{"id":9570,"date":"2017-05-03T15:45:29","date_gmt":"2017-05-03T15:45:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/prealgebra\/?post_type=chapter&#038;p=9570"},"modified":"2020-09-03T10:54:36","modified_gmt":"2020-09-03T10:54:36","slug":"solving-equations-that-contain-fractions-using-the-addition-subtraction-and-division-properties-of-equality","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/solving-equations-that-contain-fractions-using-the-addition-subtraction-and-division-properties-of-equality\/","title":{"raw":"2.7.a - Solving Equations Containing Fractions Using the Addition, Subtraction, and Division Properties of Equality","rendered":"2.7.a &#8211; Solving Equations Containing Fractions Using the Addition, Subtraction, and Division Properties of Equality"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine whether a fraction is a solution of an equation<\/li>\r\n \t<li>Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Determine Whether a Fraction is a Solution of an Equation<\/h2>\r\nAs we saw in previous lessons, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. <a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/determining-whether-a-whole-number-is-a-solution-to-an-equation\/\">In those sections, we found whole number and integer solutions to equations<\/a>. Now that we have worked with fractions, we are ready to find fraction solutions to equations.\r\n\r\nThe steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.\r\n<div class=\"textbox shaded\">\r\n<h3>Determine whether a number is a solution to an equation.<\/h3>\r\n<ol id=\"eip-id1168469777675\" class=\"stepwise\">\r\n \t<li>Substitute the number for the variable in the equation.<\/li>\r\n \t<li>Simplify the expressions on both sides of the equation.<\/li>\r\n \t<li>Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDetermine whether each of the following is a solution of [latex]x-\\Large\\frac{3}{10}=\\Large\\frac{1}{2}[\/latex]\r\n<ol>\r\n \t<li>[latex]x=1[\/latex]<\/li>\r\n \t<li>[latex]x=\\Large\\frac{4}{5}[\/latex]<\/li>\r\n \t<li>[latex]x=-\\Large\\frac{4}{5}[\/latex]<\/li>\r\n<\/ol>\r\nSolution:\r\n<table id=\"eip-id1168468728112\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says x minus 3 tenths equals one half. The next line says substitute 1 for x and shows 1 minus 3 tenths followed by an equal sign with a question mark and 1 half. The next line says, \">\r\n<tbody>\r\n<tr style=\"height: 15.46875px\">\r\n<td style=\"height: 15.46875px\">1.<\/td>\r\n<td style=\"height: 15.46875px\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\"><\/td>\r\n<td style=\"height: 15px\">[latex]x -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 49px\">\r\n<td style=\"height: 49px\">Substitute [latex]\\color{red}{1}[\/latex] for x.<\/td>\r\n<td style=\"height: 49px\">[latex]\\color{red}{1} -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 49px\">\r\n<td style=\"height: 49px\">Change to fractions with a LCD of [latex]10[\/latex].<\/td>\r\n<td style=\"height: 49px\">[latex]\\color{red}{\\Large\\frac{10}{10}} -\\Large\\frac{3}{10} =\\Large\\frac{5}{10}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 49px\">\r\n<td style=\"height: 49px\">Subtract.<\/td>\r\n<td style=\"height: 49px\">[latex]\\Large\\frac{7}{10} \\not=\\Large\\frac{5}{10}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]x=1[\/latex] does not result in a true equation, [latex]1[\/latex] is not a solution to the equation.\r\n<table id=\"eip-id1168469686493\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says x minus 3 tenths equals 1 half. The next line says substitute 4 fifths for x and shows a red 4 fifths minus 3 tenths followed by an equal sign with a question mark and 1 half. The next line shows a red 8 tenths minus 3 tenths followed by an equal sign with a question mark and 5 tenths. The last step says to subtract and shows that 5 tenths equals 5 tenths followed by a check mark.\">\r\n<tbody>\r\n<tr>\r\n<td>2.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{\\Large\\frac{4}{5}}[\/latex] for x.<\/td>\r\n<td>[latex]\\color{red}{\\Large\\frac{4}{5}} -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\color{red}{\\Large\\frac{8}{10}} -\\Large\\frac{3}{10} =\\Large\\frac{5}{10}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]\\Large\\frac{5}{10} =\\Large\\frac{5}{10}\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]x=\\Large\\frac{4}{5}[\/latex] results in a true equation, [latex]\\Large\\frac{4}{5}[\/latex] is a solution to the equation [latex]x-\\Large\\frac{3}{10}=\\Large\\frac{1}{2}[\/latex].