{"id":9037,"date":"2017-05-01T15:16:05","date_gmt":"2017-05-01T15:16:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/prealgebra\/?post_type=chapter&#038;p=9037"},"modified":"2020-10-22T09:08:42","modified_gmt":"2020-10-22T09:08:42","slug":"solving-equations-with-fraction-coefficients","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/solving-equations-with-fraction-coefficients\/","title":{"raw":"7.1.f - Solving Equations with Fraction and Decimal Coefficients","rendered":"7.1.f &#8211; Solving Equations with Fraction and Decimal Coefficients"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve equations with fraction coefficients<\/li>\r\n \t<li>Solve equations with decimal coefficients<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section, we will explore how to solve equations with fractions in them.\u00a0 To start, we will review how equations with fractions can be solved by multiplying by the reciprocal. We will then move on to learn how to use the Multiplication Property of Equality to eliminate the fractions from an equation, making it simpler to solve.\r\n<h3>Solving equations with fractions by multiplying by the reciprocal<\/h3>\r\nIn our first example, we will solve a one-step equation using the multiplication property of equality. You will see that the variable is part of a fraction in the given equation, and using the multiplication property of equality allows us to remove the variable from the fraction. Remember that fractions imply division, so you can think of this as the variable [latex]k[\/latex] is being divided by 10. To \"undo\" the division, you can use multiplication to isolate [latex]k[\/latex]. (Note that there is a negative term in the equation, so it will be important to think about the sign of each term as you work through the problem. Stop after each step you take to make sure all the terms have the correct sign.)\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]-\\frac{7}{2}=\\frac{k}{10}[\/latex] for [latex]k[\/latex].\r\n\r\n[reveal-answer q=\"471772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471772\"]\r\n\r\nWe want to isolate the [latex]k[\/latex], which is being divided by [latex]10[\/latex]. The first thing we should do is multiply both sides by [latex]10[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ Multiply the left side by 10: }\\\\\\\\\\left(10\\right)(-\\frac{7}{2})=\\frac{-10\\cdot7}{2} = \\frac{-70}{2} = -35\\\\\\\\\\text{ Now, multiply the right side by 10: }\\\\\\\\\\,\\,\\,\\frac{k}{10}\\left(10\\right) = \\frac{k\\cdot10}{10} = k\\\\\\\\\\text{ Now, replace your results in the equation: }\\\\-35=k\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\nWe write the [latex]k[\/latex] on the left side as a matter of convention.\r\n\r\n[latex]k=-35[\/latex]\r\n<p style=\"text-align: center\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nIn our next example, we are given an equation that contains a variable multiplied by a fraction. We will use a reciprocal to isolate the variable.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex]\r\n\r\n[reveal-answer q=\"444022\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"444022\"]\r\n\r\nSolution:\r\nSince the product of a number and its reciprocal is [latex]1[\/latex], our strategy will be to isolate [latex]x[\/latex] by multiplying by the reciprocal of [latex]\\Large\\frac{2}{3}[\/latex].\r\n<table id=\"eip-id1168468646645\" class=\"unnumbered unstyled\" summary=\"The first line shows two-thirds x equals 18. The next line says \">\r\n<tbody>\r\n<tr style=\"height: 37px\">\r\n<td style=\"height: 37px\">[latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 38px\">\r\n<td style=\"height: 38px\">Multiply by the reciprocal of [latex]\\Large\\frac{2}{3}[\/latex] .<\/td>\r\n<td style=\"height: 38px\">[latex]\\Large\\frac{\\color{red}{3}}{\\color{red}{2}}\\normalsize\\cdot\\Large\\frac{2}{3}\\normalsize x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 37px\">\r\n<td style=\"height: 37px\">Reciprocals multiply to one.<\/td>\r\n<td style=\"height: 37px\">[latex]1x=\\Large\\frac{3}{2}\\normalsize\\cdot\\Large\\frac{18}{1}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 18px\">\r\n<td style=\"height: 18px\">Multiply.<\/td>\r\n<td style=\"height: 18px\">[latex]x=27[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px\">\r\n<td style=\"height: 14px\">Check your answer.<\/td>\r\n<td style=\"height: 44.8594px\">[latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 44.8594px\">\r\n<td style=\"height: 14px\">Let [latex]x=27[\/latex].<\/td>\r\n<td style=\"height: 45px\">[latex]\\Large\\frac{2}{3}\\normalsize\\cdot\\color{red}{27}\\stackrel{\\text{?}}{=}18[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td><\/td>\r\n<td style=\"height: 24px\">[latex]18=18\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that we could have divided both sides of the equation [latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex] by [latex]\\Large\\frac{2}{3}[\/latex] to isolate [latex]x[\/latex]. While this would work, multiplying by the reciprocal requires fewer steps.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141871&amp;theme=oea&amp;iframe_resize_id=mom22[\/embed]\r\n\r\n<\/div>\r\nIn the following video, you will see examples of using the multiplication property of equality to solve a one-step equation involving negative fractions.\r\nhttps:\/\/www.youtube.com\/watch?v=AhBdGeUGgsI&amp;feature=youtu.be\r\n<h3>Solving equations with fractions by clearing the denominators<\/h3>\r\nYou may\u00a0feel overwhelmed\u00a0when you\u00a0see fractions in an equation, so we are going to show a method to solve equations with fractions where you use the common denominator to eliminate the fractions from an equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions.\r\n\r\nPay attention to the fact that each term in the equation gets multiplied by the least\u00a0common denominator. That's what makes it equal to the original!\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve: [latex]\\Large\\frac{1}{8}\\normalsize x+\\Large\\frac{1}{2}=\\Large\\frac{1}{4}[\/latex]\r\n\r\nSolution:\r\n<table id=\"eip-id1168467351085\" class=\"unnumbered unstyled\" summary=\"The top line says, \">\r\n<tbody>\r\n<tr style=\"height: 30px\">\r\n<td style=\"height: 30px\"><\/td>\r\n<td style=\"height: 30px\">[latex]\\Large\\frac{1}{8}\\normalsize x+\\Large\\frac{1}{2}=\\Large\\frac{1}{4}\\normalsize\\quad{LCD=8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 45px\">\r\n<td style=\"height: 45px\">Multiply both sides of the equation by that LCD, [latex]8[\/latex]. This clears the fractions.<\/td>\r\n<td style=\"height: 45px\">[latex]\\color{red}{8(}\\Large\\frac{1}{8}\\normalsize x+\\Large\\frac{1}{2}\\color{red}{)}=\\normalsize\\color{red}{8(}\\Large\\frac{1}{4}\\color{red}{)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 30.9531px\">\r\n<td style=\"height: 30.9531px\">Use the Distributive Property.<\/td>\r\n<td style=\"height: 30.9531px\">[latex]8\\cdot\\Large\\frac{1}{8}\\normalsize x+8\\cdot\\Large\\frac{1}{2}\\normalsize=8\\cdot\\Large\\frac{1}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 30px\">\r\n<td style=\"height: 30px\">Simplify \u2014 and notice, no more fractions!<\/td>\r\n<td style=\"height: 30px\">[latex]x+4=2[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 30px\">\r\n<td style=\"height: 30px\">Solve using the General Strategy for Solving Linear Equations.<\/td>\r\n<td style=\"height: 30px\">[latex]x+4\\color{red}{-4}=2\\color{red}{-4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\">Simplify.