{"id":6586,"date":"2020-10-01T16:18:00","date_gmt":"2020-10-01T16:18:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6586"},"modified":"2021-08-26T20:04:03","modified_gmt":"2021-08-26T20:04:03","slug":"1-5-solving-equations-containing-fractions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/1-5-solving-equations-containing-fractions\/","title":{"raw":"1.5 Solving Equations Containing Fractions","rendered":"1.5 Solving Equations Containing Fractions"},"content":{"raw":"
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SECTION 1.5 Learning Objectives<\/h3>\r\n1.5: Solving Equations Containing Fractions<\/strong>\r\n
    \r\n \t
  • Use the properties of equality to solve one-step equations containing fractions<\/li>\r\n \t
  • Clear fractions in an equation and then solve the equation<\/li>\r\n \t
  • Solve multi-step equations containing fractions<\/li>\r\n \t
  • Solve a basic rational equation<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n
    \r\n

    THINK ABOUT IT<\/h3>\r\nCan you determine\u00a0what you would do differently if you were asked to solve equations like these?\r\n\r\nSolve [latex]\\frac{1}{4} + y = 3[\/latex]. What makes this example different than the previous ones? Use the box below to write down a few thoughts about how you would solve this equation with a fraction.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n\r\n[reveal-answer q=\"690980\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"690980\"]\r\n\r\nUsing the Addition Property of Equality, subtract [latex]\\frac{1}{4}[\/latex] from both sides of the equation to isolate the variable, y<\/em>. You choose to\u00a0subtract [latex]\\frac{1}{4}[\/latex] as [latex]\\frac{1}{4}[\/latex] is being added to the variable, y<\/em>.\r\n

    [latex]\\displaystyle\\begin{array}{r}\\frac{1}{4} + y\\,\\,\\,=\\,\\,\\,\\,{3}\\,\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\underline{-\\frac{1}{4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\frac{1}{4}}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,\\,=\\,3-\\frac{1}{4}\\,\\,\\end{array}[\/latex]<\/p>\r\nTo subtract\u00a0[latex]\\frac{1}{4}[\/latex] from 3, you need a common denominator.\r\n\r\nMake 3 into a fraction by dividing by 1,\u00a0[latex]\\frac{3}{1}[\/latex]. \u00a0Your denominators are 1 and 4. The least common multiple is 4.\r\n

    [latex]\\begin{array}{r}\\frac{3}{1}\\cdot\\frac{4}{4}=\\frac{12}{4}\\\\\\frac{12}{4} -\\frac{1}{4} =\\frac{11}{4}\\end{array}[\/latex]<\/p>\r\nTherefore, [latex]y =3 -\\frac{1}{4} =\\frac{11}{4}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    <\/h2>\r\n

    Use the properties of equality to solve one-step equations containing fractions<\/h2>\r\n

    Recall the addition property of equality from a previous section<\/h3>\r\n
    \r\n

    Addition Property of Equality<\/h3>\r\nFor all real numbers a<\/i>, b<\/i>, and c<\/i>: If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].\r\n\r\nIf two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal.\r\n\r\n<\/div>\r\nThe next video shows how to use the addition property of equality to solve equations with fractions.\r\n\r\nhttps:\/\/youtu.be\/O7SPM7Cs8Ds\r\n\r\n \r\n\r\nWhen you follow the steps to solve an equation, you try to isolate the variable. The variable is a quantity we don't know yet. You have a solution when you get the equation x<\/i> = some value.\r\n

    Recall the multiplication property of equality from a previous section<\/h3>\r\n
    \r\n

    Multiplication Property of Equality<\/h3>\r\nFor all real numbers a<\/i>, b<\/i>, and c<\/i>: If a<\/i> = b<\/i>, then [latex]a\\cdot{c}=b\\cdot{c}[\/latex]\u00a0(or ab<\/i> = ac<\/i>).\r\n\r\nIf two expressions are equal to each other and you multiply both sides by the same number, the resulting expressions will also be equivalent.\r\n\r\n<\/div>\r\nIn the video below you will see examples of how to use the multiplication property of equality to solve one-step equations with integers and fractions.\r\nhttps:\/\/youtu.be\/BN7iVWWl2y0\r\n\r\nIn the next example, it asks us to Solve [latex]-\\frac{7}{2}=\\frac{k}{10}[\/latex] for k<\/em>. We will solve this 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 [latex]\\frac{k}{10}[\/latex] as the variable k<\/em> is being divided by 10. To \"undo\" the division, you can use multiplication to isolate k<\/em>. Lastly, 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
    \r\n