\r\n<table id=\"eip-id1168468469809\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says x minus 3 tenths equals 1 half. The next line says substitute negative 4 fifths for x and shows a red negative 4 fifths minus 3 tenths followed by an equal sign with a question mark and 1 half. The next line says to simplify and shows a red negative 8 tenths minus 3 tenths followed by an equal sign with a question mark and 5 tenths. The last step says to subtract and shows that negative 11 tenths does not equal 5 tenths.\">\r\n<tbody>\r\n<tr>\r\n<td>3.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{-\\Large\\frac{4}{5}}[\/latex] for x.<\/td>\r\n<td>[latex]\\color{red}{-\\Large\\frac{4}{5}} -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\color{red}{-\\Large\\frac{8}{10}} -\\Large\\frac{3}{10} =\\Large\\frac{5}{10}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]-\\Large\\frac{11}{10}\\not=\\Large\\frac{5}{10}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]x=-\\Large\\frac{4}{5}[\/latex] does not result in a true equation, [latex]-\\Large\\frac{4}{5}[\/latex] is not a solution to the equation.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]146128[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality<\/h2>\r\nWe also previously solved equations using the Addition, Subtraction, and Division Properties of Equality<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/solving-equations-using-the-subtraction-property-of-equality\/\">.<\/a>\u00a0We will use these same properties to solve equations with fractions.\r\n<div class=\"textbox shaded\">\r\n<h3>Addition, Subtraction, and Division Properties of Equality<\/h3>\r\nFor any numbers [latex]a,b,\\text{ and }c[\/latex],\r\n<table id=\"eip-id1172107133172\" style=\"width: 85%\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\text{if }a=b,\\text{ then }a+c=b+c[\/latex].<\/td>\r\n<td>Addition Property of Equality<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{if }a=b,\\text{ then }a-c=b-c[\/latex].<\/td>\r\n<td>Subtraction Property of Equality<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\text{if }a=b,\\text{ then }\\Large\\frac{a}{c}=\\Large\\frac{b}{c}\\normalsize ,c\\ne 0[\/latex].<\/td>\r\n<td>Division Property of Equality<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.\r\n\r\n<\/div>\r\n<h2>Solve Equations with Fractions Using the Subtraction Property of Equality<\/h2>\r\nLet's start by looking at equations that can be solved using the subtraction property of equality.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]y+\\Large\\frac{9}{16}=\\Large\\frac{5}{16}[\/latex]\r\n[reveal-answer q=\"60465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"60465\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1165720941272\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says y plus 9 over 16 equals 5 over 16. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]y +\\Large\\frac{9}{16} =\\Large\\frac{5}{16}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Subtract [latex]\\Large\\frac{9}{16}[\/latex] from each side to undo the addition.<\/td>\r\n<td>[latex]y +\\Large\\frac{9}{16}\\color{red}{-}\\color{red}{\\Large\\frac{9}{16}} =\\Large\\frac{5}{16}\\color{red}{-}\\color{red}{\\Large\\frac{9}{16}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify on each side of the equation.<\/td>\r\n<td>[latex]y + 0 = -\\Large\\frac{4}{16}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify the fraction.<\/td>\r\n<td>[latex]y = -\\Large\\frac{1}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]y +\\Large\\frac{9}{16} =\\Large\\frac{5}{16}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]y=-\\Large\\frac{1}{4}[\/latex] .<\/td>\r\n<td>[latex]\\color{red}{-\\Large\\frac{1}{4}} +\\Large\\frac{9}{16} \\stackrel{?}{=}\\Large\\frac{5}{16}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite as fractions with the LCD.<\/td>\r\n<td>[latex]\\color{red}{-\\Large\\frac{4}{16}} +\\Large\\frac{9}{16}\\stackrel{?}{=}\\Large\\frac{5}{16}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]\\Large\\frac{5}{16} =\\Large\\frac{5}{16}\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]y=-\\Large\\frac{1}{4}[\/latex] makes [latex]y+\\Large\\frac{9}{16}=\\Large\\frac{5}{16}[\/latex] a true statement, we know we have found the solution to this equation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we will solve an equation with fractions whose denominators are different. We will need to make an additional step to find the common denominator.