{"id":68,"date":"2018-03-19T17:54:44","date_gmt":"2018-03-19T17:54:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/?post_type=chapter&#038;p=68"},"modified":"2024-04-26T22:01:14","modified_gmt":"2024-04-26T22:01:14","slug":"adding-and-subtracting-fractions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/chapter\/adding-and-subtracting-fractions\/","title":{"raw":"Adding and Subtracting Fractions","rendered":"Adding and Subtracting Fractions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning OUTCOMES<\/h3>\r\n<ul>\r\n \t<li>Use addition and subtraction when evaluating expressions with fractions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Addition and Subtraction of Fractions with Common Denominators<\/h2>\r\nHow many quarters are pictured below?\r\n\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220949\/CNX_BMath_Figure_04_04_001.png\" alt=\"Three U.S. quarters are shown. One is shown on the left, and two are shown on the right.\" \/>\r\n\r\nYou can quickly count three, but if you look carefully the image represents [latex]1[\/latex] quarter plus [latex]2[\/latex] quarters equals [latex]3[\/latex] quarters.\r\n\r\nRemember, quarters are really fractions of a dollar. Quarters are another way to say fourths. So the picture of the coins shows that\r\n<p style=\"text-align: center;\">[latex]{\\Large\\frac{1}{4}}+{\\Large\\frac{2}{4}}=\\Large{\\frac{3}{4}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{one quarter }+\\text{ two quarters }=\\text{ three quarters} [\/latex]<\/p>\r\nLet\u2019s use fraction circles to model the same example, [latex]\\Large\\frac{1}{4}\\normalsize+\\Large\\frac{2}{4}[\/latex].\r\n<table id=\"eip-id1168467352669\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says, \">\r\n<tbody>\r\n<tr>\r\n<td>Start with one [latex]\\Large\\frac{1}{4}[\/latex] piece.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220951\/CNX_BMath_Figure_04_04_002_img-01.png\" alt=\"one quarter circle\" width=\"205\" height=\"93\" \/><\/td>\r\n<td>[latex]\\Large\\frac{1}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add two more [latex]\\Large\\frac{1}{4}[\/latex] pieces.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220953\/CNX_BMath_Figure_04_04_002_img-02.png\" alt=\"plus 2 quarter circles\" width=\"205\" height=\"116\" \/><\/td>\r\n<td>[latex]+\\Large\\frac{2}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The result is [latex]\\Large\\frac{3}{4}[\/latex] .<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220954\/CNX_BMath_Figure_04_04_002_img-03.png\" alt=\"3 quarter circles\" width=\"205\" height=\"185\" \/><\/td>\r\n<td>[latex]\\Large\\frac{3}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo again, we see that\r\n<p style=\"text-align: center;\">[latex]\\Large\\frac{1}{4}\\normalsize+\\Large\\frac{2}{4}\\normalsize=\\Large\\frac{3}{4}[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nUse a model to find the sum [latex]\\Large\\frac{3}{8}\\normalsize+\\Large\\frac{2}{8}[\/latex].\r\n\r\nSolution:\r\n<table id=\"eip-id1168468316426\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says, \">\r\n<tbody>\r\n<tr>\r\n<td>Start with three [latex]\\Large\\frac{1}{8}[\/latex] pieces.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220957\/CNX_BMath_Figure_04_04_003_img-01.png\" alt=\"3 eighths circles\" width=\"202\" height=\"157\" \/><\/td>\r\n<td>[latex]\\Large\\frac{3}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add two [latex]\\Large\\frac{1}{8}[\/latex] pieces.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220959\/CNX_BMath_Figure_04_04_003_img-02.png\" alt=\"plus 2 eighths circles\" width=\"202\" height=\"121\" \/><\/td>\r\n<td>[latex]+\\Large\\frac{2}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>How many [latex]\\Large\\frac{1}{8}[\/latex] pieces are there?<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221002\/CNX_BMath_Figure_04_04_003_img-03.png\" alt=\"5 eighths circles\" width=\"202\" height=\"179\" \/><\/td>\r\n<td>[latex]\\Large\\frac{5}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThere are five [latex]\\Large\\frac{1}{8}[\/latex] pieces, or five-eighths. The model shows that [latex]\\Large\\frac{3}{8}\\normalsize+\\Large\\frac{2}{8}\\normalsize=\\Large\\frac{5}{8}[\/latex].\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse a model to find each sum. Show a diagram to illustrate your model.\r\n\r\n[latex]\\Large\\frac{1}{8}\\normalsize+\\Large\\frac{4}{8}[\/latex]\r\n[reveal-answer q=\"304582\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"304582\"]\r\n\r\n[latex]\\frac{5}{8}[\/latex]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221004\/CNX_BMath_Figure_04_04_004_img.png\" alt=\"A circle divided into 8 sections, 5 of which are shaded.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nUse a model to find each sum. Show a diagram to illustrate your model.\r\n[latex]\\Large\\frac{1}{6}\\normalsize+\\Large\\frac{4}{6}[\/latex]\r\n[reveal-answer q=\"297291\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"297291\"]\r\n\r\n[latex]\\frac{5}{6}[\/latex]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221004\/CNX_BMath_Figure_04_04_005_img.png\" alt=\"A circle divided into 6 sections, 5 of which are shaded.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"270\"]146178[\/ohm_question]\r\n\r\n<\/div>\r\nThe following video shows more examples of how to use models to add fractions with like denominators (the value in the lower part of a fraction that represents how many equal parts a whole has been divided into).\r\n\r\nhttps:\/\/youtu.be\/GTkY34kl6Kw\r\n\r\nSubtracting two fractions with common denominators follows the same process as adding fractions with common denominators. Think of a pizza that was cut into [latex]12[\/latex] slices. Suppose five pieces are eaten for dinner. This means that, after dinner, there are seven pieces (or [latex]{\\Large\\frac{7}{12}}[\/latex] of the pizza) left in the box. If Leonardo eats [latex]2[\/latex] of these remaining pieces (or [latex]{\\Large\\frac{2}{12}}[\/latex] of the pizza), how much is left? There would be [latex]5[\/latex] pieces left (or [latex]{\\Large\\frac{5}{12}}[\/latex] of the pizza).\r\n<p style=\"text-align: center;\">[latex]{\\Large\\frac{7}{12}}-{\\Large\\frac{2}{12}}={\\Large\\frac{5}{12}}[\/latex]<\/p>\r\nLet\u2019s use fraction circles to model the same example, [latex]{\\Large\\frac{7}{12}}-{\\Large\\frac{2}{12}}[\/latex].\r\n\r\nStart with seven [latex]{\\Large\\frac{1}{12}}[\/latex] pieces. Take away two [latex]{\\Large\\frac{1}{12}}[\/latex] pieces. How many twelfths are left?\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221006\/CNX_BMath_Figure_04_04_006_img.png\" alt=\"The bottom reads 7 twelfths minus 2 twelfths equals 5 twelfths. Above 7 twelfths, there is a circle divided into 12 equal pieces, with 7 pieces shaded in orange. Above 2 twelfths, the same circle is shown, but 2 of the 7 pieces are shaded in grey. Above 5 twelfths, the 2 grey pieces are no longer shaded, so there is a circle divided into 12 pieces with 5 of the pieces shaded in orange.\" data-media-type=\"image\/png\" \/>\r\nAgain, we have five twelfths, [latex]{\\Large\\frac{5}{12}}[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse fraction circles to find the difference: [latex]{\\Large\\frac{4}{5}}-{\\Large\\frac{1}{5}}[\/latex]\r\n\r\nSolution:\r\nStart with four [latex]{\\Large\\frac{1}{5}}[\/latex] pieces. Take away one [latex]{\\Large\\frac{1}{5}}[\/latex] piece. Count how many fifths are left. There are three [latex]{\\Large\\frac{1}{5}}[\/latex] pieces left, or [latex]{\\Large\\frac{3}{5}}[\/latex] of the circle left.