    Example 1<\/h3>\r\nSolve [latex]-\\frac{7}{2}=\\frac{k}{10}[\/latex] for k<\/em>.\r\n\r\n[reveal-answer q=\"471772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471772\"]We want to isolate the k<\/em>, which is being divided by 10. The first thing we should do is multiply both sides by 10.\r\n\r\n \r\n

    [latex]\\begin{array}{l}\\text{ Multiply the left and right sides of the equation by 10 }\\end{array}[\/latex]<\/p>\r\n

    [latex](10)(-\\frac{7}{2})=\\frac{k}{10}(10)[\/latex]<\/p>\r\n

    [latex]\\begin{array}{l}\\text{ Simplify each side }\\end{array}[\/latex]<\/p>\r\n

    [latex]\\frac{-70}{2} = \\frac{k\\cdot10}{10}[\/latex]<\/p>\r\n

    [latex]\\begin{array}{l}\\text{ Simplify again }\\end{array}[\/latex]<\/p>\r\n

    [latex]-35 =k[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nWe write the k on the left side as a matter of convention.\r\n\r\n[latex]k=-35[\/latex]\r\n\r\n[\/hidden-answer]\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:\/\/youtu.be\/AhBdGeUGgsI\r\n

    Clear fractions in an equation and then solve the equation<\/h2>\r\nSometimes, you will encounter a multi-step equation with fractions. If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation.\r\n\r\nFinding a least<\/em> common denominator involves finding a \"Least Common Multiple\" (LCM). If you need a review on how to find a LCM, see the video below:\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=Tr75SIxNf80[\/embed]\r\n\r\nNow lets look at the example below and see how we use a common denominator to clear fractions before solving the equation.\r\n
    \r\n

    Example 2<\/h3>\r\nSolve \u00a0[latex]\\frac{1}{2}x-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.\r\n\r\n[reveal-answer q=\"129951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"129951\"]\r\n\r\nNotice the\u00a0least common multiple for the denominators of 2 and 4 is 4. We will now use this LCM to clear fractions.\r\n\r\nMultiply both sides of the equation by 4, the common denominator of the fractional coefficients.\r\n

    [latex]\\begin{array}{r}\\frac{1}{2}x-3=2-\\frac{3}{4}x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\ 4\\left(\\frac{1}{2}x-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\end{array}[\/latex]<\/p>\r\nUse the distributive property to expand the expressions on both sides.\u00a0Multiply.\r\n

    [latex]\\begin{array}{r}4\\left(\\frac{1}{2}x\\right)-4\\left(3\\right)=4\\left(2\\right)-4\\left(\\frac{3}{4}x\\right)\\\\\\\\ \\frac{4}{2}x-12=8-\\frac{12}{4}x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ 2x-12=8-3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\nAdd 3x<\/em> to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable\u00a0terms to only one side.\r\n

    [latex]\\begin{array}{r}2x-12=8-3x\\, \\\\\\underline{+3x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+3x}\\\\ 5x-12=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nAdd 12 to both sides to move the constant<\/b> terms to the other side.\r\n

    [latex]\\begin{array}{r}5x-12=8\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,+12\\,+12} \\\\5x=20\\end{array}[\/latex]<\/p>\r\nDivide to isolate the variable.\r\n

    [latex]\\begin{array}{r}\\underline{5x}=\\underline{20}\\\\ 5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\,\\,\\,\\\\ x=4\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=4[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nOf course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.\r\n

    Solving multi-step equations containing fractions<\/h2>\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 <\/span>the equation contains parenthese<\/span>s, distribute the coeffi<\/span>cient in front of the parenthese<\/span>s first, <\/span>and <\/span>then\u00a0<\/span>clear the fractions.\u00a0<\/span>In the next video, we will show an example.\r\n\r\n[embed]https:\/\/youtu.be\/HVFZdteYzXk[\/embed]\r\n
    \r\n