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]p+\\Large\\frac{1}{2}=\\Large\\frac{2}{3}[\/latex]\r\n[reveal-answer q=\"69865\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"69865\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1165720941272\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says y plus 9 over 16 equals 5 over 16. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]p+\\Large\\frac{1}{2}=\\Large\\frac{2}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Subtract [latex]\\Large\\frac{1}{2}[\/latex] from each side to undo the addition.<\/td>\r\n<td>[latex]p+\\Large\\frac{1}{2}\\color{red}{-}\\color{red}{\\Large\\frac{1}{2}} =\\Large\\frac{2}{3}\\color{red}{-}\\color{red}{\\Large\\frac{1}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify on each side of the equation.<\/td>\r\n<td>[latex]p + 0 =\\Large\\frac{2}{3}-\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify the fraction by finding a common denominator.<\/td>\r\n<td>[latex]p =\\Large\\frac{2}{3}\\cdot\\color{red}{\\Large\\frac{2}{2}}-\\Large\\frac{1}{2}\\cdot\\color{red}{\\Large\\frac{3}{3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]p =\\Large\\frac{4}{6}-\\Large\\frac{3}{6}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]p =\\Large\\frac{1}{6}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]p+\\Large\\frac{1}{2}=\\Large\\frac{2}{3}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]p=\\Large\\frac{1}{6}[\/latex] . \u00a0Rewrite as fractions with the LCD.<\/td>\r\n<td>[latex]\\color{red}{\\Large\\frac{1}{6}} +\\Large\\frac{1}{2} \\stackrel{?}{=}\\Large\\frac{2}{3}=\\color{red}{\\Large\\frac{1}{6}}+\\Large\\frac{3}{6}\\stackrel{?}{=}\\Large\\frac{4}{6}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\Large\\frac{4}{6} =\\Large\\frac{4}{6}\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe know we have found the solution to this equation since [latex]\\Large\\frac{4}{6} =\\Large\\frac{4}{6}[\/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]146492[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solve Equations with Fractions Using the Addition Property of Equality<\/h2>\r\nWe used the Subtraction Property of Equality in the example above. Now we\u2019ll use the Addition Property of Equality.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]a-\\Large\\frac{1}{4}=-\\Large\\frac{2}{3}[\/latex]\r\n[reveal-answer q=\"896044\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"896044\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1165721109263\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says a minus 5 ninths equals negative 8 ninths. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]a-\\Large\\frac{1}{4}=-\\Large\\frac{2}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Add [latex]\\Large\\frac{5}{9}[\/latex] from each side to undo the addition.<\/td>\r\n<td>[latex]a-\\Large\\frac{1}{4}\\color{red}{+}\\color{red}{\\Large\\frac{1}{4}} =-\\Large\\frac{2}{3}\\color{red}{+}\\color{red}{\\Large\\frac{1}{4}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Find a common denominator.<\/td>\r\n<td>[latex]a =-\\Large\\frac{2}{3}\\cdot\\color{red}{\\Large\\frac{4}{4}}+\\Large\\frac{1}{4}\\cdot\\color{red}{\\Large\\frac{3}{3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify on each side of the equation.<\/td>\r\n<td>[latex]a = -\\Large\\frac{8}{12}+\\Large\\frac{3}{12}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify the fraction.<\/td>\r\n<td>[latex]a = -\\Large\\frac{5}{12}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]a-\\Large\\frac{1}{4}=-\\Large\\frac{2}{3}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]a=-\\frac{5}{12}[\/latex] .<\/td>\r\n<td>[latex]\\color{red}{-\\Large\\frac{5}{12}} -\\Large\\frac{1}{4}\\stackrel{?}{=}-\\Large\\frac{2}{3}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Change to common denominator.<\/td>\r\n<td>[latex]\\color{red}{-\\Large\\frac{5}{12}} -\\Large\\frac{3}{12}\\stackrel{?}{=}-\\Large\\frac{8}{12}[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]-\\Large\\frac{8}{12} = -\\Large\\frac{8}{12}\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]a=-\\Large\\frac{5}{12}[\/latex] makes the equation true, we know that [latex]a=-\\Large\\frac{5}{12}[\/latex] is the solution to the equation.\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]146494[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video we show more examples of solving an equation with fractions where you are required to find a common denominator.\r\n\r\nhttps:\/\/youtu.be\/O7SPM7Cs8Ds\r\n<h2>Solve Equations with Fractions Using the Division Property of Equality<\/h2>\r\nNext, we will solve equations that require division to isolate the variable. First, let's consider the division property of equality again.