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221008\/CNX_BMath_Figure_04_04_007_img.png\" alt=\"The bottom reads 4 fifths minus 1 fifth equals 3 fifths. Above 4 fifths, there is a circle divided into 5 equal pieces, with 4 pieces shaded in orange. Above 1 fifth, the same circle is shown, but 1 of the 4 shaded pieces is shaded in grey. Above 3 fifths, the 1 grey piece is no longer shaded, so there is a circle divided into 5 pieces with 3 of the pieces shaded in orange.\" data-media-type=\"image\/png\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question height=\"270\"]146190[\/ohm_question]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/7CeAQcpOJw0\r\n<p data-type=\"title\">The examples above show that to add or subtract the same-size pieces\u2014meaning that the fractions have the same denominator\u2014we just add or subtract the number of pieces.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Fraction Addition<\/h3>\r\nIf [latex]a,b,\\text{ and }c[\/latex] are numbers where [latex]c\\ne 0[\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\Large\\frac{a}{c}\\normalsize+\\Large\\frac{b}{c}\\normalsize=\\Large\\frac{a+b}{c}[\/latex]<\/p>\r\nTo add fractions with a common denominators, add the numerators and place the sum over the common denominator.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Fraction Subtraction<\/h3>\r\nIf [latex]a,b,\\text{ and }c[\/latex] are numbers where [latex]c\\ne 0[\/latex], then\r\n<p style=\"text-align: center;\">[latex]{\\Large\\frac{a}{c}}-{\\Large\\frac{b}{c}}={\\Large\\frac{a-b}{c}}[\/latex]<\/p>\r\nTo subtract fractions with a common denominators,\u00a0 subtract the numerators and place the difference over the common denominator.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the difference: [latex]{\\Large\\frac{23}{24}}-{\\Large\\frac{14}{24}}[\/latex]\r\n[reveal-answer q=\"842015\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"842015\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168469826203\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\".\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td data-align=\"left\">[latex]{\\Large\\frac{23}{24}}-{\\Large\\frac{14}{24}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td data-align=\"left\">Subtract the numerators and place the difference over the common denominator.<\/td>\r\n<td data-align=\"center\">[latex]{\\Large\\frac{23 - 14}{24}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td data-align=\"left\">Simplify the numerator.<\/td>\r\n<td data-align=\"center\">[latex]{\\Large\\frac{9}{24}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td data-align=\"left\">Simplify the fraction by removing common factors.<\/td>\r\n<td data-align=\"center\">[latex]{\\Large\\frac{3}{8}}[\/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=\"270\"]146191[\/ohm_question]\r\n\r\n<\/div>\r\nNow lets do an example that involves both addition and subtraction.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify: [latex]{\\Large\\frac{3}{8}}+\\left({\\Large\\frac{7}{8}}\\right)-{\\Large\\frac{5}{8}}[\/latex]\r\n[reveal-answer q=\"637721\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"637721\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168468417178\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\".\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\Large\\frac{3}{8}+\\left(\\frac{7}{8}\\right)-\\frac{5}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine the numerators over the common denominator.<\/td>\r\n<td>[latex]{\\Large\\frac{3+\\left(7\\right)-5}{8}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerator, working left to right.<\/td>\r\n<td>[latex]{\\Large\\frac{10 - 5}{8}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract the terms in the numerator.<\/td>\r\n<td data-align=\"center\">[latex]{\\Large\\frac{5}{8}}[\/latex]<\/td>\r\n<td data-align=\"center\"><\/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=\"270\"]146250[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Addition and Subtraction of Fractions with Different Denominators<\/h2>\r\n<p data-type=\"title\">We just reviewed how to add and subtract fractions with common denominators. But how can we add and subtract fractions with unlike denominators?<\/p>\r\n<p data-type=\"title\">Let\u2019s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that\u2019s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit\u2014cents. One quarter equals [latex]25[\/latex] cents and one dime equals [latex]10[\/latex] cents, so the sum is [latex]35[\/latex] cents. See the image below.<\/p>\r\nTogether, a quarter and a dime are worth [latex]35[\/latex] cents, or [latex]{\\Large\\frac{35}{100}}[\/latex] of a dollar.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221013\/CNX_BMath_Figure_04_05_002_img.png\" alt=\"A quarter and a dime are shown. Below them, it reads 25 cents plus 10 cents. Below that, it reads 35 cents.\" data-media-type=\"image\/png\" \/>\r\nSimilarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is [latex]100[\/latex]. Since there are [latex]100[\/latex] cents in one dollar, [latex]25[\/latex] cents is [latex]\\Large\\frac{25}{100}[\/latex] and [latex]10[\/latex] cents is [latex]\\Large\\frac{10}{100}[\/latex]. So we add [latex]\\Large\\frac{25}{100}+\\Large\\frac{10}{100}[\/latex] to get [latex]\\Large\\frac{35}{100}[\/latex], which is [latex]35[\/latex] cents.\r\n\r\nYou have practiced adding and subtracting fractions with common denominators. Now let\u2019s see what you need to do with fractions that have different denominators.\r\n\r\nFirst, we will use fraction tiles to model finding the common denominator of [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex].\r\n\r\nWe\u2019ll start with one [latex]\\Large\\frac{1}{2}[\/latex] tile and [latex]\\Large\\frac{1}{3}[\/latex] tile. We want to find a common fraction tile that we can use to match <em data-effect=\"italics\">both<\/em> [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex] exactly.\r\nIf we try the [latex]\\Large\\frac{1}{4}[\/latex] pieces, [latex]2[\/latex] of them exactly match the [latex]\\Large\\frac{1}{2}[\/latex] piece, but they do not exactly match the [latex]\\Large\\frac{1}{3}[\/latex] piece.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221015\/CNX_BMath_Figure_04_05_003_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into two pieces, each labeled 1 fourth. Underneath the second rectangle are two pieces, each labeled 1 fourth. These rectangles together are longer than the rectangle labeled as 1 third.\" data-media-type=\"image\/png\" \/>\r\nIf we try the [latex]\\Large\\frac{1}{5}[\/latex] pieces, they do not exactly cover the [latex]\\Large\\frac{1}{2}[\/latex] piece or the [latex]\\Large\\frac{1}{3}[\/latex] piece.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221016\/CNX_BMath_Figure_04_05_004_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into three pieces, each labeled 1 sixth. Underneath the second rectangle is an equally sized rectangle split vertically into 2 pieces, each labeled 1 sixth.\" data-media-type=\"image\/png\" \/>\r\nIf we try the [latex]\\Large\\frac{1}{6}[\/latex] pieces, we see that exactly [latex]3[\/latex] of them cover the [latex]\\Large\\frac{1}{2}[\/latex] piece, and exactly [latex]2[\/latex] of them cover the [latex]\\Large\\frac{1}{3}[\/latex] piece.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221018\/CNX_BMath_Figure_04_05_005_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle are three smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 half rectangle. Below the 1 third rectangle are two smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 third rectangle.\" data-media-type=\"image\/png\" \/>\r\nIf we were to try the [latex]\\Large\\frac{1}{12}[\/latex] pieces, they would also work.