{"id":1178,"date":"2021-12-02T19:36:52","date_gmt":"2021-12-03T00:36:52","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/2-4-fractional-linear-equations\/"},"modified":"2023-08-30T12:14:19","modified_gmt":"2023-08-30T16:14:19","slug":"fractional-linear-equations","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/fractional-linear-equations\/","title":{"raw":"2.4 Fractional Linear Equations","rendered":"2.4 Fractional Linear Equations"},"content":{"raw":"When working with fractions built into linear equations, it is often easiest to remove the fraction in the very first step. This generally means finding the LCD of the fraction and then multiplying every term in the entire equation by the LCD.\r\n
\r\n

Example 2.4.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for [latex]x[\/latex] in the equation [latex]\\dfrac{3}{4}x - \\dfrac{7}{2} = \\dfrac{5}{6}.[\/latex]\r\n\r\nFor this equation, the\u00a0LCD is\u00a012, so every term in this equation will be multiplied by\u00a012.\r\n\r\n[latex]\\dfrac{3}{4}x(12) - \\dfrac{7}{2}(12) = \\dfrac{5}{6}(12)[\/latex]\r\n\r\nCancelling out the denominator yields:\r\n\r\n[latex]3x(3) - 7(6) = 5(2)[\/latex]\r\n\r\nMultiplying results in:\r\n\r\n[latex]\\begin{array}{rrrrr}\r\n9x&-&42&=&10 \\\\\r\n&+&42&&+42 \\\\\r\n\\hline\r\n&&\\dfrac{9x}{9}&=&\\dfrac{52}{9} \\\\ \\\\\r\n&&x&=&\\dfrac{52}{9}\r\n\\end{array}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 2.4.2<\/p>\r\n\r\n<\/header>\r\n

Solve for [latex]x[\/latex] in the equation [latex]\\dfrac{3\\left(\\dfrac{5}{9}x+\\dfrac{4}{27}\\right)}{2}=3.[\/latex]<\/div>\r\n
First, remove the outside denominator\u00a02 by multiplying both sides by\u00a02:<\/div>\r\n
[latex]\\left(2\\right)\\dfrac{3\\left(\\dfrac{5}{9}x+\\dfrac{4}{27}\\right)}{2}=3(2)[\/latex]<\/div>\r\n
[latex]3\\left(\\dfrac{5}{9}x+\\dfrac{4}{27}\\right)=6[\/latex]<\/div>\r\n
Now divide both sides by\u00a03, which leaves:<\/div>\r\n
[latex]\\dfrac{5}{9}x + \\dfrac{4}{27} = 2[\/latex]<\/div>\r\n
To remove the\u00a09\u00a0and\u00a027, multiply both sides by the\u00a0LCD, 27:<\/div>\r\n
[latex]\\dfrac{5}{9}x\\left(27\\right) + \\dfrac{4}{27}\\left(27\\right) = 2(27)[\/latex]<\/div>\r\n
This leaves:<\/div>\r\n
[latex]\\begin{array}{rrrrl}\r\n5x(3)&+&4&=&54 \\\\\r\n&-&4&&-4 \\\\\r\n\\hline\r\n&&15x&=&50 \\\\ \\\\\r\n&&x&=&\\dfrac{50}{15}\\text{ or }\\dfrac{10}{3}\r\n\\end{array}[\/latex]<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nFor questions 1 to 18, solve each linear equation.\r\n
    \r\n \t
  1. [latex]\\dfrac{3}{5}\\left(1 + p\\right) = \\dfrac{21}{20}[\/latex]<\/li>\r\n \t
  2. [latex]-\\dfrac{1}{2} = \\dfrac{3k}{2} + \\dfrac{3}{2}[\/latex]<\/li>\r\n \t
  3. [latex]0 = -\\dfrac{5}{4}\\left(x-\\dfrac{6}{5}\\right)[\/latex]<\/li>\r\n \t
  4. [latex]\\dfrac{3}{2}n - 8 = -\\dfrac{29}{12}[\/latex]<\/li>\r\n \t
  5. [latex]\\dfrac{3}{4} - \\dfrac{5}{4}m = \\dfrac{108}{24}[\/latex]<\/li>\r\n \t
  6. [latex]\\dfrac{11}{4} + \\dfrac{3}{4}r = \\dfrac{160}{32}[\/latex]<\/li>\r\n \t
  7. [latex]2b + \\dfrac{9}{5} = -\\dfrac{11}{5}[\/latex]<\/li>\r\n \t
  8. [latex]\\dfrac{3}{2} - \\dfrac{7}{4}v = -\\dfrac{9}{8}[\/latex]<\/li>\r\n \t
  9. [latex]\\dfrac{3}{2}\\left(\\dfrac{7}{3}n+1\\right) = \\dfrac{3}{2}[\/latex]<\/li>\r\n \t
  10. [latex]\\dfrac{41}{9} = \\dfrac{5}{2}\\left(x+\\dfrac{2}{3}\\right) - \\dfrac{1}{3}x[\/latex]<\/li>\r\n \t
  11. [latex]-a - \\dfrac{5}{4}\\left(-\\dfrac{8}{3}a+ 1\\right) = -\\dfrac{19}{4}[\/latex]<\/li>\r\n \t
  12. [latex]\\dfrac{1}{3}\\left(-\\dfrac{7}{4}k + 1\\right) - \\dfrac{10}{3}k = -\\dfrac{13}{8}[\/latex]<\/li>\r\n \t
  13. [latex]\\dfrac{55}{6} = -\\dfrac{5}{2}\\left(\\dfrac{3}{2}p-\\dfrac{5}{3}\\right)[\/latex]<\/li>\r\n \t
  14. [latex]-\\dfrac{1}{2}\\left(\\dfrac{2}{3}x-\\dfrac{3}{4}\\right)-\\dfrac{7}{2}x=-\\dfrac{83}{24}[\/latex]<\/li>\r\n \t
  15. [latex]-\\dfrac{5}{8}=\\dfrac{5}{4}\\left(r-\\dfrac{3}{2}\\right)[\/latex]<\/li>\r\n \t
  16. [latex]\\dfrac{1}{12}=\\dfrac{4}{3}x+\\dfrac{5}{3}\\left(x-\\dfrac{7}{4}\\right)[\/latex]<\/li>\r\n \t
  17. [latex]-\\dfrac{11}{3}+\\dfrac{3}{2}b=\\dfrac{5}{2}\\left(b-\\dfrac{5}{3}\\right)[\/latex]<\/li>\r\n \t
  18. [latex]\\dfrac{7}{6}-\\dfrac{4}{3}n=-\\dfrac{3}{2}n+2\\left(n+\\dfrac{3}{2}\\right)[\/latex]<\/li>\r\n<\/ol>\r\nAnswer Key 2.4<\/a>","rendered":"