\r\n<div class=\"textbox shaded\">\r\n<h3>The Division Property of Equality<\/h3>\r\nFor any numbers [latex]a,b[\/latex], and [latex]c[\/latex],\u00a0[latex]c\\ne 0[\/latex]\r\n<p style=\"text-align: center\">[latex]\\text{if }a=b,\\text{ then }\\Large\\frac{a}{c}=\\Large\\frac{b}{c}[\/latex].<\/p>\r\nIf you divide both sides of an equation by the same quantity, you still have equality.\r\n\r\n<\/div>\r\nLet's put this idea in practice with an example. We are looking for the number you multiply by [latex]10[\/latex] to get [latex]44[\/latex], and we can use division to find out.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]10q=44[\/latex]\r\n\r\nSolution:\r\n<table id=\"eip-id1168467284261\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]10q=44[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Divide both sides by [latex]10[\/latex] to undo the multiplication.<\/td>\r\n<td>[latex]\\Large\\frac{10q}{10}=\\Large\\frac{44}{10}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify.<\/td>\r\n<td>[latex]q=\\Large\\frac{22}{5}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Check:<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]q=\\Large\\frac{22}{5}[\/latex] into the original equation.<\/td>\r\n<td>[latex]10\\left(\\Large\\frac{22}{5}\\right)\\stackrel{?}{=}44[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]\\stackrel{2}{\\overline{)10}}\\left(\\Large\\frac{22}{\\overline{)5}}\\right)\\stackrel{?}{=}44[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]44=44\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe solution to the equation was the fraction [latex]\\Large\\frac{22}{5}[\/latex]. We leave it as an improper fraction.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]146141[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine whether a fraction is a solution of an equation<\/li>\n<li>Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality<\/li>\n<\/ul>\n<\/div>\n<h2>Determine Whether a Fraction is a Solution of an Equation<\/h2>\n<p>As we saw in previous lessons, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. <a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/determining-whether-a-whole-number-is-a-solution-to-an-equation\/\">In those sections, we found whole number and integer solutions to equations<\/a>. Now that we have worked with fractions, we are ready to find fraction solutions to equations.<\/p>\n<p>The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.<\/p>\n<div class=\"textbox shaded\">\n<h3>Determine whether a number is a solution to an equation.<\/h3>\n<ol id=\"eip-id1168469777675\" class=\"stepwise\">\n<li>Substitute the number for the variable in the equation.<\/li>\n<li>Simplify the expressions on both sides of the equation.<\/li>\n<li>Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Determine whether each of the following is a solution of [latex]x-\\Large\\frac{3}{10}=\\Large\\frac{1}{2}[\/latex]<\/p>\n<ol>\n<li>[latex]x=1[\/latex]<\/li>\n<li>[latex]x=\\Large\\frac{4}{5}[\/latex]<\/li>\n<li>[latex]x=-\\Large\\frac{4}{5}[\/latex]<\/li>\n<\/ol>\n<p>Solution:<\/p>\n<table id=\"eip-id1168468728112\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says x minus 3 tenths equals one half. The next line says substitute 1 for x and shows 1 minus 3 tenths followed by an equal sign with a question mark and 1 half. The next line says,\">\n<tbody>\n<tr style=\"height: 15.46875px\">\n<td style=\"height: 15.46875px\">1.<\/td>\n<td style=\"height: 15.46875px\"><\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\"><\/td>\n<td style=\"height: 15px\">[latex]x -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 49px\">\n<td style=\"height: 49px\">Substitute [latex]\\color{red}{1}[\/latex] for x.<\/td>\n<td style=\"height: 49px\">[latex]\\color{red}{1} -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 49px\">\n<td style=\"height: 49px\">Change to fractions with a LCD of [latex]10[\/latex].<\/td>\n<td style=\"height: 49px\">[latex]\\color{red}{\\Large\\frac{10}{10}} -\\Large\\frac{3}{10} =\\Large\\frac{5}{10}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 49px\">\n<td style=\"height: 49px\">Subtract.<\/td>\n<td style=\"height: 49px\">[latex]\\Large\\frac{7}{10} \\not=\\Large\\frac{5}{10}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]x=1[\/latex] does not result in a true equation, [latex]1[\/latex] is not a solution to the equation.<\/p>\n<table id=\"eip-id1168469686493\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says x minus 3 tenths equals 1 half. The next line says substitute 4 fifths for x and shows a red 4 fifths minus 3 tenths followed by an equal sign with a question mark and 1 half. The next line shows a red 8 tenths minus 3 tenths followed by an equal sign with a question mark and 5 tenths. The last step says to subtract and shows that 5 tenths equals 5 tenths followed by a check mark.