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221019\/CNX_BMath_Figure_04_05_006_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into 6 pieces, each labeled 1 twelfth. Underneath the second rectangle is an equally sized rectangle split vertically into 4 pieces, each labeled 1 twelfth.\" data-media-type=\"image\/png\" \/>\r\nEven smaller tiles, such as [latex]\\Large\\frac{1}{24}[\/latex] and [latex]\\Large\\frac{1}{48}[\/latex], would also exactly cover the [latex]\\Large\\frac{1}{2}[\/latex] piece and the [latex]\\Large\\frac{1}{3}[\/latex] piece.\r\n\r\nThe denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex] is [latex]6[\/latex].\r\n\r\nNotice that all of the tiles that cover [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex] have something in common: Their denominators are common multiples of [latex]2[\/latex] and [latex]3[\/latex], the denominators of [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex]. The least common multiple (LCM) of the denominators is [latex]6[\/latex], and so we say that [latex]6[\/latex] is the least common denominator (LCD) of the fractions [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex].\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\nTo find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the LCD for the fractions: [latex]\\Large\\frac{7}{12}[\/latex] and [latex]\\Large\\frac{5}{18}[\/latex]\r\n\r\nSolution:\r\n<table id=\"eip-id1168467165136\" class=\"unnumbered unstyled\" style=\"width: 95.196%; height: 340px;\" summary=\"The first line says, \" data-label=\"\">\r\n<tbody>\r\n<tr style=\"height: 159px;\">\r\n<td style=\"width: 42%; height: 159px;\">Factor each denominator into its primes.<\/td>\r\n<td style=\"width: 64.0183%; height: 159px;\"><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221021\/CNX_BMath_Figure_04_05_025_img-01.png\" alt=\"Two adjacent factoring trees. The first tree is for 12. 12 factors to 3 and 4. 4 factors to 2 and 2. The second tree is for 18. 18 factors to 3 and 6. 6 factors to 2 and 3.\" width=\"544\" height=\"188\" data-media-type=\"image\/png\" \/><\/td>\r\n<\/tr>\r\n<tr style=\"height: 49px;\">\r\n<td style=\"width: 42%; height: 49px;\">List the primes of [latex]12[\/latex] and the primes of [latex]18[\/latex] lining them up in columns when possible.<\/td>\r\n<td style=\"width: 64.0183%; height: 49px;\"><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221023\/CNX_BMath_Figure_04_05_025_img-02.png\" alt=\"The prime factorization of 12 written as 12 equals 2 times 2 times 3. Below this, the prime factorization of 18 is written as 18 equals 2 times 3 times 3. The two equations align vertically along the equals sign. The first 2 from the prime factorization of 12 and the 2 in the prime factorization of 18 aling vertically. The second 2 in the prime factorization of 12 does not align vertically with any factor from the prime factorization of 18. The 3 from the prime factorization of 12 aligns with the first 3 from the prime factorization of 18. The second 3 from the prime factorization of 18 does not align with any factor from the prime factorization of 12. Below the two equations, a horizontal line is drawn.\" width=\"544\" height=\"56\" data-media-type=\"image\/png\" \/><\/td>\r\n<\/tr>\r\n<tr style=\"height: 84px;\">\r\n<td style=\"width: 42%; height: 84px;\">Bring down the columns.<\/td>\r\n<td style=\"width: 64.0183%; height: 84px;\"><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221025\/CNX_BMath_Figure_04_05_025_img-03.png\" alt=\"Arrows are drawn through each column of factors from the prime factorizations of 12 and 18. Below the horizontal line is the equation LCM equals 2 times 2 times 3 times 3.\" width=\"527\" height=\"95\" data-media-type=\"image\/png\" \/><\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 42%; height: 12px;\">Multiply the factors. The product is the LCM.<\/td>\r\n<td style=\"width: 64.0183%; height: 12px;\">[latex]\\text{LCM}=36[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 36px;\">\r\n<td style=\"width: 42%; height: 36px;\">The LCM of [latex]12[\/latex] and [latex]18[\/latex] is [latex]36[\/latex], so the LCD of [latex]\\Large\\frac{7}{12}[\/latex] and [latex]\\Large\\frac{5}{18}[\/latex] is 36.<\/td>\r\n<td style=\"width: 64.0183%; height: 36px;\">LCD of [latex]\\Large\\frac{7}{12}[\/latex] and [latex]\\Large\\frac{5}{18}[\/latex] is 36.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question height=\"270\"]146252[\/ohm_question]\r\n\r\n<\/div>\r\nTo find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM.\r\n<div class=\"textbox shaded\">\r\n<h3>Find the least common denominator (LCD) of two fractions<\/h3>\r\n<ol id=\"eip-id1168469505824\" class=\"stepwise\" data-number-style=\"arabic\">\r\n \t<li>Factor each denominator into its primes.<\/li>\r\n \t<li>List the primes, matching primes in columns when possible.<\/li>\r\n \t<li>Bring down the columns.<\/li>\r\n \t<li>Multiply the factors. The product is the LCM of the denominators.<\/li>\r\n \t<li>The LCM of the denominators is the LCD of the fractions.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the least common denominator for the fractions: [latex]\\Large\\frac{8}{15}[\/latex] and [latex]\\Large\\frac{11}{24}[\/latex]\r\n[reveal-answer q=\"954069\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"954069\"]\r\n\r\nSolution:\r\nTo find the LCD, we find the LCM of the denominators.\r\nFind the LCM of [latex]15[\/latex] and [latex]24[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221026\/CNX_BMath_Figure_04_05_016_img-01.png\" alt=\"The top line shows 15 equals 3 times 5. The next line shows 24 equals 2 times 2 times 2 times 3. The 3s are lined up vertically. The next line shows LCM equals 2 times 2 times 2 times 3 times 5. The last line shows LCM equals 120.\" data-media-type=\"image\/png\" \/>\r\nThe LCM of [latex]15[\/latex] and [latex]24[\/latex] is [latex]120[\/latex]. So, the LCD of [latex]\\Large\\frac{8}{15}[\/latex] and [latex]\\Large\\frac{11}{24}[\/latex] is [latex]120[\/latex].\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=\"270\"]146251[\/ohm_question]\r\n\r\n<\/div>\r\n<p data-type=\"title\">Earlier, we used fraction tiles to see that the LCD of [latex]\\Large\\frac{1}{4}\\normalsize\\text{and}\\Large\\frac{1}{6}[\/latex] is [latex]12[\/latex]. We saw that three [latex]\\Large\\frac{1}{12}[\/latex] pieces exactly covered [latex]\\Large\\frac{1}{4}[\/latex] and two [latex]\\Large\\frac{1}{12}[\/latex] pieces exactly covered [latex]\\Large\\frac{1}{6}[\/latex], so<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large\\frac{1}{4}=\\Large\\frac{3}{12}\\normalsize\\text{ and }\\Large\\frac{1}{6}=\\Large\\frac{2}{12}[\/latex].<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221027\/CNX_BMath_Figure_04_05_026_img.png\" alt=\"On the left is a rectangle labeled 1 fourth. Below it is an identical rectangle split vertically into 3 equal pieces, each labeled 1 twelfth. On the right is a rectangle labeled 1 sixth. Below it is an identical rectangle split vertically into 2 equal pieces, each labeled 1 twelfth.\" data-media-type=\"image\/png\" \/>\r\nWe say that [latex]\\Large\\frac{1}{4}\\normalsize\\text{ and }\\Large\\frac{3}{12}[\/latex] are equivalent fractions and also that [latex]\\Large\\frac{1}{6}\\normalsize\\text{ and }\\Large\\frac{2}{12}[\/latex] are equivalent fractions.\r\n\r\nWe 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 Equivalent Fractions Property is repeated below for reference.\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\nTo add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let\u2019s see how to change [latex]\\Large\\frac{1}{4}\\normalsize\\text{ and }\\Large\\frac{1}{6}[\/latex] to equivalent fractions with denominator [latex]12[\/latex] without using models.