    When working with fractions built into linear equations, it is often easiest to remove the fraction in the very first step. This generally means finding the LCD of the fraction and then multiplying every term in the entire equation by the LCD.<\/p>\n

    \n
    \n

    Example 2.4.1<\/p>\n<\/header>\n

    \n

    Solve for [latex]x[\/latex] in the equation [latex]\\dfrac{3}{4}x - \\dfrac{7}{2} = \\dfrac{5}{6}.[\/latex]<\/p>\n

    For this equation, the\u00a0LCD is\u00a012, so every term in this equation will be multiplied by\u00a012.<\/p>\n

    [latex]\\dfrac{3}{4}x(12) - \\dfrac{7}{2}(12) = \\dfrac{5}{6}(12)[\/latex]<\/p>\n

    Cancelling out the denominator yields:<\/p>\n

    [latex]3x(3) - 7(6) = 5(2)[\/latex]<\/p>\n

    Multiplying results in:<\/p>\n

    [latex]\\begin{array}{rrrrr} 9x&-&42&=&10 \\\\ &+&42&&+42 \\\\ \\hline &&\\dfrac{9x}{9}&=&\\dfrac{52}{9} \\\\ \\\\ &&x&=&\\dfrac{52}{9} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 2.4.2<\/p>\n<\/header>\n