<\/td>\r\n<td style=\"height: 15px\">[latex]x=-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 340px\">\r\n<td style=\"height: 340px\">Check: Let [latex]x=-2[\/latex][latex]\r\n\r\n\\Large\\frac{1}{8}\\normalsize x+\r\n\r\n\\Large\\frac{1}{2}=\r\n\r\n\\Large\\frac{1}{4}[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{1}{8}\\normalsize(\\color{red}{-2})+\r\n\r\n\\Large\\frac{1}{2}\\normalsize\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{4}[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{-2}{8}+\r\n\r\n\\Large\\frac{1}{2}\\normalsize\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{4}[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{-2}{8}+\r\n\r\n\\Large\\frac{4}{8}\\normalsize\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{4}[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{2}{8}\\normalsize\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{4}[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{1}{4}=\r\n\r\n\\Large\\frac{1}{4}\\quad\\checkmark[\/latex]\r\n\r\n&nbsp;<\/td>\r\n<td style=\"height: 340px\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nIn the example above, the least common denominator was [latex]8[\/latex]. Now it's your turn to find an LCD, and clear the fractions before you solve these linear equations.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=71948&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\nNotice\u00a0that once we cleared the equation of fractions, the equation was like those we learned how to solve earlier. We changed the problem to one we already knew how to solve!\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Solve equations by clearing the Denominators<\/h3>\r\n<ol id=\"eip-id1168466140919\" class=\"stepwise\">\r\n \t<li>Find the least common denominator of <em>all<\/em> the fractions in the equation.<\/li>\r\n \t<li>Multiply both sides of the equation by that LCD. This clears the fractions.<\/li>\r\n \t<li>Isolate the variable terms on one side, and the constant terms on the other side.<\/li>\r\n \t<li>Simplify both sides.<\/li>\r\n \t<li>Use the multiplication or division property to make the coefficient on the variable equal to [latex]1[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nHere's an example where you have\u00a0three variable terms. After you clear fractions with the LCD, you will simplify the three variable terms, then isolate the variable.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]7=\\Large\\frac{1}{2}\\normalsize x+\\Large\\frac{3}{4}\\normalsize x-\\Large\\frac{2}{3}\\normalsize x[\/latex]\r\n<p class=\"p1\">[reveal-answer q=\"190190\"]Show Solution[\/reveal-answer]<\/p>\r\n<p class=\"p1\">[hidden-answer a=\"190190\"]<\/p>\r\nSolution:\r\nWe want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.\r\n<table id=\"eip-id1168468606496\" class=\"unnumbered unstyled\" summary=\"The top line says \">\r\n<tbody>\r\n<tr>\r\n<td>Find the least common denominator of <em>all<\/em> the fractions in the equation.<\/td>\r\n<td>[latex]7=\\Large\\frac{1}{2}\\normalsize x+\\Large\\frac{3}{4}\\normalsize x-\\Large\\frac{2}{3}\\normalsize x\\quad{LCD=12}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply both sides of the equation by [latex]12[\/latex].<\/td>\r\n<td>[latex]\\color{red}{12}(7)=\\color{red}{12}\\cdot\\Large(\\frac{1}{2}\\normalsize x+\\Large\\frac{3}{4}\\normalsize x-\\Large\\frac{2}{3}\\normalsize x\\Large)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute.<\/td>\r\n<td>[latex]12(7)=12\\cdot\\Large\\frac{1}{2}\\normalsize x+12\\cdot\\Large\\frac{3}{4}\\normalsize x-12\\cdot\\Large\\frac{2}{3}\\normalsize x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify \u2014 and notice, no more fractions!<\/td>\r\n<td>[latex]84=6x+9x-8x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]84=7x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide by [latex]7[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{84}{\\color{red}{7}}=\\Large\\frac{7x}{\\color{red}{7}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]12=x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check: Let [latex]x=12[\/latex].<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]7=\\Large\\frac{1}{2}\\normalsize x+\r\n\r\n\\Large\\frac{3}{4}\\normalsize x-\r\n\r\n\\Large\\frac{2}{3}\\normalsize x[\/latex]\r\n\r\n[latex]7\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{2}\\normalsize(\\color{red}{12})+\r\n\r\n\\Large\\frac{3}{4}\\normalsize(\\color{red}{12})-\r\n\r\n\\Large\\frac{2}{3}\\normalsize(\\color{red}{12})[\/latex]\r\n\r\n[latex]7\\stackrel{\\text{?}}{=}6+9-8[\/latex]\r\n\r\n[latex]7=7\\quad\\checkmark[\/latex]\r\n\r\n&nbsp;<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow here's a similar problem for you to try. Clear the fractions, simplify, then solve.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=71948&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox shaded\">\r\n<h3>Caution!<\/h3>\r\nOne of the most common mistakes when you clear fractions is forgetting to multiply BOTH sides of the equation by the LCD. If your answer doesn't check, make sure you have multiplied both sides of the equation by the LCD.\r\n\r\n<\/div>\r\nIn the next example, we\u2019ll have variables and fractions on both sides of the equation. After you clear the fractions using the LCD, you will see that this equation is similar to ones with variables on both sides that we solved previously. Remember to choose a variable side and a constant side to help you organize your work.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]x+\\Large\\frac{1}{3}=\\Large\\frac{1}{6}\\normalsize x-\\Large\\frac{1}{2}[\/latex]\r\n<p class=\"p1\">[reveal-answer q=\"888555\"]Show Solution[\/reveal-answer]<\/p>\r\n<p class=\"p1\">[hidden-answer a=\"888555\"]<\/p>\r\nSolution:\r\n<table id=\"eip-id1168468770578\" class=\"unnumbered unstyled\" summary=\"The top line says, \">\r\n<tbody>\r\n<tr>\r\n<td>Find the LCD of all the fractions in the equation.<\/td>\r\n<td>[latex]x+\\Large\\frac{1}{3}=\\Large\\frac{1}{6}\\normalsize x-\\Large\\frac{1}{2}\\normalsize,\\quad{LCD=6}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply both sides by the LCD.<\/td>\r\n<td>[latex]\\color{red}{6}(x+\\Large\\frac{1}{3}\\normalsize)=\\color{red}{6}(\\Large\\frac{1}{6}\\normalsize x-\\Large\\frac{1}{2})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute.<\/td>\r\n<td>[latex]6\\cdot{x}+6\\cdot\\Large\\frac{1}{3}\\normalsize=6\\cdot\\Large\\frac{1}{6}\\normalsize x-6\\cdot\\Large\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify \u2014 no more fractions!<\/td>\r\n<td>[latex]6x+2=x-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract [latex]x[\/latex] from both sides.<\/td>\r\n<td>[latex]6x-\\color{red}{x}+2=x-\\color{red}{x}-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]5x+2=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract 2 from both sides.<\/td>\r\n<td>[latex]5x+2\\color{red}{-2}=-3\\color{red}{-2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]5x=-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide by [latex]5[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{5x}{\\color{red}{5}}=\\Large\\frac{-5}{\\color{red}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x=-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check: Substitute [latex]x=-1[\/latex].<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x+\\Large\\frac{1}{3}=\r\n\r\n\\Large\\frac{1}{6}\\normalsize x-\r\n\r\n\\Large\\frac{1}{2}[\/latex]\r\n\r\n[latex](\\color{red}{-1})+\r\n\r\n\\Large\\frac{1}{3}\\normalsize\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{6}\\normalsize(\\color{red}{-1})-\r\n\r\n\\Large\\frac{1}{2}[\/latex]\r\n\r\n[latex](-1)+\r\n\r\n\\Large\\frac{1}{3}\\normalsize\\stackrel{\\text{?