    Example 3<\/h3>\r\nSolve the equation [latex]\\frac{3}{2}(\\frac{5}{9}x + \\frac{4}{27})=\\frac{32}{9}[\/latex]\r\n\r\n[reveal-answer q=\"159711\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"159711\"]\r\n

    [latex]\\frac{3}{2}(\\frac{5}{9}x + \\frac{4}{27})=\\frac{32}{9}[\/latex]<\/p>\r\nDistribute [latex]\\frac{3}{2}[\/latex] through parentheses and simplify\r\n

    [latex]\\frac{5}{6}x + \\frac{2}{9}=\\frac{32}{9}[\/latex]<\/p>\r\nThe LCD =18\r\n\r\nMultiply each term by [latex](\\frac{18}{1})[\/latex]\r\n

    [latex](\\frac{18}{1})\\frac{5}{6}x + (\\frac{18}{1})\\frac{2}{9}=(\\frac{18}{1})\\frac{32}{9}[\/latex]<\/p>\r\n

    [latex]15x+4=64[\/latex]<\/p>\r\nSubtract 4 from both sides of equation\r\n

    [latex]\\begin{array}{r}15x+4=64\\,\\,\\\\ \\underline{\\,\\,\\,\\,-4\\,\\,-4\\,\\,} \\\\15x=60\\end{array}[\/latex]<\/p>\r\nDivide both sides of equation by 15\r\n

    [latex]\\begin{array}{r}\\underline{15x}=\\underline{60}\\\\ \\,\\,\\,15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,15\\,\\\\ x=4\\end{array}[\/latex]<\/p>\r\nAnswer<\/strong>\r\n\r\n[latex]x=4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere are some steps to follow when you solve multi-step equations.\r\n

    \r\n

    Steps for Solving Multi-Step Equations<\/span><\/h3>\r\n1. Simplify each side by clearing parentheses and combining like terms.\r\n\r\n2.\u00a0(Optional) Multiply to clear any fractions or decimals.\r\n\r\n3. Add or subtract to isolate the variable term\u2014you may have to move a term with the variable.\r\n\r\n4. Multiply or divide to isolate the variable.\r\n\r\n5. Check the solution.\r\n\r\n<\/div>\r\n

    Solve a basic rational equation<\/h2>\r\n

    Rational Equations<\/h3>\r\nEquations that contain fractional expressions are sometimes called\u00a0rational equations<\/strong>. For example, [latex] \\frac{2x+1}{4}=\\frac{x}{3}[\/latex] is a rational equation.\u00a0Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of\u00a0proportional relationships.\r\n\r\nThe difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions.\u00a0 In the next examples, we will clear the denominators of a rational equation with a term that has a polynomial in the numerator.\u00a0 Note: We will discuss polynomials more in depth in a later module.\u00a0 In the following example, [latex]{x+5}[\/latex] is the polynomial being referred to.\r\n
    \r\n

    Example 4<\/h3>\r\nSolve the equation [latex] \\frac{x+5}{8}=\\frac{7}{4}[\/latex].\r\n[reveal-answer q=\"425621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425621\"]\r\n\r\nFind the least common denominator of\u00a0[latex]4[\/latex] and\u00a0[latex]8[\/latex]. Remember, to find the LCD, identify the greatest number of times each factor appears in each factorization. Here,\u00a0[latex]2[\/latex] appears\u00a0[latex]3[\/latex] times, so [latex]2\\cdot2\\cdot2[\/latex], or\u00a0[latex]8[\/latex], will be the LCD.\r\n\r\nMultiply both sides of the equation by the common denominator,\u00a0[latex]8[\/latex], to keep the equation balanced and to eliminate the denominators.\r\n

    [latex]\\begin{array}{r}8\\cdot \\frac{x+5}{8}=\\frac{7}{4}\\cdot 8\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8(x+5)}{8}=\\frac{7(8)}{4}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8}{8}\\cdot (x+5)=\\frac{7(4\\cdot 2)}{4}\\\\\\\\\\frac{8}{8}\\cdot (x+5)=7\\cdot 2\\cdot \\frac{4}{4}\\\\\\\\1\\cdot (x+5)=14\\cdot 1\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSimplify and solve for x<\/i>.\r\n