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert [latex]\\Large\\frac{1}{4}\\normalsize\\text{ and }\\Large\\frac{1}{6}[\/latex] to equivalent fractions with denominator [latex]12[\/latex], their LCD.\r\n\r\nSolution:\r\n<table id=\"eip-id1168467209573\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says, \" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td>Find the LCD.<\/td>\r\n<td>The LCD of [latex]\\Large\\frac{1}{4}[\/latex] and [latex]\\Large\\frac{1}{6}[\/latex] is [latex]12[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the number to multiply [latex]4[\/latex] to get [latex]12[\/latex].<\/td>\r\n<td>[latex]4\\cdot\\color{red}{3}=12[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the number to multiply [latex]6[\/latex] to get [latex]12[\/latex].<\/td>\r\n<td>[latex]6\\cdot\\color{red}{2}=12[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the Equivalent Fractions Property to convert each fraction to an equivalent fraction with the LCD, multiplying both the numerator and denominator of each fraction by the same number.<\/td>\r\n<td>[latex]\r\n\r\n\\Large\\frac{1}{4}[\/latex] \u00a0 \u00a0 \u00a0[latex]\r\n\r\n\\Large\\frac{1}{6}[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{1\\cdot\\color{red}{3}}{4\\cdot\\color{red}{3}}[\/latex] \u00a0 \u00a0 \u00a0[latex]\r\n\r\n\\Large\\frac{1\\cdot\\color{red}{2}}{6\\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}{12}[\/latex] \u00a0 [latex]\\Large\\frac{2}{12}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nWe do not reduce the resulting fractions. If we did, we would get back to our original fractions and lose the common denominator.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question height=\"270\"]146254[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Convert two fractions to equivalent fractions with their LCD as the common denominator<\/h3>\r\n<ol id=\"eip-id1168468227692\" class=\"stepwise\" data-number-style=\"arabic\">\r\n \t<li>Find the LCD.<\/li>\r\n \t<li>For each fraction, determine the number needed to multiply the denominator to get the LCD.<\/li>\r\n \t<li>Use the Equivalent Fractions Property to multiply both the numerator and denominator by the number you found in Step 2.<\/li>\r\n \t<li>Simplify the numerator and denominator.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConvert [latex]\\Large\\frac{8}{15}[\/latex] and [latex]\\Large\\frac{11}{24}[\/latex] to equivalent fractions with denominator [latex]120[\/latex], their LCD.\r\n[reveal-answer q=\"831064\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"831064\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168466215186\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says, \" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td>The LCD is [latex]120[\/latex]. We will start at Step 2.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the number that must multiply [latex]15[\/latex] to get [latex]120[\/latex].<\/td>\r\n<td>[latex]15\\cdot\\color{red}{8}=120[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Find the number that must multiply [latex]24[\/latex] to get [latex]120[\/latex].<\/td>\r\n<td>[latex]24\\cdot\\color{red}{5}=120[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the Equivalent Fractions Property.<\/td>\r\n<td>[latex]\\Large\\frac{8\\cdot\\color{red}{8}}{15\\cdot\\color{red}{8}}[\/latex] \u00a0 \u00a0 \u00a0 \u00a0[latex]\\Large\\frac{11\\cdot\\color{red}{5}}{24\\cdot\\color{red}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerators and denominators.<\/td>\r\n<td>[latex]\\Large\\frac{64}{120}[\/latex] \u00a0 \u00a0 \u00a0 \u00a0[latex]\\Large\\frac{55}{120}[\/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=\"270\"]146255[\/ohm_question]\r\n\r\n<\/div>\r\nIn our next video we show two more examples of how to use the column method to find the least common denominator of two fractions.\r\n\r\nhttps:\/\/youtu.be\/JsHF9CW_SUM\r\n<p data-type=\"title\">Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators.<\/p>\r\n\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\" data-number-style=\"arabic\">\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, \" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td data-align=\"center\">[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=\"alignnone\" 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=\"The prime factorization of 2 written as 2 equals 2. Below this is the prime factorization of 3 written as 3 equals 3. The two equations align vertically along the equals sign. Both the 2 and the 3 from the prime factorizations do not align vertically with each other. Below the equations is a horizontal line. Below this line is the equation LCD equals 2 times 3, which simplifies to LCD equals 6.\" width=\"81\" height=\"110\" data-media-type=\"image\/png\" \/><\/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 data-align=\"center\">[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 data-align=\"center\">[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\nAdd: [latex]\\Large\\frac{7}{12}+\\Large\\frac{5}{18}[\/latex]\r\n[reveal-answer q=\"826911\"]Show Answer[\/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, \" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td data-align=\"center\">[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=\"alignnone\" 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=\"The prime factorization of 12 written as 12 equals 2 times 2 times 3. Below this, the prime factorization of 18 is written as 18 equals 2 times 3 times 3. The two equations align vertically along the equals sign. The first 2 from the prime factorization of 12 and the 2 in the prime factorization of 18 aling vertically. The second 2 in the prime factorization of 12 does not align vertically with any factor from the prime factorization of 18. The 3 from the prime factorization of 12 aligns with the first 3 from the prime factorization of 18. The second 3 from the prime factorization of 18 does not align with any factor from the prime factorization of 12. Below the two equations, a horizontal line is drawn. Below this line is the equation LCD equals 2 times 2 times 3 times 3, which simplifies to LCD equals 36.\" width=\"146\" height=\"75\" data-media-type=\"image\/png\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\r\n<td data-align=\"center\">[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 data-align=\"center\">[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 data-align=\"center\">[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\"]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 key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question height=\"230\"]146265[\/ohm_question]\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>Use addition and subtraction when evaluating expressions with fractions<\/li>\n<\/ul>\n<\/div>\n<h2>Addition and Subtraction of Fractions with Common Denominators<\/h2>\n<p>How many quarters are pictured below?<\/p>\n<p><img decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220949\/CNX_BMath_Figure_04_04_001.png\" alt=\"Three U.S. quarters are shown. One is shown on the left, and two are shown on the right.\" \/><\/p>\n<p>You can quickly count three, but if you look carefully the image represents [latex]1[\/latex] quarter plus [latex]2[\/latex] quarters equals [latex]3[\/latex] quarters.<\/p>\n<p>Remember, quarters are really fractions of a dollar. Quarters are another way to say fourths. So the picture of the coins shows that<\/p>\n<p style=\"text-align: center;\">[latex]{\\Large\\frac{1}{4}}+{\\Large\\frac{2}{4}}=\\Large{\\frac{3}{4}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\text{one quarter }+\\text{ two quarters }=\\text{ three quarters}[\/latex]<\/p>\n<p>Let\u2019s use fraction circles to model the same example, [latex]\\Large\\frac{1}{4}\\normalsize+\\Large\\frac{2}{4}[\/latex].