<\/td>\r\n<td><img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221043\/CNX_BMath_Figure_04_05_031_img-01.png\" alt=\".\" width=\"146\" height=\"75\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\r\n<td>[latex]\\Large\\frac{7\\cdot\\color{red}{3}}{12\\cdot\\color{red}{3}} +\\Large\\frac{5\\cdot\\color{red}{2}}{18\\cdot\\color{red}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerators and denominators.<\/td>\r\n<td>[latex]\\Large\\frac{21}{36}+\\Large\\frac{10}{36}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]\\Large\\frac{31}{36}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nBecause [latex]31[\/latex] is a prime number, it has no factors in common with [latex]36[\/latex]. The answer is simplified.\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 height=\"230\"]146265[\/ohm_question]\r\n\r\n<\/div>\r\nWhen we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The \"missing\" factors of each denominator are the numbers you need.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221045\/CNX_BMath_Figure_04_05_015.png\" alt=\"The first line says 12 equals 2 times 2 times 3. There is a blank space next to the 3. The next line says 18 equals 2 times 3 times 3. There is a blank space between the 2 and the first 3. There are red lines drawn from the blank spaces. This is labeled as missing factors. There is a horizontal line. Below the line, it says LCD equals 2 times 2 times 3 times 3. Below this, it says LCD equals 36.\" \/>\r\nThe LCD, [latex]36[\/latex], has [latex]2[\/latex] factors of [latex]2[\/latex] and [latex]2[\/latex] factors of [latex]3[\/latex].\r\n\r\nTwelve has two factors of [latex]2[\/latex], but only one of [latex]3[\/latex] \u2014so it is \u2018missing\u2018 one [latex]3[\/latex]. We multiplied the numerator and denominator of [latex]\\Large\\frac{7}{12}[\/latex] by [latex]3[\/latex] to get an equivalent fraction with denominator [latex]36[\/latex].\r\n\r\nEighteen is missing one factor of [latex]2[\/latex] \u2014so you multiply the numerator and denominator [latex]\\Large\\frac{5}{18}[\/latex] by [latex]2[\/latex] to get an equivalent fraction with denominator [latex]36[\/latex]. We will apply this method as we subtract the fractions in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSubtract: [latex]\\Large\\frac{7}{15}-\\Large\\frac{19}{24}[\/latex]\r\n[reveal-answer q=\"169999\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"169999\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168466319808\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 7 over 15 minus 19 over 24. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\Large\\frac{7}{15}-\\Large\\frac{19}{24}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the LCD.\r\n\r\n&nbsp;<\/td>\r\n<td><img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221046\/CNX_BMath_Figure_04_05_032_img-01.png\" alt=\".\" width=\"170\" height=\"80\" \/>\r\n\r\n[latex]15[\/latex] is 'missing' three factors of [latex]2[\/latex]\r\n\r\n[latex]24[\/latex] is 'missing' a factor of [latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\r\n<td>[latex]\\Large\\frac{7\\cdot\\color{red}{8}}{15\\cdot\\color{red}{8}} -\\Large\\frac{19\\cdot\\color{red}{5}}{24\\cdot\\color{red}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify each numerator and denominator.<\/td>\r\n<td>[latex]\\Large\\frac{56}{120}-\\Large\\frac{95}{120}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]-\\Large\\frac{39}{120}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite showing the common factor of [latex]3[\/latex].<\/td>\r\n<td>[latex]-\\Large\\frac{13\\cdot 3}{40\\cdot 3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Remove the common factor to simplify.<\/td>\r\n<td>[latex]-\\Large\\frac{13}{40}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\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 height=\"230\"]146266[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAdd: [latex]-\\Large\\frac{11}{30}+\\Large\\frac{23}{42}[\/latex]\r\n[reveal-answer q=\"808641\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"808641\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168466855270\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says negative 11 over 30 plus 23 over 42. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td>[latex]-\\Large\\frac{11}{30}+\\Large\\frac{23}{42}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the LCD.\r\n\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221049\/CNX_BMath_Figure_04_05_033_img-01.png\" alt=\".\" width=\"151\" height=\"82\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\r\n<td>[latex]-\\Large\\frac{11\\cdot\\color{red}{7}}{30\\cdot\\color{red}{7}} +\\Large\\frac{23\\cdot\\color{red}{5}}{42\\cdot\\color{red}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify each numerator and denominator.