}}{=}-\r\n\r\n\\Large\\frac{1}{6}-\r\n\r\n\\Large\\frac{1}{2}[\/latex]\r\n\r\n[latex]-\r\n\r\n\\Large\\frac{3}{3}+\r\n\r\n\\Large\\frac{1}{3}\\normalsize\\stackrel{\\text{?}}{=}-\r\n\r\n\\Large\\frac{1}{6}-\r\n\r\n\\Large\\frac{3}{6}[\/latex]\r\n\r\n[latex]-\r\n\r\n\\Large\\frac{2}{3}\\normalsize\\stackrel{\\text{?}}{=}-\r\n\r\n\\Large\\frac{4}{6}[\/latex]\r\n\r\n[latex]-\r\n\r\n\\Large\\frac{2}{3}=-\r\n\r\n\\Large\\frac{2}{3}\\quad\\checkmark[\/latex]\r\n\r\n&nbsp;<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow you can try solving an equation with fractions that has variables on both sides of the equal sign. The answer may be a fraction.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142514&amp;theme=oea&amp;iframe_resize_id=mom3[\/embed]\r\n\r\n<\/div>\r\nIn the following video, we show another example of how to solve an equation that contains fractions and variables on both sides of the equal sign.\r\n\r\nhttps:\/\/youtu.be\/G5R9jySFMpw\r\n\r\nIn the next example, we start with an equation where the variable term is locked up in parentheses and multiplied by a fraction. You can clear the fraction, or if you use the distributive property, it will eliminate the fraction. \u00a0Can you see why?\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve: [latex]1=\\Large\\frac{1}{2}\\normalsize\\left(4x+2\\right)[\/latex]\r\n<p class=\"p1\">[reveal-answer q=\"912234\"]Show Solution[\/reveal-answer]<\/p>\r\n<p class=\"p1\">[hidden-answer a=\"912234\"]<\/p>\r\nSolution:\r\n<table id=\"eip-id1168469839370\" class=\"unnumbered unstyled\" summary=\"The top line says 1 equals one-half times parentheses 4x plus 2. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]1=\\Large\\frac{1}{2}\\normalsize(4x+2)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute.<\/td>\r\n<td>[latex]1=\\Large\\frac{1}{2}\\normalsize\\cdot4x+\\Large\\frac{1}{2}\\normalsize\\cdot2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify. Now there are no fractions to clear!<\/td>\r\n<td>[latex]1=2x+1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract 1 from both sides.<\/td>\r\n<td>[latex]1\\color{red}{-1}=2x+1\\color{red}{-1}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]0=2x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide by [latex]2[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{0}{\\color{red}{2}}=\\Large\\frac{2x}{\\color{red}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]0=x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check: Let [latex]x=0[\/latex].<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1=\\Large\\frac{1}{2}\\normalsize(4x+2)[\/latex]\r\n\r\n[latex]1\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{2}\\normalsize(4(\\color{red}{0})+2)[\/latex]\r\n\r\n[latex]1\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{1}{2}\\normalsize(2)[\/latex]\r\n\r\n[latex]1\\stackrel{\\text{?}}{=}\r\n\r\n\\Large\\frac{2}{2}[\/latex]\r\n\r\n[latex]1=1\\quad\\checkmark[\/latex]\r\n\r\n&nbsp;<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow you can try solving an equation that has the variable term in parentheses that are multiplied by a fraction.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142542&amp;theme=oea&amp;iframe_resize_id=mom25[\/embed]\r\n\r\n<span style=\"color: #ff0000\">\u00a0<\/span>\r\n\r\n<\/div>\r\nIn the following video, we show how to solve a multi-step equation with fractions.\r\n\r\nhttps:\/\/youtu.be\/AvJTPeACTY0\r\n\r\nIf you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.\u00a0\u00a0Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation.\r\n<h2>Solving Equations By Clearing Decimals<\/h2>\r\nSome equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, [latex]0.3=\\Large\\frac{3}{10}[\/latex] and [latex]0.17=\\Large\\frac{17}{100}[\/latex]. So, when we have an equation with decimals, we can use the same process we used to clear fractions\u2014multiply both sides of the equation by the least common denominator.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]0.8x - 5=7[\/latex]\r\n\r\nSolution:\r\nThe only decimal in the equation is [latex]0.8[\/latex]. Since [latex]0.8=\\Large\\frac{8}{10}[\/latex], the LCD is [latex]10[\/latex]. We can multiply both sides by [latex]10[\/latex] to clear the decimal.\r\n<table id=\"eip-id1168467123212\" class=\"unnumbered unstyled\" summary=\"The first line says 0.8x minus 5 equals 7. The next line says, \">\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\"><\/td>\r\n<td style=\"height: 15px\">[latex]0.8x-5=7[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\">Multiply both sides by the LCD.<\/td>\r\n<td style=\"height: 15px\">[latex]\\color{red}{10}(0.8x-5)=\\color{red}{10}(7)[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">Distribute.<\/td>\r\n<td style=\"height: 24px\">[latex]10(0.8x)-10(5)=10(7)[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 23px\">\r\n<td style=\"height: 23px\">Multiply, and notice: no more decimals!<\/td>\r\n<td style=\"height: 23px\">[latex]8x-50=70[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 20px\">\r\n<td style=\"height: 20px\">Add 50 to get all constants to the right.<\/td>\r\n<td style=\"height: 20px\">[latex]8x-50\\color{red}{+50}=70\\color{red}{+50}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 20.75px\">\r\n<td style=\"height: 20.75px\">Simplify.<\/td>\r\n<td style=\"height: 20.75px\">[latex]8x=120[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px\">\r\n<td style=\"height: 42px\">Divide both sides by [latex]8[\/latex].<\/td>\r\n<td style=\"height: 42px\">[latex]\\Large\\frac{8x}{\\color{red}{8}}\\normalsize =\\Large\\frac{120}{\\color{red}{8}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px\">\r\n<td style=\"height: 19px\">Simplify.<\/td>\r\n<td style=\"height: 19px\">[latex]x=15[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\">Check: Let [latex]x=15[\/latex].<\/td>\r\n<td style=\"height: 15px\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 103px\">\r\n<td style=\"height: 103px\">[latex]0.8(\\color{red}{15})-5\\stackrel{\\text{?}}{=}7[\/latex][latex]12-5\\stackrel{\\text{?}}{=}7[\/latex]\r\n\r\n[latex]7=7\\quad\\checkmark[\/latex]<\/td>\r\n<td style=\"height: 103px\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question]3555[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]0.06x+0.02=0.25x - 1.5[\/latex]\r\n<p class=\"p1\">[reveal-answer q=\"190834\"]Show Solution[\/reveal-answer]<\/p>\r\n<p class=\"p1\">[hidden-answer a=\"190834\"]<\/p>\r\nSolution:\r\nLook at the decimals and think of the equivalent fractions.\r\n[latex]0.06=\\Large\\frac{6}{100}\\normalsize ,0.02=\\Large\\frac{2}{100}\\normalsize ,0.25=\\Large\\frac{25}{100}\\normalsize ,1.5=1\\Large\\frac{5}{10}[\/latex]\r\nNotice, the LCD is [latex]100[\/latex].\r\nBy multiplying by the LCD we will clear the decimals.\r\n<table id=\"eip-id1168466076129\" class=\"unnumbered unstyled\" summary=\"The top line says 0.06x plus 0.02 equals 0.25x minus 1.5. The next step says, \">\r\n<tbody>\r\n<tr style=\"height: 23px\">\r\n<td style=\"height: 23px\"><\/td>\r\n<td style=\"height: 23px\">[latex]0.06x+0.02=0.25x-1.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">Multiply both sides by 100.<\/td>\r\n<td style=\"height: 24px\">[latex]\\color{red}{100}(0.06x+0.02)=\\color{red}{100}(0.25x-1.5)[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 32px\">\r\n<td style=\"height: 32px\">Distribute.