<\/p>\n<table id=\"eip-id1168467352669\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says,\">\n<tbody>\n<tr>\n<td>Start with one [latex]\\Large\\frac{1}{4}[\/latex] piece.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220951\/CNX_BMath_Figure_04_04_002_img-01.png\" alt=\"one quarter circle\" width=\"205\" height=\"93\" \/><\/td>\n<td>[latex]\\Large\\frac{1}{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add two more [latex]\\Large\\frac{1}{4}[\/latex] pieces.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220953\/CNX_BMath_Figure_04_04_002_img-02.png\" alt=\"plus 2 quarter circles\" width=\"205\" height=\"116\" \/><\/td>\n<td>[latex]+\\Large\\frac{2}{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The result is [latex]\\Large\\frac{3}{4}[\/latex] .<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220954\/CNX_BMath_Figure_04_04_002_img-03.png\" alt=\"3 quarter circles\" width=\"205\" height=\"185\" \/><\/td>\n<td>[latex]\\Large\\frac{3}{4}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So again, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\Large\\frac{1}{4}\\normalsize+\\Large\\frac{2}{4}\\normalsize=\\Large\\frac{3}{4}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Use a model to find the sum [latex]\\Large\\frac{3}{8}\\normalsize+\\Large\\frac{2}{8}[\/latex].<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168468316426\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says,\">\n<tbody>\n<tr>\n<td>Start with three [latex]\\Large\\frac{1}{8}[\/latex] pieces.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220957\/CNX_BMath_Figure_04_04_003_img-01.png\" alt=\"3 eighths circles\" width=\"202\" height=\"157\" \/><\/td>\n<td>[latex]\\Large\\frac{3}{8}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add two [latex]\\Large\\frac{1}{8}[\/latex] pieces.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220959\/CNX_BMath_Figure_04_04_003_img-02.png\" alt=\"plus 2 eighths circles\" width=\"202\" height=\"121\" \/><\/td>\n<td>[latex]+\\Large\\frac{2}{8}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>How many [latex]\\Large\\frac{1}{8}[\/latex] pieces are there?<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221002\/CNX_BMath_Figure_04_04_003_img-03.png\" alt=\"5 eighths circles\" width=\"202\" height=\"179\" \/><\/td>\n<td>[latex]\\Large\\frac{5}{8}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>There are five [latex]\\Large\\frac{1}{8}[\/latex] pieces, or five-eighths. The model shows that [latex]\\Large\\frac{3}{8}\\normalsize+\\Large\\frac{2}{8}\\normalsize=\\Large\\frac{5}{8}[\/latex].<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use a model to find each sum. Show a diagram to illustrate your model.<\/p>\n<p>[latex]\\Large\\frac{1}{8}\\normalsize+\\Large\\frac{4}{8}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q304582\">Show Answer<\/span><\/p>\n<div id=\"q304582\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{5}{8}[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221004\/CNX_BMath_Figure_04_04_004_img.png\" alt=\"A circle divided into 8 sections, 5 of which are shaded.\" \/><\/p>\n<\/div>\n<\/div>\n<p>Use a model to find each sum. Show a diagram to illustrate your model.<br \/>\n[latex]\\Large\\frac{1}{6}\\normalsize+\\Large\\frac{4}{6}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q297291\">Show Answer<\/span><\/p>\n<div id=\"q297291\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{5}{6}[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221004\/CNX_BMath_Figure_04_04_005_img.png\" alt=\"A circle divided into 6 sections, 5 of which are shaded.\" \/><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm146178\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146178&theme=oea&iframe_resize_id=ohm146178&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<p>The following video shows more examples of how to use models to add fractions with like denominators (the value in the lower part of a fraction that represents how many equal parts a whole has been divided into).<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Add Fractions with Like Denominators\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GTkY34kl6Kw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Subtracting two fractions with common denominators follows the same process as adding fractions with common denominators. Think of a pizza that was cut into [latex]12[\/latex] slices. Suppose five pieces are eaten for dinner. This means that, after dinner, there are seven pieces (or [latex]{\\Large\\frac{7}{12}}[\/latex] of the pizza) left in the box. If Leonardo eats [latex]2[\/latex] of these remaining pieces (or [latex]{\\Large\\frac{2}{12}}[\/latex] of the pizza), how much is left? There would be [latex]5[\/latex] pieces left (or [latex]{\\Large\\frac{5}{12}}[\/latex] of the pizza).<\/p>\n<p style=\"text-align: center;\">[latex]{\\Large\\frac{7}{12}}-{\\Large\\frac{2}{12}}={\\Large\\frac{5}{12}}[\/latex]<\/p>\n<p>Let\u2019s use fraction circles to model the same example, [latex]{\\Large\\frac{7}{12}}-{\\Large\\frac{2}{12}}[\/latex].<\/p>\n<p>Start with seven [latex]{\\Large\\frac{1}{12}}[\/latex] pieces. Take away two [latex]{\\Large\\frac{1}{12}}[\/latex] pieces. How many twelfths are left?<\/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\/24221006\/CNX_BMath_Figure_04_04_006_img.png\" alt=\"The bottom reads 7 twelfths minus 2 twelfths equals 5 twelfths. Above 7 twelfths, there is a circle divided into 12 equal pieces, with 7 pieces shaded in orange. Above 2 twelfths, the same circle is shown, but 2 of the 7 pieces are shaded in grey. Above 5 twelfths, the 2 grey pieces are no longer shaded, so there is a circle divided into 12 pieces with 5 of the pieces shaded in orange.\" data-media-type=\"image\/png\" \/><br \/>\nAgain, we have five twelfths, [latex]{\\Large\\frac{5}{12}}[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use fraction circles to find the difference: [latex]{\\Large\\frac{4}{5}}-{\\Large\\frac{1}{5}}[\/latex]<\/p>\n<p>Solution:<br \/>\nStart with four [latex]{\\Large\\frac{1}{5}}[\/latex] pieces. Take away one [latex]{\\Large\\frac{1}{5}}[\/latex] piece. Count how many fifths are left. There are three [latex]{\\Large\\frac{1}{5}}[\/latex] pieces left, or [latex]{\\Large\\frac{3}{5}}[\/latex] of the circle left.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221008\/CNX_BMath_Figure_04_04_007_img.png\" alt=\"The bottom reads 4 fifths minus 1 fifth equals 3 fifths. Above 4 fifths, there is a circle divided into 5 equal pieces, with 4 pieces shaded in orange. Above 1 fifth, the same circle is shown, but 1 of the 4 shaded pieces is shaded in grey. Above 3 fifths, the 1 grey piece is no longer shaded, so there is a circle divided into 5 pieces with 3 of the pieces shaded in orange.\" data-media-type=\"image\/png\" \/><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146190\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146190&theme=oea&iframe_resize_id=ohm146190&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Subtract Fractions with Like Denominators\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/7CeAQcpOJw0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p data-type=\"title\">The examples above show that to add or subtract the same-size pieces\u2014meaning that the fractions have the same denominator\u2014we just add or subtract the number of pieces.<\/p>\n<div class=\"textbox shaded\">\n<h3>Fraction Addition<\/h3>\n<p>If [latex]a,b,\\text{ and }c[\/latex] are numbers where [latex]c\\ne 0[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\Large\\frac{a}{c}\\normalsize+\\Large\\frac{b}{c}\\normalsize=\\Large\\frac{a+b}{c}[\/latex]<\/p>\n<p>To add fractions with a common denominators, add the numerators and place the sum over the common denominator.