<\/td>\r\n<td>[latex]-\\Large\\frac{77}{210}+\\Large\\frac{115}{210}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]\\Large\\frac{38}{210}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite showing the common factor of [latex]2[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{19\\cdot 2}{105\\cdot 2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Remove the common factor to simplify.<\/td>\r\n<td>[latex]\\Large\\frac{19}{105}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\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 height=\"230\"]146267[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Add or subtract fractions with different denominators<\/li>\n<li>Identify and use fraction operations<\/li>\n<\/ul>\n<\/div>\n<p>To add or subtract fractions with different denominators, we first must write them as equivalent fractions having the same denominator. We&#8217;ll use the techniques from the previous section to find the LCM of the denominators of the fractions. Recall that we call this the LCD (the least common denominator).\u00a0We only use the denominators of the fractions, not the numerators, when finding the LCD.<\/p>\n<p>Then we can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The steps for finding the LCD and the Equivalent Fractions Property are repeated below for reference.<\/p>\n<div class=\"textbox shaded\">\n<h3>Least Common Denominator<\/h3>\n<p>The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox shaded\">\n<h3>Equivalent Fractions Property<\/h3>\n<p>If [latex]a,b,c[\/latex] are whole numbers where [latex]b\\ne 0,c\\ne 0,\\text{then}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\Large\\frac{a}{b}=\\Large\\frac{a\\cdot c}{b\\cdot c}\\normalsize\\text{ and }\\Large\\frac{a\\cdot c}{b\\cdot c}=\\Large\\frac{a}{b}[\/latex]<\/p>\n<\/div>\n<p>Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators. Try the examples and practice problems below to refresh these skills.<\/p>\n<div class=\"textbox shaded\">\n<h3>Add or subtract fractions with different denominators<\/h3>\n<ol id=\"eip-id1168468303196\" class=\"stepwise\">\n<li>Find the LCD.<\/li>\n<li>Convert each fraction to an equivalent form with the LCD as the denominator.<\/li>\n<li>Add or subtract the fractions.<\/li>\n<li>Write the result in simplified form.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Add: [latex]\\Large\\frac{1}{2}+\\Large\\frac{1}{3}[\/latex]<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168466144855\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 1 half plus 1 third. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\Large\\frac{1}{2}+\\Large\\frac{1}{3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the LCD of [latex]2[\/latex], [latex]3[\/latex].<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221040\/CNX_BMath_Figure_04_05_029_img-01.png\" alt=\".\" width=\"81\" height=\"110\" \/><\/td>\n<\/tr>\n<tr>\n<td>Change into equivalent fractions with the LCD [latex]6[\/latex].<\/td>\n<td>[latex]\\Large\\frac{1\\cdot\\color{red}{3}}{2\\cdot\\color{red}{3}} +\\Large\\frac{1\\cdot\\color{red}{2}}{3\\cdot\\color{red}{2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerators and denominators.<\/td>\n<td>[latex]\\Large\\frac{3}{6}+\\Large\\frac{2}{6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]\\Large\\frac{5}{6}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Remember, always check to see if the answer can be simplified. Since [latex]5[\/latex] and [latex]6[\/latex] have no common factors, the fraction [latex]\\Large\\frac{5}{6}[\/latex] cannot be reduced.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146262\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146262&theme=oea&iframe_resize_id=ohm146262&show_question_numbers\" width=\"100%\" height=\"230\"><\/iframe><\/p>\n<\/div>\n<p>Watch the following video to see more examples and explanation about how to add two fractions with unlike denominators.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Add Fractions with Unlike Denominators (Basic with Model)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zV4q7j1-89I?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Subtract: [latex]\\Large\\frac{1}{2}-\\left(-\\Large\\frac{1}{4}\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q841760\">Show Solution<\/span><\/p>\n<div id=\"q841760\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168466426391\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 1 half minus negative 1 fourth. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\Large\\frac{1}{2}-\\left(-\\Large\\frac{1}{4}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the LCD of [latex]2[\/latex] and [latex]4[\/latex].<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221041\/CNX_BMath_Figure_04_05_030_img-01.png\" alt=\".\" width=\"100\" height=\"76\" \/><\/td>\n<\/tr>\n<tr>\n<td>Rewrite as equivalent fractions using the LCD [latex]4[\/latex].