<\/td>\r\n<td style=\"height: 32px\">[latex]100(0.06x)+100(0.02)=100(0.25x)-100(1.5)[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px\">\r\n<td style=\"height: 14px\">Multiply, and now, no more decimals.<\/td>\r\n<td style=\"height: 14px\">[latex]6x+2=25x-150[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 23px\">\r\n<td style=\"height: 23px\">Collect the variables to the right.<\/td>\r\n<td style=\"height: 23px\">[latex]6x\\color{red}{-6x}+2=25x\\color{red}{-6x}-150[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px\">\r\n<td style=\"height: 14px\">Simplify.<\/td>\r\n<td style=\"height: 14px\">[latex]2=19x-150[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 25px\">\r\n<td style=\"height: 25px\">Collect the constants to the left.<\/td>\r\n<td style=\"height: 25px\">[latex]2\\color{red}{+150}=19x-150\\color{red}{+150}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px\">\r\n<td style=\"height: 14px\">Simplify.<\/td>\r\n<td style=\"height: 14px\">[latex]152=19x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 43px\">\r\n<td style=\"height: 43px\">Divide by [latex]19[\/latex].<\/td>\r\n<td style=\"height: 43px\">[latex]\\Large\\frac{152}{\\color{red}{19}}\\normalsize =\\Large\\frac{19x}{\\color{red}{19}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px\">\r\n<td style=\"height: 14px\">Simplify.<\/td>\r\n<td style=\"height: 14px\">[latex]8=x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px\">\r\n<td style=\"height: 14px\">Check: Let [latex]x=8[\/latex].<\/td>\r\n<td style=\"height: 14px\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 87.6426px\">\r\n<td style=\"height: 87.6426px\">[latex]0.06(\\color{red}{8})+0.02=0.25(\\color{red}{8})-1.5[\/latex][latex]0.48+0.02=2.00-1.5[\/latex]\r\n\r\n[latex]0.50=0.50\\quad\\checkmark[\/latex]<\/td>\r\n<td style=\"height: 87.6426px\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p class=\"p1\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3y+10.5=6.5+2.5y[\/latex] by clearing the decimals in the equation first.\r\n\r\n[reveal-answer q=\"159951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"159951\"]\r\n\r\nSince the smallest decimal place represented in the equation is\u00a0[latex]0.10[\/latex], we want to multiply by [latex]10[\/latex] to make \u00a0[latex]1.0[\/latex]\u00a0and clear\u00a0the decimals from the equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}3y+10.5=6.5+2.5y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 10\\left(3y+10.5\\right)=10\\left(6.5+2.5y\\right)\\end{array}[\/latex]<\/p>\r\nUse the distributive property to expand the expressions on both sides.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}10\\left(3y\\right)+10\\left(10.5\\right)=10\\left(6.5\\right)+10\\left(2.5y\\right)\\end{array}[\/latex]<\/p>\r\nMultiply.\r\n<p style=\"text-align: center\">[latex]30y+105=65+25y[\/latex]<\/p>\r\nMove the smaller variable term, [latex]25y[\/latex], by subtracting it from both sides.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}30y+105=65+25y\\,\\,\\\\ \\underline{-25y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-25y} \\\\5y+105=65\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubtract \u00a0[latex]105[\/latex] from both sides to isolate the term with the variable.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}5y+105=\\,\\,65\\,\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,-105\\,\\,\\,-105} \\\\5y=-40\\end{array}[\/latex]<\/p>\r\nDivide both sides by 5 to isolate the [latex]y[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\underline{5y}=\\underline{-40}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\\\ \\,\\,\\,y=-8\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]y=-8[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=71955&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\nIn the following video, we present another example of how to solve an equation that contains decimals and variable terms on both sides of the equal sign.\r\n\r\nhttps:\/\/youtu.be\/pZWTJvua-P8\r\n\r\nThe next example uses an equation that is typical of the ones we will see in the money applications. Note that we will distribute the decimal first, before we clear all decimals in the equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]0.25x+0.05\\left(x+3\\right)=2.85[\/latex]\r\n<p class=\"p1\">[reveal-answer q=\"777666\"]Show Solution[\/reveal-answer]<\/p>\r\n<p class=\"p1\">[hidden-answer a=\"777666\"]<\/p>\r\nSolution:\r\n<table id=\"eip-id1168466031358\" class=\"unnumbered unstyled\" summary=\"The top line says 0.25x plus 0.05 times parentheses x plus 3 equals 2.85. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]0.25x+0.05(x+3)=2.85[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute first.<\/td>\r\n<td>[latex]0.25x+0.05x+0.15=2.85[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]0.30x+0.15=2.85[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>To clear decimals, multiply by [latex]100[\/latex].<\/td>\r\n<td>[latex]\\color{red}{100}(0.30x+0.15)=\\color{red}{100}(2.85)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute.<\/td>\r\n<td>[latex]30x+15=285[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract [latex]15[\/latex] from both sides.<\/td>\r\n<td>[latex]30x+15\\color{red}{-15}=285\\color{red}{-15}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]30x=270[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide by [latex]30[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{30x}{\\color{red}{30}}\\normalsize =\\Large\\frac{270}{\\color{red}{30}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x=9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check: Let [latex]x=9[\/latex].<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0.25x+0.05(x+3)=2.85[\/latex][latex]0.25(\\color{red}{9})+0.05(\\color{red}{9}+3)\\stackrel{\\text{?}}{=}2.85[\/latex]\r\n\r\n[latex]2.25+0.05(12)\\stackrel{\\text{?}}{=}2.85[\/latex]\r\n\r\n[latex]2.85=2.85\\quad\\checkmark[\/latex]\r\n\r\n&nbsp;<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p class=\"p1\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question]140292[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve equations with fraction coefficients<\/li>\n<li>Solve equations with decimal coefficients<\/li>\n<\/ul>\n<\/div>\n<p>In this section, we will explore how to solve equations with fractions in them.\u00a0 To start, we will review how equations with fractions can be solved by multiplying by the reciprocal. We will then move on to learn how to use the Multiplication Property of Equality to eliminate the fractions from an equation, making it simpler to solve.<\/p>\n<h3>Solving equations with fractions by multiplying by the reciprocal<\/h3>\n<p>In our first example, we will solve a one-step equation using the multiplication property of equality. You will see that the variable is part of a fraction in the given equation, and using the multiplication property of equality allows us to remove the variable from the fraction. Remember that fractions imply division, so you can think of this as the variable [latex]k[\/latex] is being divided by 10. To &#8220;undo&#8221; the division, you can use multiplication to isolate [latex]k[\/latex]. (Note that there is a negative term in the equation, so it will be important to think about the sign of each term as you work through the problem. Stop after each step you take to make sure all the terms have the correct sign.)