\">\n<tbody>\n<tr>\n<td>2.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]x -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{\\Large\\frac{4}{5}}[\/latex] for x.<\/td>\n<td>[latex]\\color{red}{\\Large\\frac{4}{5}} -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]\\color{red}{\\Large\\frac{8}{10}} -\\Large\\frac{3}{10} =\\Large\\frac{5}{10}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]\\Large\\frac{5}{10} =\\Large\\frac{5}{10}\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]x=\\Large\\frac{4}{5}[\/latex] results in a true equation, [latex]\\Large\\frac{4}{5}[\/latex] is a solution to the equation [latex]x-\\Large\\frac{3}{10}=\\Large\\frac{1}{2}[\/latex].<\/p>\n<table id=\"eip-id1168468469809\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says x minus 3 tenths equals 1 half. The next line says substitute negative 4 fifths for x and shows a red negative 4 fifths minus 3 tenths followed by an equal sign with a question mark and 1 half. The next line says to simplify and shows a red negative 8 tenths minus 3 tenths followed by an equal sign with a question mark and 5 tenths. The last step says to subtract and shows that negative 11 tenths does not equal 5 tenths.\">\n<tbody>\n<tr>\n<td>3.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]x -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{-\\Large\\frac{4}{5}}[\/latex] for x.<\/td>\n<td>[latex]\\color{red}{-\\Large\\frac{4}{5}} -\\Large\\frac{3}{10} =\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]\\color{red}{-\\Large\\frac{8}{10}} -\\Large\\frac{3}{10} =\\Large\\frac{5}{10}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]-\\Large\\frac{11}{10}\\not=\\Large\\frac{5}{10}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]x=-\\Large\\frac{4}{5}[\/latex] does not result in a true equation, [latex]-\\Large\\frac{4}{5}[\/latex] is not a solution to the equation.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146128\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146128&theme=oea&iframe_resize_id=ohm146128&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality<\/h2>\n<p>We also previously solved equations using the Addition, Subtraction, and Division Properties of Equality<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/solving-equations-using-the-subtraction-property-of-equality\/\">.<\/a>\u00a0We will use these same properties to solve equations with fractions.<\/p>\n<div class=\"textbox shaded\">\n<h3>Addition, Subtraction, and Division Properties of Equality<\/h3>\n<p>For any numbers [latex]a,b,\\text{ and }c[\/latex],<\/p>\n<table id=\"eip-id1172107133172\" style=\"width: 85%\" summary=\".\">\n<tbody>\n<tr>\n<td>[latex]\\text{if }a=b,\\text{ then }a+c=b+c[\/latex].<\/td>\n<td>Addition Property of Equality<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{if }a=b,\\text{ then }a-c=b-c[\/latex].<\/td>\n<td>Subtraction Property of Equality<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\text{if }a=b,\\text{ then }\\Large\\frac{a}{c}=\\Large\\frac{b}{c}\\normalsize ,c\\ne 0[\/latex].<\/td>\n<td>Division Property of Equality<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.<\/p>\n<\/div>\n<h2>Solve Equations with Fractions Using the Subtraction Property of Equality<\/h2>\n<p>Let&#8217;s start by looking at equations that can be solved using the subtraction property of equality.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]y+\\Large\\frac{9}{16}=\\Large\\frac{5}{16}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q60465\">Show Solution<\/span><\/p>\n<div id=\"q60465\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1165720941272\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says y plus 9 over 16 equals 5 over 16. The next line says,\">\n<tbody>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]y +\\Large\\frac{9}{16} =\\Large\\frac{5}{16}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Subtract [latex]\\Large\\frac{9}{16}[\/latex] from each side to undo the addition.<\/td>\n<td>[latex]y +\\Large\\frac{9}{16}\\color{red}{-}\\color{red}{\\Large\\frac{9}{16}} =\\Large\\frac{5}{16}\\color{red}{-}\\color{red}{\\Large\\frac{9}{16}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify on each side of the equation.<\/td>\n<td>[latex]y + 0 = -\\Large\\frac{4}{16}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify the fraction.