{"id":654,"date":"2019-04-29T16:22:31","date_gmt":"2019-04-29T20:22:31","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=654"},"modified":"2019-12-05T11:33:35","modified_gmt":"2019-12-05T16:33:35","slug":"8-6-solving-complex-fractions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/8-6-solving-complex-fractions\/","title":{"raw":"8.6 Solving Complex Fractions","rendered":"8.6 Solving Complex Fractions"},"content":{"raw":"[latexpage]\r\n\r\nWhen solving two or more equated fractions, the easiest solution is to first remove all fractions by multiplying both sides of the equations by the LCD. This strategy is shown in the next examples.\r\n
\r\n

Example 8.6.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve \\(\\dfrac{x+3}{4}=\\dfrac{2}{3}\\).\r\n\r\nFor these two fractions, the LCD is 3\u00a0\u00d7 4 = 12. Therefore, we multiply both sides of the equation by 12:\r\n

\\(12\\left(\\dfrac{x+3}{4}\\right)=\\left(\\dfrac{2}{3}\\right)12\\)<\/p>\r\nThis reduces the complex fraction to:\r\n

\\(3(x+3)=2(4)\\)<\/p>\r\nMultiplying this out yields:\r\n

\\(3x + 9\u00a0 =\u00a0 8\\)<\/p>\r\nNow just isolate and solve for \\(x\\):\r\n

\\(\\begin{array}{rrrrr}\r\n3x&+&9&=&8 \\\\\r\n&-&9&&-9 \\\\\r\n\\midrule\r\n&&3x&=&-1 \\\\ \\\\\r\n&&x&=&-\\dfrac{1}{3}\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 8.6.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve \\(\\dfrac{2x-3}{3x+4} = \\dfrac{2}{5}\\).\r\n\r\nFor these two fractions, the LCD is \\(5(3x + 4)\\). Therefore, both sides of the equation are multiplied by \\(5(3x + 4)\\):\r\n

\\(5(3x+4)\\left(\\dfrac{2x-3}{3x+4}\\right)=\\left(\\dfrac{2}{5}\\right)5(3x+4)\\)<\/p>\r\nThis reduces the complex fraction to:\r\n

\\(5(2x-3)=2(3x+4)\\)<\/p>\r\nMultiplying this out yields:\r\n

\\(10x - 15\u00a0 =\u00a0 6x + 8\\)<\/p>\r\nNow isolate and solve for \\(x\\):\r\n

\\(\\begin{array}{rrrrrrr}\r\n10x&-&15&=&6x&+&8 \\\\\r\n-6x&+&15&&-6x&+&15 \\\\\r\n\\midrule\r\n&&4x&=&23&& \\\\ \\\\\r\n&&x&=&\\dfrac{23}{4}&&\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 8.6.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve \\(\\dfrac{k+3}{3}= \\dfrac{8}{k-2}\\).\r\n\r\nFor these two fractions, the LCD is \\(3(k-2)\\). Therefore,\u00a0 multiply both sides of the equation by \\(3(k-2)\\):\r\n

\\(3(k-2)\\left(\\dfrac{k+3}{3}\\right)=\\left(\\dfrac{8}{k-2}\\right)3(k-2)\\)<\/p>\r\nThis reduces the complex fraction to:\r\n

\\((k - 2) (k + 3)\u00a0 =\u00a0 8 (3)\\)<\/p>\r\nThis multiplies out to:\r\n

\\(k^2 + k - 6\u00a0 =\u00a0 24\\)<\/p>\r\nNow subtract 24 from both sides of the equation to turn this into an equation that can be easily factored:\r\n

\\(\\begin{array}{rrrrrrr}\r\nk^2&+&k&-&6&=&24 \\\\\r\n&&&-&24&&-24 \\\\\r\n\\midrule\r\nk^2&+&k&-&30&=&0\r\n\\end{array}\\)<\/p>\r\nThis equation factors to:\r\n

\\((k + 6)(k - 5) = 0\\)<\/p>\r\nThe solutions are:\r\n

\\(k = -6\\) and \\(k=5\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nSolve each of the following complex fractions.\r\n
    \r\n \t
  1. \\(\\dfrac{m-1}{5}=\\dfrac{8}{2}\\)<\/li>\r\n \t
  2. \\(\\dfrac{8}{2}=\\dfrac{8}{x-8}\\)<\/li>\r\n \t
  3. \\(\\dfrac{2}{9}=\\dfrac{10}{p-4}\\)<\/li>\r\n \t
  4. \\(\\dfrac{9}{n+2}=\\dfrac{3}{9}\\)<\/li>\r\n \t
  5. \\(\\dfrac{3}{10}=\\dfrac{a}{a+2}\\)<\/li>\r\n \t
  6. \\(\\dfrac{x+1}{3}=\\dfrac{x+3}{4}\\)<\/li>\r\n \t
  7. \\(\\dfrac{2}{p+4}=\\dfrac{p+5}{3}\\)<\/li>\r\n \t
  8. \\(\\dfrac{5}{n+1}=\\dfrac{n-4}{10}\\)<\/li>\r\n \t
  9. \\(\\dfrac{x+5}{5}=\\dfrac{6}{x-2}\\)<\/li>\r\n \t
  10. \\(\\dfrac{4}{x-3}=\\dfrac{x+5}{5}\\)<\/li>\r\n \t
  11. \\(\\dfrac{m+3}{4}=\\dfrac{11}{m-4}\\)<\/li>\r\n \t
  12. \\(\\dfrac{x-5}{8}=\\dfrac{4}{x-1}\\)<\/li>\r\n<\/ol>\r\nAnswer Key 8.6<\/a>","rendered":"