<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]-\\frac{7}{2}=\\frac{k}{10}[\/latex] for [latex]k[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471772\">Show Solution<\/span><\/p>\n<div id=\"q471772\" class=\"hidden-answer\" style=\"display: none\">\n<p>We want to isolate the [latex]k[\/latex], which is being divided by [latex]10[\/latex]. The first thing we should do is multiply both sides by [latex]10[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ Multiply the left side by 10: }\\\\\\\\\\left(10\\right)(-\\frac{7}{2})=\\frac{-10\\cdot7}{2} = \\frac{-70}{2} = -35\\\\\\\\\\text{ Now, multiply the right side by 10: }\\\\\\\\\\,\\,\\,\\frac{k}{10}\\left(10\\right) = \\frac{k\\cdot10}{10} = k\\\\\\\\\\text{ Now, replace your results in the equation: }\\\\-35=k\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>We write the [latex]k[\/latex] on the left side as a matter of convention.<\/p>\n<p>[latex]k=-35[\/latex]<\/p>\n<p style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<p>In our next example, we are given an equation that contains a variable multiplied by a fraction. We will use a reciprocal to isolate the variable.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444022\">Show Solution<\/span><\/p>\n<div id=\"q444022\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nSince the product of a number and its reciprocal is [latex]1[\/latex], our strategy will be to isolate [latex]x[\/latex] by multiplying by the reciprocal of [latex]\\Large\\frac{2}{3}[\/latex].<\/p>\n<table id=\"eip-id1168468646645\" class=\"unnumbered unstyled\" summary=\"The first line shows two-thirds x equals 18. The next line says\">\n<tbody>\n<tr style=\"height: 37px\">\n<td style=\"height: 37px\">[latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 38px\">\n<td style=\"height: 38px\">Multiply by the reciprocal of [latex]\\Large\\frac{2}{3}[\/latex] .<\/td>\n<td style=\"height: 38px\">[latex]\\Large\\frac{\\color{red}{3}}{\\color{red}{2}}\\normalsize\\cdot\\Large\\frac{2}{3}\\normalsize x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 37px\">\n<td style=\"height: 37px\">Reciprocals multiply to one.<\/td>\n<td style=\"height: 37px\">[latex]1x=\\Large\\frac{3}{2}\\normalsize\\cdot\\Large\\frac{18}{1}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 18px\">\n<td style=\"height: 18px\">Multiply.<\/td>\n<td style=\"height: 18px\">[latex]x=27[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px\">\n<td style=\"height: 14px\">Check your answer.<\/td>\n<td style=\"height: 44.8594px\">[latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 44.8594px\">\n<td style=\"height: 14px\">Let [latex]x=27[\/latex].<\/td>\n<td style=\"height: 45px\">[latex]\\Large\\frac{2}{3}\\normalsize\\cdot\\color{red}{27}\\stackrel{\\text{?}}{=}18[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td><\/td>\n<td style=\"height: 24px\">[latex]18=18\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that we could have divided both sides of the equation [latex]\\Large\\frac{2}{3}\\normalsize x=18[\/latex] by [latex]\\Large\\frac{2}{3}[\/latex] to isolate [latex]x[\/latex]. While this would work, multiplying by the reciprocal requires fewer steps.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141871\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141871&#38;theme=oea&#38;iframe_resize_id=ohm141871&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video, you will see examples of using the multiplication property of equality to solve a one-step equation involving negative fractions.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solving One Step Equations Using Multiplication (Fractions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/AhBdGeUGgsI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Solving equations with fractions by clearing the denominators<\/h3>\n<p>You may\u00a0feel overwhelmed\u00a0when you\u00a0see fractions in an equation, so we are going to show a method to solve equations with fractions where you use the common denominator to eliminate the fractions from an equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions.<\/p>\n<p>Pay attention to the fact that each term in the equation gets multiplied by the least\u00a0common denominator. That&#8217;s what makes it equal to the original!<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve: [latex]\\Large\\frac{1}{8}\\normalsize x+\\Large\\frac{1}{2}=\\Large\\frac{1}{4}[\/latex]<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168467351085\" class=\"unnumbered unstyled\" summary=\"The top line says,\">\n<tbody>\n<tr style=\"height: 30px\">\n<td style=\"height: 30px\"><\/td>\n<td style=\"height: 30px\">[latex]\\Large\\frac{1}{8}\\normalsize x+\\Large\\frac{1}{2}=\\Large\\frac{1}{4}\\normalsize\\quad{LCD=8}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 45px\">\n<td style=\"height: 45px\">Multiply both sides of the equation by that LCD, [latex]8[\/latex]. This clears the fractions.<\/td>\n<td style=\"height: 45px\">[latex]\\color{red}{8(}\\Large\\frac{1}{8}\\normalsize x+\\Large\\frac{1}{2}\\color{red}{)}=\\normalsize\\color{red}{8(}\\Large\\frac{1}{4}\\color{red}{)}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 30.9531px\">\n<td style=\"height: 30.9531px\">Use the Distributive Property.<\/td>\n<td style=\"height: 30.9531px\">[latex]8\\cdot\\Large\\frac{1}{8}\\normalsize x+8\\cdot\\Large\\frac{1}{2}\\normalsize=8\\cdot\\Large\\frac{1}{4}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 30px\">\n<td style=\"height: 30px\">Simplify \u2014 and notice, no more fractions!<\/td>\n<td style=\"height: 30px\">[latex]x+4=2[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 30px\">\n<td style=\"height: 30px\">Solve using the General Strategy for Solving Linear Equations.<\/td>\n<td style=\"height: 30px\">[latex]x+4\\color{red}{-4}=2\\color{red}{-4}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\">Simplify.<\/td>\n<td style=\"height: 15px\">[latex]x=-2[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 340px\">\n<td style=\"height: 340px\">Check: Let [latex]x=-2[\/latex][latex]\\Large\\frac{1}{8}\\normalsize x+    \\Large\\frac{1}{2}=    \\Large\\frac{1}{4}[\/latex]<\/p>\n<p>[latex]\\Large\\frac{1}{8}\\normalsize(\\color{red}{-2})+    \\Large\\frac{1}{2}\\normalsize\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{4}[\/latex]<\/p>\n<p>[latex]\\Large\\frac{-2}{8}+    \\Large\\frac{1}{2}\\normalsize\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{4}[\/latex]<\/p>\n<p>[latex]\\Large\\frac{-2}{8}+    \\Large\\frac{4}{8}\\normalsize\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{4}[\/latex]<\/p>\n<p>[latex]\\Large\\frac{2}{8}\\normalsize\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{4}[\/latex]<\/p>\n<p>[latex]\\Large\\frac{1}{4}=    \\Large\\frac{1}{4}\\quad\\checkmark[\/latex]<\/p>\n<p>&nbsp;<\/td>\n<td style=\"height: 340px\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>In the example above, the least common denominator was [latex]8[\/latex]. Now it&#8217;s your turn to find an LCD, and clear the fractions before you solve these linear equations.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm71948\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=71948&#38;theme=oea&#38;iframe_resize_id=ohm71948&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Notice\u00a0that once we cleared the equation of fractions, the equation was like those we learned how to solve earlier. We changed the problem to one we already knew how to solve!<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Solve equations by clearing the Denominators<\/h3>\n<ol id=\"eip-id1168466140919\" class=\"stepwise\">\n<li>Find the least common denominator of <em>all<\/em> the fractions in the equation.<\/li>\n<li>Multiply both sides of the equation by that LCD. This clears the fractions.