{"id":3844,"date":"2020-02-09T19:59:03","date_gmt":"2020-02-09T19:59:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/mathforlibscoreq\/?post_type=chapter&#038;p=3844"},"modified":"2021-02-05T23:55:36","modified_gmt":"2021-02-05T23:55:36","slug":"solving-equations-that-contain-fractions-using-the-addition-and-subtraction-properties-of-equality","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/solving-equations-that-contain-fractions-using-the-addition-and-subtraction-properties-of-equality\/","title":{"raw":"Solving Equations That Contain Fractions Using the Addition and Subtraction Properties of Equality","rendered":"Solving Equations That Contain Fractions Using the Addition and Subtraction 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 to an equation<\/li>\r\n \t<li>Use the equality properties of addition and subtraction to solve equations that contain fractions<\/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. In those sections, we found whole number and integer solutions to equations. 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 and Subtraction Properties of Equality<\/h2>\r\nWe also solved equations using the Addition, Subtraction, and Division Properties of Equality. We 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<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\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}\\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","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine whether a fraction is a solution to an equation<\/li>\n<li>Use the equality properties of addition and subtraction to solve equations that contain fractions<\/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. In those sections, we found whole number and integer solutions to equations. 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 and Subtraction Properties of Equality<\/h2>\n<p>We also solved equations using the Addition, Subtraction, and Division Properties of Equality. We 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<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<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}\\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\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-3844\">\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":25777,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757\"},{\"type\":\"cc\",\"description\":\"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":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3844","chapter","type-chapter","status-web-only","hentry"],"part":356,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3844","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/users\/25777"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3844\/revisions"}],"predecessor-version":[{"id":3849,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3844\/revisions\/3849"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/parts\/356"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3844\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/media?parent=3844"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=3844"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/contributor?post=3844"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/license?post=3844"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}