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Fraction Subtraction<\/h3>\n<p>If [latex]a,b,\\text{ and }c[\/latex] are numbers where [latex]c\\ne 0[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]{\\Large\\frac{a}{c}}-{\\Large\\frac{b}{c}}={\\Large\\frac{a-b}{c}}[\/latex]<\/p>\n<p>To subtract fractions with a common denominators,\u00a0 subtract the numerators and place the difference over the common denominator.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the difference: [latex]{\\Large\\frac{23}{24}}-{\\Large\\frac{14}{24}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q842015\">Show Answer<\/span><\/p>\n<div id=\"q842015\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168469826203\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\".\" data-label=\"\">\n<tbody>\n<tr>\n<td data-align=\"left\">[latex]{\\Large\\frac{23}{24}}-{\\Large\\frac{14}{24}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td data-align=\"left\">Subtract the numerators and place the difference over the common denominator.<\/td>\n<td data-align=\"center\">[latex]{\\Large\\frac{23 - 14}{24}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td data-align=\"left\">Simplify the numerator.<\/td>\n<td data-align=\"center\">[latex]{\\Large\\frac{9}{24}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td data-align=\"left\">Simplify the fraction by removing common factors.<\/td>\n<td data-align=\"center\">[latex]{\\Large\\frac{3}{8}}[\/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=\"ohm146191\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146191&theme=oea&iframe_resize_id=ohm146191&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<p>Now lets do an example that involves both addition and subtraction.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify: [latex]{\\Large\\frac{3}{8}}+\\left({\\Large\\frac{7}{8}}\\right)-{\\Large\\frac{5}{8}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q637721\">Show Answer<\/span><\/p>\n<div id=\"q637721\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168468417178\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\".\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\Large\\frac{3}{8}+\\left(\\frac{7}{8}\\right)-\\frac{5}{8}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine the numerators over the common denominator.<\/td>\n<td>[latex]{\\Large\\frac{3+\\left(7\\right)-5}{8}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerator, working left to right.<\/td>\n<td>[latex]{\\Large\\frac{10 - 5}{8}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract the terms in the numerator.<\/td>\n<td data-align=\"center\">[latex]{\\Large\\frac{5}{8}}[\/latex]<\/td>\n<td data-align=\"center\"><\/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=\"ohm146250\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146250&theme=oea&iframe_resize_id=ohm146250&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<h2>Addition and Subtraction of Fractions with Different Denominators<\/h2>\n<p data-type=\"title\">We just reviewed how to add and subtract fractions with common denominators. But how can we add and subtract fractions with unlike denominators?<\/p>\n<p data-type=\"title\">Let\u2019s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that\u2019s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit\u2014cents. One quarter equals [latex]25[\/latex] cents and one dime equals [latex]10[\/latex] cents, so the sum is [latex]35[\/latex] cents. See the image below.<\/p>\n<p>Together, a quarter and a dime are worth [latex]35[\/latex] cents, or [latex]{\\Large\\frac{35}{100}}[\/latex] of a dollar.<\/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\/24221013\/CNX_BMath_Figure_04_05_002_img.png\" alt=\"A quarter and a dime are shown. Below them, it reads 25 cents plus 10 cents. Below that, it reads 35 cents.\" data-media-type=\"image\/png\" \/><br \/>\nSimilarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is [latex]100[\/latex]. Since there are [latex]100[\/latex] cents in one dollar, [latex]25[\/latex] cents is [latex]\\Large\\frac{25}{100}[\/latex] and [latex]10[\/latex] cents is [latex]\\Large\\frac{10}{100}[\/latex]. So we add [latex]\\Large\\frac{25}{100}+\\Large\\frac{10}{100}[\/latex] to get [latex]\\Large\\frac{35}{100}[\/latex], which is [latex]35[\/latex] cents.<\/p>\n<p>You have practiced adding and subtracting fractions with common denominators. Now let\u2019s see what you need to do with fractions that have different denominators.<\/p>\n<p>First, we will use fraction tiles to model finding the common denominator of [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex].<\/p>\n<p>We\u2019ll start with one [latex]\\Large\\frac{1}{2}[\/latex] tile and [latex]\\Large\\frac{1}{3}[\/latex] tile. We want to find a common fraction tile that we can use to match <em data-effect=\"italics\">both<\/em> [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex] exactly.<br \/>\nIf we try the [latex]\\Large\\frac{1}{4}[\/latex] pieces, [latex]2[\/latex] of them exactly match the [latex]\\Large\\frac{1}{2}[\/latex] piece, but they do not exactly match the [latex]\\Large\\frac{1}{3}[\/latex] piece.<\/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\/24221015\/CNX_BMath_Figure_04_05_003_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into two pieces, each labeled 1 fourth. Underneath the second rectangle are two pieces, each labeled 1 fourth. These rectangles together are longer than the rectangle labeled as 1 third.\" data-media-type=\"image\/png\" \/><br \/>\nIf we try the [latex]\\Large\\frac{1}{5}[\/latex] pieces, they do not exactly cover the [latex]\\Large\\frac{1}{2}[\/latex] piece or the [latex]\\Large\\frac{1}{3}[\/latex] piece.<\/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\/24221016\/CNX_BMath_Figure_04_05_004_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into three pieces, each labeled 1 sixth. Underneath the second rectangle is an equally sized rectangle split vertically into 2 pieces, each labeled 1 sixth.\" data-media-type=\"image\/png\" \/><br \/>\nIf we try the [latex]\\Large\\frac{1}{6}[\/latex] pieces, we see that exactly [latex]3[\/latex] of them cover the [latex]\\Large\\frac{1}{2}[\/latex] piece, and exactly [latex]2[\/latex] of them cover the [latex]\\Large\\frac{1}{3}[\/latex] piece.<\/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\/24221018\/CNX_BMath_Figure_04_05_005_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle are three smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 half rectangle. Below the 1 third rectangle are two smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 third rectangle.\" data-media-type=\"image\/png\" \/><br \/>\nIf we were to try the [latex]\\Large\\frac{1}{12}[\/latex] pieces, they would also work.<\/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\/24221019\/CNX_BMath_Figure_04_05_006_img.png\" alt=\"Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into 6 pieces, each labeled 1 twelfth. Underneath the second rectangle is an equally sized rectangle split vertically into 4 pieces, each labeled 1 twelfth.\" data-media-type=\"image\/png\" \/><br \/>\nEven smaller tiles, such as [latex]\\Large\\frac{1}{24}[\/latex] and [latex]\\Large\\frac{1}{48}[\/latex], would also exactly cover the [latex]\\Large\\frac{1}{2}[\/latex] piece and the [latex]\\Large\\frac{1}{3}[\/latex] piece.<\/p>\n<p>The denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex] is [latex]6[\/latex].