<\/li>\n<li>Isolate the variable terms on one side, and the constant terms on the other side.<\/li>\n<li>Simplify both sides.<\/li>\n<li>Use the multiplication or division property to make the coefficient on the variable equal to [latex]1[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>Here&#8217;s an example where you have\u00a0three variable terms. After you clear fractions with the LCD, you will simplify the three variable terms, then isolate the variable.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]7=\\Large\\frac{1}{2}\\normalsize x+\\Large\\frac{3}{4}\\normalsize x-\\Large\\frac{2}{3}\\normalsize x[\/latex]<\/p>\n<p class=\"p1\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q190190\">Show Solution<\/span><\/p>\n<p class=\"p1\">\n<div id=\"q190190\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nWe want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.<\/p>\n<table id=\"eip-id1168468606496\" class=\"unnumbered unstyled\" summary=\"The top line says\">\n<tbody>\n<tr>\n<td>Find the least common denominator of <em>all<\/em> the fractions in the equation.<\/td>\n<td>[latex]7=\\Large\\frac{1}{2}\\normalsize x+\\Large\\frac{3}{4}\\normalsize x-\\Large\\frac{2}{3}\\normalsize x\\quad{LCD=12}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply both sides of the equation by [latex]12[\/latex].<\/td>\n<td>[latex]\\color{red}{12}(7)=\\color{red}{12}\\cdot\\Large(\\frac{1}{2}\\normalsize x+\\Large\\frac{3}{4}\\normalsize x-\\Large\\frac{2}{3}\\normalsize x\\Large)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute.<\/td>\n<td>[latex]12(7)=12\\cdot\\Large\\frac{1}{2}\\normalsize x+12\\cdot\\Large\\frac{3}{4}\\normalsize x-12\\cdot\\Large\\frac{2}{3}\\normalsize x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify \u2014 and notice, no more fractions!<\/td>\n<td>[latex]84=6x+9x-8x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]84=7x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide by [latex]7[\/latex].<\/td>\n<td>[latex]\\Large\\frac{84}{\\color{red}{7}}=\\Large\\frac{7x}{\\color{red}{7}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]12=x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check: Let [latex]x=12[\/latex].<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]7=\\Large\\frac{1}{2}\\normalsize x+    \\Large\\frac{3}{4}\\normalsize x-    \\Large\\frac{2}{3}\\normalsize x[\/latex]<\/p>\n<p>[latex]7\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{2}\\normalsize(\\color{red}{12})+    \\Large\\frac{3}{4}\\normalsize(\\color{red}{12})-    \\Large\\frac{2}{3}\\normalsize(\\color{red}{12})[\/latex]<\/p>\n<p>[latex]7\\stackrel{\\text{?}}{=}6+9-8[\/latex]<\/p>\n<p>[latex]7=7\\quad\\checkmark[\/latex]<\/p>\n<p>&nbsp;<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Now here&#8217;s a similar problem for you to try. Clear the fractions, simplify, then solve.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm71948\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=71948&#38;theme=oea&#38;iframe_resize_id=ohm71948&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox shaded\">\n<h3>Caution!<\/h3>\n<p>One of the most common mistakes when you clear fractions is forgetting to multiply BOTH sides of the equation by the LCD. If your answer doesn&#8217;t check, make sure you have multiplied both sides of the equation by the LCD.<\/p>\n<\/div>\n<p>In the next example, we\u2019ll have variables and fractions on both sides of the equation. After you clear the fractions using the LCD, you will see that this equation is similar to ones with variables on both sides that we solved previously. Remember to choose a variable side and a constant side to help you organize your work.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]x+\\Large\\frac{1}{3}=\\Large\\frac{1}{6}\\normalsize x-\\Large\\frac{1}{2}[\/latex]<\/p>\n<p class=\"p1\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888555\">Show Solution<\/span><\/p>\n<p class=\"p1\">\n<div id=\"q888555\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168468770578\" class=\"unnumbered unstyled\" summary=\"The top line says,\">\n<tbody>\n<tr>\n<td>Find the LCD of all the fractions in the equation.<\/td>\n<td>[latex]x+\\Large\\frac{1}{3}=\\Large\\frac{1}{6}\\normalsize x-\\Large\\frac{1}{2}\\normalsize,\\quad{LCD=6}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply both sides by the LCD.<\/td>\n<td>[latex]\\color{red}{6}(x+\\Large\\frac{1}{3}\\normalsize)=\\color{red}{6}(\\Large\\frac{1}{6}\\normalsize x-\\Large\\frac{1}{2})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute.<\/td>\n<td>[latex]6\\cdot{x}+6\\cdot\\Large\\frac{1}{3}\\normalsize=6\\cdot\\Large\\frac{1}{6}\\normalsize x-6\\cdot\\Large\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify \u2014 no more fractions!<\/td>\n<td>[latex]6x+2=x-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract [latex]x[\/latex] from both sides.<\/td>\n<td>[latex]6x-\\color{red}{x}+2=x-\\color{red}{x}-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]5x+2=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract 2 from both sides.<\/td>\n<td>[latex]5x+2\\color{red}{-2}=-3\\color{red}{-2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]5x=-5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide by [latex]5[\/latex].<\/td>\n<td>[latex]\\Large\\frac{5x}{\\color{red}{5}}=\\Large\\frac{-5}{\\color{red}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x=-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check: Substitute [latex]x=-1[\/latex].<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]x+\\Large\\frac{1}{3}=    \\Large\\frac{1}{6}\\normalsize x-    \\Large\\frac{1}{2}[\/latex]<\/p>\n<p>[latex](\\color{red}{-1})+    \\Large\\frac{1}{3}\\normalsize\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{6}\\normalsize(\\color{red}{-1})-    \\Large\\frac{1}{2}[\/latex]<\/p>\n<p>[latex](-1)+    \\Large\\frac{1}{3}\\normalsize\\stackrel{\\text{?}}{=}-    \\Large\\frac{1}{6}-    \\Large\\frac{1}{2}[\/latex]<\/p>\n<p>[latex]-    \\Large\\frac{3}{3}+    \\Large\\frac{1}{3}\\normalsize\\stackrel{\\text{?}}{=}-    \\Large\\frac{1}{6}-    \\Large\\frac{3}{6}[\/latex]<\/p>\n<p>[latex]-    \\Large\\frac{2}{3}\\normalsize\\stackrel{\\text{?}}{=}-    \\Large\\frac{4}{6}[\/latex]<\/p>\n<p>[latex]-    \\Large\\frac{2}{3}=-    \\Large\\frac{2}{3}\\quad\\checkmark[\/latex]<\/p>\n<p>&nbsp;<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Now you can try solving an equation with fractions that has variables on both sides of the equal sign. The answer may be a fraction.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm142514\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142514&#38;theme=oea&#38;iframe_resize_id=ohm142514&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video, we show another example of how to solve an equation that contains fractions and variables on both sides of the equal sign.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Solve an Equation with Fractions with Variable Terms on Both Sides\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/G5R9jySFMpw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we start with an equation where the variable term is locked up in parentheses and multiplied by a fraction. You can clear the fraction, or if you use the distributive property, it will eliminate the fraction. \u00a0Can you see why?