<\/td>\n<td>[latex]\\Large\\frac{1\\cdot\\color{red}{2}}{2\\cdot\\color{red}{2}} - (--\\Large\\frac{1}{4})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the first fraction.<\/td>\n<td>[latex]\\Large\\frac{2}{4}-\\left(-\\Large\\frac{1}{4}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]\\Large\\frac{2-\\left(-1\\right)}{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]\\Large\\frac{3}{4}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>One of the fractions already had the least common denominator, so we only had to convert the other fraction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146264\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146264&theme=oea&iframe_resize_id=ohm146264&show_question_numbers\" width=\"100%\" height=\"230\"><\/iframe><\/p>\n<\/div>\n<p>The following video provides two more examples of how to subtract two fractions with unlike denominators.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Example:  Subtract Fractions with Unlike Denominators\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/aXlkygPPzQ8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Add: [latex]\\Large\\frac{7}{12}+\\Large\\frac{5}{18}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q826911\">Show Solution<\/span><\/p>\n<div id=\"q826911\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168468285390\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 7 over 12 plus 5 over 18. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\Large\\frac{7}{12}+\\Large\\frac{5}{18}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the LCD of [latex]12[\/latex] and [latex]18[\/latex].<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221043\/CNX_BMath_Figure_04_05_031_img-01.png\" alt=\".\" width=\"146\" height=\"75\" \/><\/td>\n<\/tr>\n<tr>\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\n<td>[latex]\\Large\\frac{7\\cdot\\color{red}{3}}{12\\cdot\\color{red}{3}} +\\Large\\frac{5\\cdot\\color{red}{2}}{18\\cdot\\color{red}{2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerators and denominators.<\/td>\n<td>[latex]\\Large\\frac{21}{36}+\\Large\\frac{10}{36}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]\\Large\\frac{31}{36}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Because [latex]31[\/latex] is a prime number, it has no factors in common with [latex]36[\/latex]. The answer is simplified.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146265\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146265&theme=oea&iframe_resize_id=ohm146265&show_question_numbers\" width=\"100%\" height=\"230\"><\/iframe><\/p>\n<\/div>\n<p>When we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The &#8220;missing&#8221; factors of each denominator are the numbers you need.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221045\/CNX_BMath_Figure_04_05_015.png\" alt=\"The first line says 12 equals 2 times 2 times 3. There is a blank space next to the 3. The next line says 18 equals 2 times 3 times 3. There is a blank space between the 2 and the first 3. There are red lines drawn from the blank spaces. This is labeled as missing factors. There is a horizontal line. Below the line, it says LCD equals 2 times 2 times 3 times 3. Below this, it says LCD equals 36.\" \/><br \/>\nThe LCD, [latex]36[\/latex], has [latex]2[\/latex] factors of [latex]2[\/latex] and [latex]2[\/latex] factors of [latex]3[\/latex].<\/p>\n<p>Twelve has two factors of [latex]2[\/latex], but only one of [latex]3[\/latex] \u2014so it is \u2018missing\u2018 one [latex]3[\/latex]. We multiplied the numerator and denominator of [latex]\\Large\\frac{7}{12}[\/latex] by [latex]3[\/latex] to get an equivalent fraction with denominator [latex]36[\/latex].<\/p>\n<p>Eighteen is missing one factor of [latex]2[\/latex] \u2014so you multiply the numerator and denominator [latex]\\Large\\frac{5}{18}[\/latex] by [latex]2[\/latex] to get an equivalent fraction with denominator [latex]36[\/latex]. We will apply this method as we subtract the fractions in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Subtract: [latex]\\Large\\frac{7}{15}-\\Large\\frac{19}{24}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q169999\">Show Solution<\/span><\/p>\n<div id=\"q169999\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168466319808\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says 7 over 15 minus 19 over 24. The next line says\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\Large\\frac{7}{15}-\\Large\\frac{19}{24}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the LCD.<\/p>\n<p>&nbsp;<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221046\/CNX_BMath_Figure_04_05_032_img-01.png\" alt=\".\" width=\"170\" height=\"80\" \/><\/p>\n<p>[latex]15[\/latex] is &#8216;missing&#8217; three factors of [latex]2[\/latex]<\/p>\n<p>[latex]24[\/latex] is &#8216;missing&#8217; a factor of [latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\n<td>[latex]\\Large\\frac{7\\cdot\\color{red}{8}}{15\\cdot\\color{red}{8}} -\\Large\\frac{19\\cdot\\color{red}{5}}{24\\cdot\\color{red}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify each numerator and denominator.