<\/td>\n<td>[latex]y = -\\Large\\frac{1}{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]y +\\Large\\frac{9}{16} =\\Large\\frac{5}{16}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]y=-\\Large\\frac{1}{4}[\/latex] .<\/td>\n<td>[latex]\\color{red}{-\\Large\\frac{1}{4}} +\\Large\\frac{9}{16} \\stackrel{?}{=}\\Large\\frac{5}{16}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Rewrite as fractions with the LCD.<\/td>\n<td>[latex]\\color{red}{-\\Large\\frac{4}{16}} +\\Large\\frac{9}{16}\\stackrel{?}{=}\\Large\\frac{5}{16}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]\\Large\\frac{5}{16} =\\Large\\frac{5}{16}\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]y=-\\Large\\frac{1}{4}[\/latex] makes [latex]y+\\Large\\frac{9}{16}=\\Large\\frac{5}{16}[\/latex] a true statement, we know we have found the solution to this equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we will solve an equation with fractions whose denominators are different. We will need to make an additional step to find the common denominator.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]p+\\Large\\frac{1}{2}=\\Large\\frac{2}{3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q69865\">Show Solution<\/span><\/p>\n<div id=\"q69865\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1165720941272\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says y plus 9 over 16 equals 5 over 16. The next line says,\">\n<tbody>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]p+\\Large\\frac{1}{2}=\\Large\\frac{2}{3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Subtract [latex]\\Large\\frac{1}{2}[\/latex] from each side to undo the addition.<\/td>\n<td>[latex]p+\\Large\\frac{1}{2}\\color{red}{-}\\color{red}{\\Large\\frac{1}{2}} =\\Large\\frac{2}{3}\\color{red}{-}\\color{red}{\\Large\\frac{1}{2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify on each side of the equation.<\/td>\n<td>[latex]p + 0 =\\Large\\frac{2}{3}-\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify the fraction by finding a common denominator.<\/td>\n<td>[latex]p =\\Large\\frac{2}{3}\\cdot\\color{red}{\\Large\\frac{2}{2}}-\\Large\\frac{1}{2}\\cdot\\color{red}{\\Large\\frac{3}{3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]p =\\Large\\frac{4}{6}-\\Large\\frac{3}{6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]p =\\Large\\frac{1}{6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]p+\\Large\\frac{1}{2}=\\Large\\frac{2}{3}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]p=\\Large\\frac{1}{6}[\/latex] . \u00a0Rewrite as fractions with the LCD.<\/td>\n<td>[latex]\\color{red}{\\Large\\frac{1}{6}} +\\Large\\frac{1}{2} \\stackrel{?}{=}\\Large\\frac{2}{3}=\\color{red}{\\Large\\frac{1}{6}}+\\Large\\frac{3}{6}\\stackrel{?}{=}\\Large\\frac{4}{6}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]\\Large\\frac{4}{6} =\\Large\\frac{4}{6}\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We know we have found the solution to this equation since [latex]\\Large\\frac{4}{6} =\\Large\\frac{4}{6}[\/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=\"ohm146492\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146492&theme=oea&iframe_resize_id=ohm146492&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Equations with Fractions Using the Addition Property of Equality<\/h2>\n<p>We used the Subtraction Property of Equality in the example above. Now we\u2019ll use the Addition Property of Equality.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]a-\\Large\\frac{1}{4}=-\\Large\\frac{2}{3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q896044\">Show Solution<\/span><\/p>\n<div id=\"q896044\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1165721109263\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\"The first line says a minus 5 ninths equals negative 8 ninths. The next line says,\">\n<tbody>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]a-\\Large\\frac{1}{4}=-\\Large\\frac{2}{3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Add [latex]\\Large\\frac{5}{9}[\/latex] from each side to undo the addition.<\/td>\n<td>[latex]a-\\Large\\frac{1}{4}\\color{red}{+}\\color{red}{\\Large\\frac{1}{4}} =-\\Large\\frac{2}{3}\\color{red}{+}\\color{red}{\\Large\\frac{1}{4}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Find a common denominator.<\/td>\n<td>[latex]a =-\\Large\\frac{2}{3}\\cdot\\color{red}{\\Large\\frac{4}{4}}+\\Large\\frac{1}{4}\\cdot\\color{red}{\\Large\\frac{3}{3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify on each side of the equation.<\/td>\n<td>[latex]a = -\\Large\\frac{8}{12}+\\Large\\frac{3}{12}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify the fraction.