<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve: [latex]1=\\Large\\frac{1}{2}\\normalsize\\left(4x+2\\right)[\/latex]<\/p>\n<p class=\"p1\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q912234\">Show Solution<\/span><\/p>\n<p class=\"p1\">\n<div id=\"q912234\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168469839370\" class=\"unnumbered unstyled\" summary=\"The top line says 1 equals one-half times parentheses 4x plus 2. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]1=\\Large\\frac{1}{2}\\normalsize(4x+2)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute.<\/td>\n<td>[latex]1=\\Large\\frac{1}{2}\\normalsize\\cdot4x+\\Large\\frac{1}{2}\\normalsize\\cdot2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify. Now there are no fractions to clear!<\/td>\n<td>[latex]1=2x+1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract 1 from both sides.<\/td>\n<td>[latex]1\\color{red}{-1}=2x+1\\color{red}{-1}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]0=2x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide by [latex]2[\/latex].<\/td>\n<td>[latex]\\Large\\frac{0}{\\color{red}{2}}=\\Large\\frac{2x}{\\color{red}{2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]0=x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check: Let [latex]x=0[\/latex].<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]1=\\Large\\frac{1}{2}\\normalsize(4x+2)[\/latex]<\/p>\n<p>[latex]1\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{2}\\normalsize(4(\\color{red}{0})+2)[\/latex]<\/p>\n<p>[latex]1\\stackrel{\\text{?}}{=}    \\Large\\frac{1}{2}\\normalsize(2)[\/latex]<\/p>\n<p>[latex]1\\stackrel{\\text{?}}{=}    \\Large\\frac{2}{2}[\/latex]<\/p>\n<p>[latex]1=1\\quad\\checkmark[\/latex]<\/p>\n<p>&nbsp;<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Now you can try solving an equation that has the variable term in parentheses that are multiplied by a fraction.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm142542\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142542&#38;theme=oea&#38;iframe_resize_id=ohm142542&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><span style=\"color: #ff0000\">\u00a0<\/span><\/p>\n<\/div>\n<p>In the following video, we show how to solve a multi-step equation with fractions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Solving an Equation with Fractions (Clear Fractions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/AvJTPeACTY0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>If you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.\u00a0\u00a0Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest! Remember to check your answer by substituting your solution into the original equation.<\/p>\n<h2>Solving Equations By Clearing Decimals<\/h2>\n<p>Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, [latex]0.3=\\Large\\frac{3}{10}[\/latex] and [latex]0.17=\\Large\\frac{17}{100}[\/latex]. So, when we have an equation with decimals, we can use the same process we used to clear fractions\u2014multiply both sides of the equation by the least common denominator.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]0.8x - 5=7[\/latex]<\/p>\n<p>Solution:<br \/>\nThe only decimal in the equation is [latex]0.8[\/latex]. Since [latex]0.8=\\Large\\frac{8}{10}[\/latex], the LCD is [latex]10[\/latex]. We can multiply both sides by [latex]10[\/latex] to clear the decimal.<\/p>\n<table id=\"eip-id1168467123212\" class=\"unnumbered unstyled\" summary=\"The first line says 0.8x minus 5 equals 7. The next line says,\">\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\"><\/td>\n<td style=\"height: 15px\">[latex]0.8x-5=7[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\">Multiply both sides by the LCD.<\/td>\n<td style=\"height: 15px\">[latex]\\color{red}{10}(0.8x-5)=\\color{red}{10}(7)[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">Distribute.<\/td>\n<td style=\"height: 24px\">[latex]10(0.8x)-10(5)=10(7)[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 23px\">\n<td style=\"height: 23px\">Multiply, and notice: no more decimals!<\/td>\n<td style=\"height: 23px\">[latex]8x-50=70[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 20px\">\n<td style=\"height: 20px\">Add 50 to get all constants to the right.<\/td>\n<td style=\"height: 20px\">[latex]8x-50\\color{red}{+50}=70\\color{red}{+50}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 20.75px\">\n<td style=\"height: 20.75px\">Simplify.<\/td>\n<td style=\"height: 20.75px\">[latex]8x=120[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 42px\">\n<td style=\"height: 42px\">Divide both sides by [latex]8[\/latex].<\/td>\n<td style=\"height: 42px\">[latex]\\Large\\frac{8x}{\\color{red}{8}}\\normalsize =\\Large\\frac{120}{\\color{red}{8}}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 19px\">\n<td style=\"height: 19px\">Simplify.<\/td>\n<td style=\"height: 19px\">[latex]x=15[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\">Check: Let [latex]x=15[\/latex].<\/td>\n<td style=\"height: 15px\"><\/td>\n<\/tr>\n<tr style=\"height: 103px\">\n<td style=\"height: 103px\">[latex]0.8(\\color{red}{15})-5\\stackrel{\\text{?}}{=}7[\/latex][latex]12-5\\stackrel{\\text{?}}{=}7[\/latex]<\/p>\n<p>[latex]7=7\\quad\\checkmark[\/latex]<\/td>\n<td style=\"height: 103px\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm3555\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3555&theme=oea&iframe_resize_id=ohm3555&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]0.06x+0.02=0.25x - 1.5[\/latex]<\/p>\n<p class=\"p1\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q190834\">Show Solution<\/span><\/p>\n<p class=\"p1\">\n<div id=\"q190834\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nLook at the decimals and think of the equivalent fractions.<br \/>\n[latex]0.06=\\Large\\frac{6}{100}\\normalsize ,0.02=\\Large\\frac{2}{100}\\normalsize ,0.25=\\Large\\frac{25}{100}\\normalsize ,1.5=1\\Large\\frac{5}{10}[\/latex]<br \/>\nNotice, the LCD is [latex]100[\/latex].<br \/>\nBy multiplying by the LCD we will clear the decimals.<\/p>\n<table id=\"eip-id1168466076129\" class=\"unnumbered unstyled\" summary=\"The top line says 0.06x plus 0.02 equals 0.25x minus 1.5. The next step says,\">\n<tbody>\n<tr style=\"height: 23px\">\n<td style=\"height: 23px\"><\/td>\n<td style=\"height: 23px\">[latex]0.06x+0.02=0.25x-1.5[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">Multiply both sides by 100.<\/td>\n<td style=\"height: 24px\">[latex]\\color{red}{100}(0.06x+0.02)=\\color{red}{100}(0.25x-1.5)[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 32px\">\n<td style=\"height: 32px\">Distribute.<\/td>\n<td style=\"height: 32px\">[latex]100(0.06x)+100(0.02)=100(0.25x)-100(1.5)[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px\">\n<td style=\"height: 14px\">Multiply, and now, no more decimals.<\/td>\n<td style=\"height: 14px\">[latex]6x+2=25x-150[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 23px\">\n<td style=\"height: 23px\">Collect the variables to the right.<\/td>\n<td style=\"height: 23px\">[latex]6x\\color{red}{-6x}+2=25x\\color{red}{-6x}-150[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px\">\n<td style=\"height: 14px\">Simplify.<\/td>\n<td style=\"height: 14px\">[latex]2=19x-150[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 25px\">\n<td style=\"height: 25px\">Collect the constants to the left.<\/td>\n<td style=\"height: 25px\">[latex]2\\color{red}{+150}=19x-150\\color{red}{+150}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px\">\n<td style=\"height: 14px\">Simplify.<\/td>\n<td style=\"height: 14px\">[latex]152=19x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 43px\">\n<td style=\"height: 43px\">Divide by [latex]19[\/latex].