    When solving two or more equated fractions, the easiest solution is to first remove all fractions by multiplying both sides of the equations by the LCD. This strategy is shown in the next examples.<\/p>\n

    \n
    \n

    Example 8.6.1<\/p>\n<\/header>\n

    \n

    Solve \"\dfrac{x+3}{4}=\dfrac{2}{3}\".<\/p>\n

    For these two fractions, the LCD is 3\u00a0\u00d7 4 = 12. Therefore, we multiply both sides of the equation by 12:<\/p>\n

    \"12\left(\dfrac{x+3}{4}\right)=\left(\dfrac{2}{3}\right)12\"<\/p>\n

    This reduces the complex fraction to:<\/p>\n

    \"3(x+3)=2(4)\"<\/p>\n

    Multiplying this out yields:<\/p>\n

    \"3x + 9 = 8\"<\/p>\n

    Now just isolate and solve for \"x\":<\/p>\n

    \"\begin{array}{rrrrr}<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 8.6.2<\/p>\n<\/header>\n

    \n

    Solve \"\dfrac{2x-3}{3x+4} = \dfrac{2}{5}\".<\/p>\n

    For these two fractions, the LCD is \"5(3x + 4)\". Therefore, both sides of the equation are multiplied by \"5(3x + 4)\":<\/p>\n

    \"5(3x+4)\left(\dfrac{2x-3}{3x+4}\right)=\left(\dfrac{2}{5}\right)5(3x+4)\"<\/p>\n

    This reduces the complex fraction to:<\/p>\n

    \"5(2x-3)=2(3x+4)\"<\/p>\n

    Multiplying this out yields:<\/p>\n

    \"10x - 15 = 6x + 8\"<\/p>\n

    Now isolate and solve for \"x\":<\/p>\n

    \"\begin{array}{rrrrrrr}<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 8.6.3<\/p>\n<\/header>\n

    \n

    Solve \"\dfrac{k+3}{3}= \dfrac{8}{k-2}\".<\/p>\n

    For these two fractions, the LCD is \"3(k-2)\". Therefore,\u00a0 multiply both sides of the equation by \"3(k-2)\":<\/p>\n

    \"3(k-2)\left(\dfrac{k+3}{3}\right)=\left(\dfrac{8}{k-2}\right)3(k-2)\"<\/p>\n

    This reduces the complex fraction to:<\/p>\n

    \"(k - 2) (k + 3) = 8 (3)\"<\/p>\n

    This multiplies out to:<\/p>\n

    \"k^2 + k - 6 = 24\"<\/p>\n

    Now subtract 24 from both sides of the equation to turn this into an equation that can be easily factored:<\/p>\n

    \"\begin{array}{rrrrrrr}<\/p>\n

    This equation factors to:<\/p>\n

    \"(k + 6)(k - 5) = 0\"<\/p>\n

    The solutions are:<\/p>\n

    \"k = -6\" and \"k=5\"<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Solve each of the following complex fractions.<\/p>\n

      \n
    1. \"\dfrac{m-1}{5}=\dfrac{8}{2}\"<\/li>\n
    2. \"\dfrac{8}{2}=\dfrac{8}{x-8}\"<\/li>\n
    3. \"\dfrac{2}{9}=\dfrac{10}{p-4}\"<\/li>\n
    4. \"\dfrac{9}{n+2}=\dfrac{3}{9}\"<\/li>\n
    5. \"\dfrac{3}{10}=\dfrac{a}{a+2}\"<\/li>\n
    6. \"\dfrac{x+1}{3}=\dfrac{x+3}{4}\"<\/li>\n
    7. \"\dfrac{2}{p+4}=\dfrac{p+5}{3}\"<\/li>\n
    8. \"\dfrac{5}{n+1}=\dfrac{n-4}{10}\"<\/li>\n
    9. \"\dfrac{x+5}{5}=\dfrac{6}{x-2}\"<\/li>\n
    10. \"\dfrac{4}{x-3}=\dfrac{x+5}{5}\"<\/li>\n
    11. \"\dfrac{m+3}{4}=\dfrac{11}{m-4}\"<\/li>\n
    12. \"\dfrac{x-5}{8}=\dfrac{4}{x-1}\"<\/li>\n<\/ol>\n

      Answer Key 8.6<\/a><\/p>\n","protected":false},"author":540,"menu_order":22,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-654","chapter","type-chapter","status-publish","hentry"],"part":386,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/654","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":9,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/654\/revisions"}],"predecessor-version":[{"id":3631,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/654\/revisions\/3631"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/386"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/654\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=654"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=654"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=654"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}