<\/p>\n<p>Notice that all of the tiles that cover [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex] have something in common: Their denominators are common multiples of [latex]2[\/latex] and [latex]3[\/latex], the denominators of [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex]. The least common multiple (LCM) of the denominators is [latex]6[\/latex], and so we say that [latex]6[\/latex] is the least common denominator (LCD) of the fractions [latex]\\Large\\frac{1}{2}[\/latex] and [latex]\\Large\\frac{1}{3}[\/latex].<\/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>To find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the LCD for the fractions: [latex]\\Large\\frac{7}{12}[\/latex] and [latex]\\Large\\frac{5}{18}[\/latex]<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168467165136\" class=\"unnumbered unstyled\" style=\"width: 95.196%; height: 340px;\" summary=\"The first line says,\" data-label=\"\">\n<tbody>\n<tr style=\"height: 159px;\">\n<td style=\"width: 42%; height: 159px;\">Factor each denominator into its primes.<\/td>\n<td style=\"width: 64.0183%; height: 159px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221021\/CNX_BMath_Figure_04_05_025_img-01.png\" alt=\"Two adjacent factoring trees. The first tree is for 12. 12 factors to 3 and 4. 4 factors to 2 and 2. The second tree is for 18. 18 factors to 3 and 6. 6 factors to 2 and 3.\" width=\"544\" height=\"188\" data-media-type=\"image\/png\" \/><\/td>\n<\/tr>\n<tr style=\"height: 49px;\">\n<td style=\"width: 42%; height: 49px;\">List the primes of [latex]12[\/latex] and the primes of [latex]18[\/latex] lining them up in columns when possible.<\/td>\n<td style=\"width: 64.0183%; height: 49px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221023\/CNX_BMath_Figure_04_05_025_img-02.png\" alt=\"The prime factorization of 12 written as 12 equals 2 times 2 times 3. Below this, the prime factorization of 18 is written as 18 equals 2 times 3 times 3. The two equations align vertically along the equals sign. The first 2 from the prime factorization of 12 and the 2 in the prime factorization of 18 aling vertically. The second 2 in the prime factorization of 12 does not align vertically with any factor from the prime factorization of 18. The 3 from the prime factorization of 12 aligns with the first 3 from the prime factorization of 18. The second 3 from the prime factorization of 18 does not align with any factor from the prime factorization of 12. Below the two equations, a horizontal line is drawn.\" width=\"544\" height=\"56\" data-media-type=\"image\/png\" \/><\/td>\n<\/tr>\n<tr style=\"height: 84px;\">\n<td style=\"width: 42%; height: 84px;\">Bring down the columns.<\/td>\n<td style=\"width: 64.0183%; height: 84px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221025\/CNX_BMath_Figure_04_05_025_img-03.png\" alt=\"Arrows are drawn through each column of factors from the prime factorizations of 12 and 18. Below the horizontal line is the equation LCM equals 2 times 2 times 3 times 3.\" width=\"527\" height=\"95\" data-media-type=\"image\/png\" \/><\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 42%; height: 12px;\">Multiply the factors. The product is the LCM.<\/td>\n<td style=\"width: 64.0183%; height: 12px;\">[latex]\\text{LCM}=36[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 36px;\">\n<td style=\"width: 42%; height: 36px;\">The LCM of [latex]12[\/latex] and [latex]18[\/latex] is [latex]36[\/latex], so the LCD of [latex]\\Large\\frac{7}{12}[\/latex] and [latex]\\Large\\frac{5}{18}[\/latex] is 36.<\/td>\n<td style=\"width: 64.0183%; height: 36px;\">LCD of [latex]\\Large\\frac{7}{12}[\/latex] and [latex]\\Large\\frac{5}{18}[\/latex] is 36.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146252\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146252&theme=oea&iframe_resize_id=ohm146252&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<p>To find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM.<\/p>\n<div class=\"textbox shaded\">\n<h3>Find the least common denominator (LCD) of two fractions<\/h3>\n<ol id=\"eip-id1168469505824\" class=\"stepwise\" data-number-style=\"arabic\">\n<li>Factor each denominator into its primes.<\/li>\n<li>List the primes, matching primes in columns when possible.<\/li>\n<li>Bring down the columns.<\/li>\n<li>Multiply the factors. The product is the LCM of the denominators.<\/li>\n<li>The LCM of the denominators is the LCD of the fractions.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the least common denominator for the fractions: [latex]\\Large\\frac{8}{15}[\/latex] and [latex]\\Large\\frac{11}{24}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q954069\">Show Answer<\/span><\/p>\n<div id=\"q954069\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nTo find the LCD, we find the LCM of the denominators.<br \/>\nFind the LCM of [latex]15[\/latex] and [latex]24[\/latex].<\/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\/24221026\/CNX_BMath_Figure_04_05_016_img-01.png\" alt=\"The top line shows 15 equals 3 times 5. The next line shows 24 equals 2 times 2 times 2 times 3. The 3s are lined up vertically. The next line shows LCM equals 2 times 2 times 2 times 3 times 5. The last line shows LCM equals 120.\" data-media-type=\"image\/png\" \/><br \/>\nThe LCM of [latex]15[\/latex] and [latex]24[\/latex] is [latex]120[\/latex]. So, the LCD of [latex]\\Large\\frac{8}{15}[\/latex] and [latex]\\Large\\frac{11}{24}[\/latex] is [latex]120[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146251\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146251&theme=oea&iframe_resize_id=ohm146251&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<p data-type=\"title\">Earlier, we used fraction tiles to see that the LCD of [latex]\\Large\\frac{1}{4}\\normalsize\\text{and}\\Large\\frac{1}{6}[\/latex] is [latex]12[\/latex]. We saw that three [latex]\\Large\\frac{1}{12}[\/latex] pieces exactly covered [latex]\\Large\\frac{1}{4}[\/latex] and two [latex]\\Large\\frac{1}{12}[\/latex] pieces exactly covered [latex]\\Large\\frac{1}{6}[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]\\Large\\frac{1}{4}=\\Large\\frac{3}{12}\\normalsize\\text{ and }\\Large\\frac{1}{6}=\\Large\\frac{2}{12}[\/latex].<\/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\/24221027\/CNX_BMath_Figure_04_05_026_img.png\" alt=\"On the left is a rectangle labeled 1 fourth. Below it is an identical rectangle split vertically into 3 equal pieces, each labeled 1 twelfth. On the right is a rectangle labeled 1 sixth. Below it is an identical rectangle split vertically into 2 equal pieces, each labeled 1 twelfth.\" data-media-type=\"image\/png\" \/><br \/>\nWe say that [latex]\\Large\\frac{1}{4}\\normalsize\\text{ and }\\Large\\frac{3}{12}[\/latex] are equivalent fractions and also that [latex]\\Large\\frac{1}{6}\\normalsize\\text{ and }\\Large\\frac{2}{12}[\/latex] are equivalent fractions.<\/p>\n<p>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 Equivalent Fractions Property is repeated below for reference.<\/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>To add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let\u2019s see how to change [latex]\\Large\\frac{1}{4}\\normalsize\\text{ and }\\Large\\frac{1}{6}[\/latex] to equivalent fractions with denominator [latex]12[\/latex] without using models.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert [latex]\\Large\\frac{1}{4}\\normalsize\\text{ and }\\Large\\frac{1}{6}[\/latex] to equivalent fractions with denominator [latex]12[\/latex], their LCD.<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168467209573\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says,\" data-label=\"\">\n<tbody>\n<tr>\n<td>Find the LCD.<\/td>\n<td>The LCD of [latex]\\Large\\frac{1}{4}[\/latex] and [latex]\\Large\\frac{1}{6}[\/latex] is [latex]12[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>Find the number to multiply [latex]4[\/latex] to get [latex]12[\/latex].