<\/td>\n<td>[latex]\\Large\\frac{56}{120}-\\Large\\frac{95}{120}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]-\\Large\\frac{39}{120}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Rewrite showing the common factor of [latex]3[\/latex].<\/td>\n<td>[latex]-\\Large\\frac{13\\cdot 3}{40\\cdot 3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Remove the common factor to simplify.<\/td>\n<td>[latex]-\\Large\\frac{13}{40}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146266\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146266&theme=oea&iframe_resize_id=ohm146266&show_question_numbers\" width=\"100%\" height=\"230\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Add: [latex]-\\Large\\frac{11}{30}+\\Large\\frac{23}{42}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q808641\">Show Solution<\/span><\/p>\n<div id=\"q808641\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168466855270\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says negative 11 over 30 plus 23 over 42. The next line says,\">\n<tbody>\n<tr>\n<td>[latex]-\\Large\\frac{11}{30}+\\Large\\frac{23}{42}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the LCD.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221049\/CNX_BMath_Figure_04_05_033_img-01.png\" alt=\".\" width=\"151\" height=\"82\" \/><\/td>\n<\/tr>\n<tr>\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\n<td>[latex]-\\Large\\frac{11\\cdot\\color{red}{7}}{30\\cdot\\color{red}{7}} +\\Large\\frac{23\\cdot\\color{red}{5}}{42\\cdot\\color{red}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify each numerator and denominator.<\/td>\n<td>[latex]-\\Large\\frac{77}{210}+\\Large\\frac{115}{210}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]\\Large\\frac{38}{210}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Rewrite showing the common factor of [latex]2[\/latex].<\/td>\n<td>[latex]\\Large\\frac{19\\cdot 2}{105\\cdot 2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Remove the common factor to simplify.<\/td>\n<td>[latex]\\Large\\frac{19}{105}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146267\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146267&theme=oea&iframe_resize_id=ohm146267&show_question_numbers\" width=\"100%\" height=\"230\"><\/iframe><\/p>\n<\/div>\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-4132\">\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>Question ID: 146262, 146264, 146265, 146266, 146267, 146268. <strong>Authored by<\/strong>: Alyson Day. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: Add Fractions with Unlike Denominators (Basic with Model). <strong>Authored by<\/strong>: James Sousa (mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zV4q7j1-89I\">https:\/\/youtu.be\/zV4q7j1-89I<\/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>Example: Subtract Fractions with Unlike Denominators. <strong>Authored by<\/strong>: James Sousa (mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/aXlkygPPzQ8\">https:\/\/youtu.be\/aXlkygPPzQ8<\/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, Specific attribution<\/div><ul class=\"citation-list\"><li>Prealgebra. <strong>Provided by<\/strong>: OpenStax. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757<\/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":25777,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757\"},{\"type\":\"cc\",\"description\":\"Ex: Add Fractions with Unlike Denominators (Basic with Model)\",\"author\":\"James Sousa (mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/zV4q7j1-89I\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Example: Subtract Fractions with Unlike Denominators\",\"author\":\"James Sousa (mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/aXlkygPPzQ8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Question ID: 146262, 146264, 146265, 146266, 146267, 146268\",\"author\":\"Alyson Day\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4132","chapter","type-chapter","status-web-only","hentry"],"part":329,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/4132","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/users\/25777"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/4132\/revisions"}],"predecessor-version":[{"id":5263,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/4132\/revisions\/5263"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/parts\/329"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/4132\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/media?parent=4132"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=4132"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/contributor?post=4132"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/license?post=4132"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}