    [latex]\\begin{array}{r}x+5=14\\\\x=9\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck the solution by substituting\u00a0[latex]9[\/latex] for x<\/i> in the original equation.\r\n

    [latex]\\begin{array}{r}\\frac{x+5}{8}=\\frac{7}{4}\\\\\\\\\\frac{9+5}{8}=\\frac{7}{4}\\\\\\\\\\frac{14}{8}=\\frac{7}{4}\\\\\\\\\\frac{7}{4}=\\frac{7}{4}\\end{array}[\/latex]<\/p>\r\nTherefore, [latex]x=9[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we show how to solve a rational equation with a variables on both sides of the equation.\r\n

    \r\n

    Example 5<\/h3>\r\nSolve the equation [latex] \\frac{x+3}{5}=\\frac{x+8}{3}[\/latex].\r\n\r\n[reveal-answer q=\"331190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"331190\"]\r\n\r\nFind the least common denominator of\u00a0[latex]5[\/latex] and\u00a0[latex]3[\/latex].\u00a0 Here,\u00a0[latex]5[\/latex] and [latex]3[\/latex] don't have any common factors so\u00a0 [latex]5\\cdot3=15[\/latex], will be the LCD.\r\n\r\nMultiply both sides of the equation by the common denominator,\u00a0[latex]15[\/latex], to keep the equation balanced and to eliminate the denominators.\r\n

    [latex]\\begin{array}{r}15\\cdot \\frac{x+3}{5}=\\frac{x+8}{3}\\cdot 15\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{15(x+3)}{5}=\\frac{(x+8)15}{3}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{15}{5}\\cdot (x+3)=(x+8)\\cdot\\frac{15}{3}\\\\\\\\3\\cdot (x+3)=(x+8)\\cdot 5\\\\\\\\3x+9=5x+40,\\,\\,\\end{array}[\/latex]<\/p>\r\nSimplify and start to solve for x by collecting x terms to one side.\r\n

    [latex]\\begin{array}{r}3x+9=5x+40\\,\\\\\\underline{-5x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-5x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\ -2x+9=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,40\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubtract 9 on both sides to move constants to the other side of the equal sign.\r\n

    [latex]\\begin{array}{r}-2x+9=40\\,\\,\\\\ \\underline{\\,\\,\\,\\,\\,\\,-9\\,\\,\\,\\,-9}\\, \\\\-2x=31\\end{array}[\/latex]<\/p>\r\nDivide by -2 on both sides to isolate the x term.\r\n

    [latex]\\begin{array}{r}\\frac{-2x}{-2}=\\frac{31}{-2}\\end{array}[\/latex]<\/p>\r\n

    [latex]x=-\\frac{31}{2}=-15.5[\/latex]<\/p>\r\nCheck the solution by substituting\u00a0[latex]-15.5[\/latex] for x<\/i> in the original equation.\r\n

    [latex]\\begin{array}{r}\\frac{x+3}{5}=\\frac{x+8}{3}\\\\\\\\\\frac{-15.5+3}{5}=\\frac{-15.5+8}{3}\\\\\\\\\\frac{-12.5}{5}=\\frac{-7.5}{3}\\\\\\\\2.5=2.5\\end{array}[\/latex]<\/p>\r\nTherefore, [latex]x=-15.5[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"

    \n

    SECTION 1.5 Learning Objectives<\/h3>\n

    1.5: Solving Equations Containing Fractions<\/strong><\/p>\n

      \n
    • Use the properties of equality to solve one-step equations containing fractions<\/li>\n
    • Clear fractions in an equation and then solve the equation<\/li>\n
    • Solve multi-step equations containing fractions<\/li>\n
    • Solve a basic rational equation<\/li>\n<\/ul>\n<\/div>\n

       <\/p>\n

      \n

      THINK ABOUT IT<\/h3>\n

      Can you determine\u00a0what you would do differently if you were asked to solve equations like these?<\/p>\n

      Solve [latex]\\frac{1}{4} + y = 3[\/latex]. What makes this example different than the previous ones? Use the box below to write down a few thoughts about how you would solve this equation with a fraction.<\/p>\n