<\/td>\n<td>[latex]4\\cdot\\color{red}{3}=12[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the number to multiply [latex]6[\/latex] to get [latex]12[\/latex].<\/td>\n<td>[latex]6\\cdot\\color{red}{2}=12[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the Equivalent Fractions Property to convert each fraction to an equivalent fraction with the LCD, multiplying both the numerator and denominator of each fraction by the same number.<\/td>\n<td>[latex]\\Large\\frac{1}{4}[\/latex] \u00a0 \u00a0 \u00a0[latex]\\Large\\frac{1}{6}[\/latex]<\/p>\n<p>[latex]\\Large\\frac{1\\cdot\\color{red}{3}}{4\\cdot\\color{red}{3}}[\/latex] \u00a0 \u00a0 \u00a0[latex]\\Large\\frac{1\\cdot\\color{red}{2}}{6\\cdot\\color{red}{2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerators and denominators.<\/td>\n<td>[latex]\\Large\\frac{3}{12}[\/latex] \u00a0 [latex]\\Large\\frac{2}{12}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>We do not reduce the resulting fractions. If we did, we would get back to our original fractions and lose the common denominator.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146254\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146254&theme=oea&iframe_resize_id=ohm146254&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Convert two fractions to equivalent fractions with their LCD as the common denominator<\/h3>\n<ol id=\"eip-id1168468227692\" class=\"stepwise\" data-number-style=\"arabic\">\n<li>Find the LCD.<\/li>\n<li>For each fraction, determine the number needed to multiply the denominator to get the LCD.<\/li>\n<li>Use the Equivalent Fractions Property to multiply both the numerator and denominator by the number you found in Step 2.<\/li>\n<li>Simplify the numerator and denominator.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Convert [latex]\\Large\\frac{8}{15}[\/latex] and [latex]\\Large\\frac{11}{24}[\/latex] to equivalent fractions with denominator [latex]120[\/latex], their LCD.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q831064\">Show Answer<\/span><\/p>\n<div id=\"q831064\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168466215186\" class=\"unnumbered unstyled\" style=\"width: 85%;\" summary=\"The first line says,\" data-label=\"\">\n<tbody>\n<tr>\n<td>The LCD is [latex]120[\/latex]. We will start at Step 2.<\/td>\n<\/tr>\n<tr>\n<td>Find the number that must multiply [latex]15[\/latex] to get [latex]120[\/latex].<\/td>\n<td>[latex]15\\cdot\\color{red}{8}=120[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Find the number that must multiply [latex]24[\/latex] to get [latex]120[\/latex].<\/td>\n<td>[latex]24\\cdot\\color{red}{5}=120[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the Equivalent Fractions Property.<\/td>\n<td>[latex]\\Large\\frac{8\\cdot\\color{red}{8}}{15\\cdot\\color{red}{8}}[\/latex] \u00a0 \u00a0 \u00a0 \u00a0[latex]\\Large\\frac{11\\cdot\\color{red}{5}}{24\\cdot\\color{red}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerators and denominators.<\/td>\n<td>[latex]\\Large\\frac{64}{120}[\/latex] \u00a0 \u00a0 \u00a0 \u00a0[latex]\\Large\\frac{55}{120}[\/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=\"ohm146255\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146255&theme=oea&iframe_resize_id=ohm146255&show_question_numbers\" width=\"100%\" height=\"270\"><\/iframe><\/p>\n<\/div>\n<p>In our next video we show two more examples of how to use the column method to find the least common denominator of two fractions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Determine the Least Common Denominator of Two Fractions (Column Method)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/JsHF9CW_SUM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p data-type=\"title\">Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators.<\/p>\n<div class=\"textbox shaded\">\n<h3>Add or subtract fractions with different denominators<\/h3>\n<ol id=\"eip-id1168468303196\" class=\"stepwise\" data-number-style=\"arabic\">\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,\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td data-align=\"center\">[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=\"alignnone\" 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=\"The prime factorization of 2 written as 2 equals 2. Below this is the prime factorization of 3 written as 3 equals 3. The two equations align vertically along the equals sign. Both the 2 and the 3 from the prime factorizations do not align vertically with each other. Below the equations is a horizontal line. Below this line is the equation LCD equals 2 times 3, which simplifies to LCD equals 6.\" width=\"81\" height=\"110\" data-media-type=\"image\/png\" \/><\/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 data-align=\"center\">[latex]\\Large\\frac{3}{6}+\\Large\\frac{2}{6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td data-align=\"center\">[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-4\" 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>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 Answer<\/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,\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td data-align=\"center\">[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=\"alignnone\" 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=\"The prime factorization of 12 written as 12 equals 2 times 2 times 3. Below this, the prime factorization of 18 is written as 18 equals 2 times 3 times 3. The two equations align vertically along the equals sign. The first 2 from the prime factorization of 12 and the 2 in the prime factorization of 18 aling vertically. The second 2 in the prime factorization of 12 does not align vertically with any factor from the prime factorization of 18. The 3 from the prime factorization of 12 aligns with the first 3 from the prime factorization of 18. The second 3 from the prime factorization of 18 does not align with any factor from the prime factorization of 12. Below the two equations, a horizontal line is drawn. Below this line is the equation LCD equals 2 times 2 times 3 times 3, which simplifies to LCD equals 36.\" width=\"146\" height=\"75\" data-media-type=\"image\/png\" \/><\/td>\n<\/tr>\n<tr>\n<td>Rewrite as equivalent fractions with the LCD.<\/td>\n<td data-align=\"center\">[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 data-align=\"center\">[latex]\\Large\\frac{21}{36}+\\Large\\frac{10}{36}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td data-align=\"center\">[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=\"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-5\" 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 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<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-68\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/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":17533,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"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\"}]","CANDELA_OUTCOMES_GUID":"1f1dd0a7-dca2-43f3-9757-33d4e41a9363, fa634827-e59a-4e9d-87a8-9b589ebee4a9","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-68","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/68","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/68\/revisions"}],"predecessor-version":[{"id":3982,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/68\/revisions\/3982"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/68\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/media?parent=68"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapter-type?post=68"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/contributor?post=68"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/license?post=68"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}