<\/td>\n<td style=\"height: 43px\">[latex]\\Large\\frac{152}{\\color{red}{19}}\\normalsize =\\Large\\frac{19x}{\\color{red}{19}}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px\">\n<td style=\"height: 14px\">Simplify.<\/td>\n<td style=\"height: 14px\">[latex]8=x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px\">\n<td style=\"height: 14px\">Check: Let [latex]x=8[\/latex].<\/td>\n<td style=\"height: 14px\"><\/td>\n<\/tr>\n<tr style=\"height: 87.6426px\">\n<td style=\"height: 87.6426px\">[latex]0.06(\\color{red}{8})+0.02=0.25(\\color{red}{8})-1.5[\/latex][latex]0.48+0.02=2.00-1.5[\/latex]<\/p>\n<p>[latex]0.50=0.50\\quad\\checkmark[\/latex]<\/td>\n<td style=\"height: 87.6426px\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p class=\"p1\"><\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]3y+10.5=6.5+2.5y[\/latex] by clearing the decimals in the equation first.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q159951\">Show Solution<\/span><\/p>\n<div id=\"q159951\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the smallest decimal place represented in the equation is\u00a0[latex]0.10[\/latex], we want to multiply by [latex]10[\/latex] to make \u00a0[latex]1.0[\/latex]\u00a0and clear\u00a0the decimals from the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}3y+10.5=6.5+2.5y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 10\\left(3y+10.5\\right)=10\\left(6.5+2.5y\\right)\\end{array}[\/latex]<\/p>\n<p>Use the distributive property to expand the expressions on both sides.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}10\\left(3y\\right)+10\\left(10.5\\right)=10\\left(6.5\\right)+10\\left(2.5y\\right)\\end{array}[\/latex]<\/p>\n<p>Multiply.<\/p>\n<p style=\"text-align: center\">[latex]30y+105=65+25y[\/latex]<\/p>\n<p>Move the smaller variable term, [latex]25y[\/latex], by subtracting it from both sides.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}30y+105=65+25y\\,\\,\\\\ \\underline{-25y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-25y} \\\\5y+105=65\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Subtract \u00a0[latex]105[\/latex] from both sides to isolate the term with the variable.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}5y+105=\\,\\,65\\,\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,-105\\,\\,\\,-105} \\\\5y=-40\\end{array}[\/latex]<\/p>\n<p>Divide both sides by 5 to isolate the [latex]y[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\underline{5y}=\\underline{-40}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\\\ \\,\\,\\,y=-8\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=-8[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm71955\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=71955&#38;theme=oea&#38;iframe_resize_id=ohm71955&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video, we present another example of how to solve an equation that contains decimals and variable terms on both sides of the equal sign.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Solve a Linear Equation With Decimals and Variables on Both Sides\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pZWTJvua-P8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The next example uses an equation that is typical of the ones we will see in the money applications. Note that we will distribute the decimal first, before we clear all decimals in the equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]0.25x+0.05\\left(x+3\\right)=2.85[\/latex]<\/p>\n<p class=\"p1\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q777666\">Show Solution<\/span><\/p>\n<p class=\"p1\">\n<div id=\"q777666\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168466031358\" class=\"unnumbered unstyled\" summary=\"The top line says 0.25x plus 0.05 times parentheses x plus 3 equals 2.85. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]0.25x+0.05(x+3)=2.85[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute first.<\/td>\n<td>[latex]0.25x+0.05x+0.15=2.85[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]0.30x+0.15=2.85[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>To clear decimals, multiply by [latex]100[\/latex].<\/td>\n<td>[latex]\\color{red}{100}(0.30x+0.15)=\\color{red}{100}(2.85)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute.<\/td>\n<td>[latex]30x+15=285[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract [latex]15[\/latex] from both sides.<\/td>\n<td>[latex]30x+15\\color{red}{-15}=285\\color{red}{-15}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]30x=270[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide by [latex]30[\/latex].<\/td>\n<td>[latex]\\Large\\frac{30x}{\\color{red}{30}}\\normalsize =\\Large\\frac{270}{\\color{red}{30}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x=9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check: Let [latex]x=9[\/latex].<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]0.25x+0.05(x+3)=2.85[\/latex][latex]0.25(\\color{red}{9})+0.05(\\color{red}{9}+3)\\stackrel{\\text{?}}{=}2.85[\/latex]<\/p>\n<p>[latex]2.25+0.05(12)\\stackrel{\\text{?}}{=}2.85[\/latex]<\/p>\n<p>[latex]2.85=2.85\\quad\\checkmark[\/latex]<\/p>\n<p>&nbsp;<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p class=\"p1\"><\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm140292\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=140292&theme=oea&iframe_resize_id=ohm140292&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-9037\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Question ID 142514, 142542. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License, CC-BY + GPL<\/li><li>Solve a Linear Equation with Parentheses and a Fraction 2\/3(9x-12)=8+2x. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/1dmEoG7DkN4\">https:\/\/youtu.be\/1dmEoG7DkN4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/G5R9jySFMpw\">https:\/\/youtu.be\/G5R9jySFMpw<\/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 71948. <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":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Question ID 142514, 142542\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License, CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/G5R9jySFMpw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solve a Linear Equation with Parentheses and a Fraction 2\/3(9x-12)=8+2x\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/1dmEoG7DkN4\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"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\":\"Question ID 71948\",\"author\":\"Alyson Day\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License, CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"0bdb5d89bfce4d0ebafbee429e80baf4, 232c3dffd03041118adbcb83e373cb93, 695604f5e92443f397fb6aae0d339946","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-9037","chapter","type-chapter","status-publish","hentry"],"part":7476,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/9037","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":53,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/9037\/revisions"}],"predecessor-version":[{"id":20301,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/9037\/revisions\/20301"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/7476"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/9037\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/media?parent=9037"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=9037"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=9037"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/license?post=9037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}