    Solve for [latex]x[\/latex] in the equation [latex]\\dfrac{3\\left(\\dfrac{5}{9}x+\\dfrac{4}{27}\\right)}{2}=3.[\/latex]<\/div>\n
    First, remove the outside denominator\u00a02 by multiplying both sides by\u00a02:<\/div>\n
    [latex]\\left(2\\right)\\dfrac{3\\left(\\dfrac{5}{9}x+\\dfrac{4}{27}\\right)}{2}=3(2)[\/latex]<\/div>\n
    [latex]3\\left(\\dfrac{5}{9}x+\\dfrac{4}{27}\\right)=6[\/latex]<\/div>\n
    Now divide both sides by\u00a03, which leaves:<\/div>\n
    [latex]\\dfrac{5}{9}x + \\dfrac{4}{27} = 2[\/latex]<\/div>\n
    To remove the\u00a09\u00a0and\u00a027, multiply both sides by the\u00a0LCD, 27:<\/div>\n
    [latex]\\dfrac{5}{9}x\\left(27\\right) + \\dfrac{4}{27}\\left(27\\right) = 2(27)[\/latex]<\/div>\n
    This leaves:<\/div>\n
    [latex]\\begin{array}{rrrrl} 5x(3)&+&4&=&54 \\\\ &-&4&&-4 \\\\ \\hline &&15x&=&50 \\\\ \\\\ &&x&=&\\dfrac{50}{15}\\text{ or }\\dfrac{10}{3} \\end{array}[\/latex]<\/div>\n<\/div>\n

    Questions<\/h1>\n

    For questions 1 to 18, solve each linear equation.<\/p>\n

      \n
    1. [latex]\\dfrac{3}{5}\\left(1 + p\\right) = \\dfrac{21}{20}[\/latex]<\/li>\n
    2. [latex]-\\dfrac{1}{2} = \\dfrac{3k}{2} + \\dfrac{3}{2}[\/latex]<\/li>\n
    3. [latex]0 = -\\dfrac{5}{4}\\left(x-\\dfrac{6}{5}\\right)[\/latex]<\/li>\n
    4. [latex]\\dfrac{3}{2}n - 8 = -\\dfrac{29}{12}[\/latex]<\/li>\n
    5. [latex]\\dfrac{3}{4} - \\dfrac{5}{4}m = \\dfrac{108}{24}[\/latex]<\/li>\n
    6. [latex]\\dfrac{11}{4} + \\dfrac{3}{4}r = \\dfrac{160}{32}[\/latex]<\/li>\n
    7. [latex]2b + \\dfrac{9}{5} = -\\dfrac{11}{5}[\/latex]<\/li>\n
    8. [latex]\\dfrac{3}{2} - \\dfrac{7}{4}v = -\\dfrac{9}{8}[\/latex]<\/li>\n
    9. [latex]\\dfrac{3}{2}\\left(\\dfrac{7}{3}n+1\\right) = \\dfrac{3}{2}[\/latex]<\/li>\n
    10. [latex]\\dfrac{41}{9} = \\dfrac{5}{2}\\left(x+\\dfrac{2}{3}\\right) - \\dfrac{1}{3}x[\/latex]<\/li>\n
    11. [latex]-a - \\dfrac{5}{4}\\left(-\\dfrac{8}{3}a+ 1\\right) = -\\dfrac{19}{4}[\/latex]<\/li>\n
    12. [latex]\\dfrac{1}{3}\\left(-\\dfrac{7}{4}k + 1\\right) - \\dfrac{10}{3}k = -\\dfrac{13}{8}[\/latex]<\/li>\n
    13. [latex]\\dfrac{55}{6} = -\\dfrac{5}{2}\\left(\\dfrac{3}{2}p-\\dfrac{5}{3}\\right)[\/latex]<\/li>\n
    14. [latex]-\\dfrac{1}{2}\\left(\\dfrac{2}{3}x-\\dfrac{3}{4}\\right)-\\dfrac{7}{2}x=-\\dfrac{83}{24}[\/latex]<\/li>\n
    15. [latex]-\\dfrac{5}{8}=\\dfrac{5}{4}\\left(r-\\dfrac{3}{2}\\right)[\/latex]<\/li>\n
    16. [latex]\\dfrac{1}{12}=\\dfrac{4}{3}x+\\dfrac{5}{3}\\left(x-\\dfrac{7}{4}\\right)[\/latex]<\/li>\n
    17. [latex]-\\dfrac{11}{3}+\\dfrac{3}{2}b=\\dfrac{5}{2}\\left(b-\\dfrac{5}{3}\\right)[\/latex]<\/li>\n
    18. [latex]\\dfrac{7}{6}-\\dfrac{4}{3}n=-\\dfrac{3}{2}n+2\\left(n+\\dfrac{3}{2}\\right)[\/latex]<\/li>\n<\/ol>\n

      Answer Key 2.4<\/a><\/p>\n","protected":false},"author":90,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1178","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1170,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1178","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1178\/revisions"}],"predecessor-version":[{"id":2075,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1178\/revisions\/2075"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1170"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1178\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1178"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1178"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1178"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1178"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}