<\/td>\n<td>[latex]a = -\\Large\\frac{5}{12}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]a-\\Large\\frac{1}{4}=-\\Large\\frac{2}{3}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]a=-\\frac{5}{12}[\/latex] .<\/td>\n<td>[latex]\\color{red}{-\\Large\\frac{5}{12}} -\\Large\\frac{1}{4}\\stackrel{?}{=}-\\Large\\frac{2}{3}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Change to common denominator.<\/td>\n<td>[latex]\\color{red}{-\\Large\\frac{5}{12}} -\\Large\\frac{3}{12}\\stackrel{?}{=}-\\Large\\frac{8}{12}[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]-\\Large\\frac{8}{12} = -\\Large\\frac{8}{12}\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]a=-\\Large\\frac{5}{12}[\/latex] makes the equation true, we know that [latex]a=-\\Large\\frac{5}{12}[\/latex] is the solution to the equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146494\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146494&theme=oea&iframe_resize_id=ohm146494&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video we show more examples of solving an equation with fractions where you are required to find a common denominator.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solving One Step Equations Using Addition and Subtraction (Fractions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/O7SPM7Cs8Ds?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Solve Equations with Fractions Using the Division Property of Equality<\/h2>\n<p>Next, we will solve equations that require division to isolate the variable. First, let&#8217;s consider the division property of equality again.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Division Property of Equality<\/h3>\n<p>For any numbers [latex]a,b[\/latex], and [latex]c[\/latex],\u00a0[latex]c\\ne 0[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\text{if }a=b,\\text{ then }\\Large\\frac{a}{c}=\\Large\\frac{b}{c}[\/latex].<\/p>\n<p>If you divide both sides of an equation by the same quantity, you still have equality.<\/p>\n<\/div>\n<p>Let&#8217;s put this idea in practice with an example. We are looking for the number you multiply by [latex]10[\/latex] to get [latex]44[\/latex], and we can use division to find out.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]10q=44[\/latex]<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168467284261\" class=\"unnumbered unstyled\" style=\"width: 85%\" summary=\".\">\n<tbody>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]10q=44[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Divide both sides by [latex]10[\/latex] to undo the multiplication.<\/td>\n<td>[latex]\\Large\\frac{10q}{10}=\\Large\\frac{44}{10}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify.<\/td>\n<td>[latex]q=\\Large\\frac{22}{5}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Check:<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]q=\\Large\\frac{22}{5}[\/latex] into the original equation.<\/td>\n<td>[latex]10\\left(\\Large\\frac{22}{5}\\right)\\stackrel{?}{=}44[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]\\stackrel{2}{\\overline{)10}}\\left(\\Large\\frac{22}{\\overline{)5}}\\right)\\stackrel{?}{=}44[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]44=44\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The solution to the equation was the fraction [latex]\\Large\\frac{22}{5}[\/latex]. We leave it as an improper fraction.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146141\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146141&theme=oea&iframe_resize_id=ohm146141&show_question_numbers\" width=\"100%\" height=\"150\"><\/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-9570\">\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>Solving One Step Equations Using Addition and Subtraction (Fractions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/O7SPM7Cs8Ds\">https:\/\/youtu.be\/O7SPM7Cs8Ds<\/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>Question ID: 146128, 146492, 146494. <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><\/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":17533,"menu_order":34,"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\":\"Solving One Step Equations Using Addition and Subtraction (Fractions)\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/O7SPM7Cs8Ds\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID: 146128, 146492, 146494\",\"author\":\"Alyson Day\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"160d148a7a4048f78f1b5c95e96ec4b9, c9fc4bcf956641e8b09a177d97d12174, 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