Answer Key 2.4

  1. \phantom{1}
    \left(\dfrac{3}{5}\left(1 + p\right) = \dfrac{21}{20}\right)(20) \\
    \dfrac{3}{\cancel{5}1}\cdot \cancel{20 }4(1 + p) = \dfrac{21}{\cancel{20}1}\cdot \cancel{20} \\
    \begin{array}{rrrrr} 12&+&12p&=&\phantom{-}21 \\ -12&&&&-12 \\ \midrule &&\dfrac{12p}{12}&=&\dfrac{9}{12} \\ \\ &&p&=&\dfrac{3}{4} \end{array}
  2. \phantom{1}
    \left(-\dfrac{1}{2} = \dfrac{3k}{2} + \dfrac{3}{2}\right)(2) \\
    \begin{array}{rrrrr} -1&=&3k&+&3 \\ -3&&&-&3 \\ \midrule \dfrac{-4}{3}&=&\dfrac{3k}{3}&& \\ \\ k&=&-\dfrac{4}{3}&& \end{array}
  3. \phantom{1}
    \left(0 = -\dfrac{5}{4}x+\dfrac{6}{4}\right)(4) \\
    \begin{array}{rrlrr} 0&=&-5x&+&6 \\ +5x&&+5x&& \\ \midrule \dfrac{5x}{5}&=&\dfrac{6}{5}&& \\ \\ x&=&\dfrac{6}{5}&& \end{array}
  4. \phantom{1}
    \left(\dfrac{3n}{2} - 8 = -\dfrac{29}{12}\right)(12) \\
    \begin{array}{rrrrr} 18n&-&96&=&-29 \\ &+&96&&+96 \\ \midrule &&\dfrac{18n}{18}&=&\dfrac{67}{18} \\ \\ &&n&=&\dfrac{67}{18} \end{array}
  5. \phantom{1}
    \left(\dfrac{3}{4} - \dfrac{5}{4}m = \dfrac{108}{24}\right)(24) \\
    \begin{array}{rrrrr} 18&-&30m&=&108 \\ -18&&&&-18 \\ \midrule &&\dfrac{-30m}{-30}&=&\dfrac{90}{-30} \\ \\ &&m&=&-3 \end{array}
  6. \phantom{1}
    \left(\dfrac{11}{4} + \dfrac{3}{4}r = \dfrac{160}{32}\right)(32) \\
    \begin{array}{crcrr} 11\cdot 8&+&3r\cdot 8&=&160 \\ \phantom{-}88&+&24r&=&160 \\ -88&&&&-88 \\ \midrule &&\dfrac{24r}{24}&=&\dfrac{72}{24} \\ \\ &&r&=&3 \end{array}
  7. \phantom{1}
    \left(2b + \dfrac{9}{5} = -\dfrac{11}{5}\right)(5) \\
    \begin{array}{rrrrr} 10b&+&9&=&-11 \\ &-&9&&-9 \\ \midrule &&\dfrac{10b}{10}&=&\dfrac{-20}{10} \\ \\ &&b&=&-2 \end{array}
  8. \phantom{1}
    \left(\dfrac{3}{2} - \dfrac{7v}{4} = -\dfrac{9}{8}\right)(8) \\
    \begin{array}{rrrrr} 3\cdot 4&-&7r\cdot 2&=&-9 \\ \phantom{-}12&-&14r&=&-9 \\ -12&&&&-12 \\ \midrule &&\dfrac{-14r}{-14}&=&\dfrac{-21}{-14} \\ \\ &&r&=&\dfrac{3}{2} \end{array}
  9. \phantom{1}
    \left(\dfrac{21n}{6}+\dfrac{3}{2} = \dfrac{3}{2}\right)(6) \\
    \begin{array}{rrrrr} \dfrac{21n}{6}&+&3\cdot 3&=&3\cdot 3 \\ &-&9&&-9 \\ \midrule &&\dfrac{21n}{6}&=&0 \\ \\ &&n&=&0 \end{array}
  10. \phantom{1}
    \left(\dfrac{41}{9} = \dfrac{5}{2}x+\dfrac{10}{6} - \dfrac{1}{3}x\right)(18) \\
    \begin{array}{rrlrrrr} 41\cdot 2&=&5x\cdot 9&+&10\cdot 3&-&x\cdot 6 \\ 82&=&45x&+&30&-&6x \\ -30&&&-&30&& \\ \midrule \dfrac{52}{39}&=&\dfrac{39x}{39}&&&& \\ \\ x&=&\dfrac{4}{3}&&&& \end{array}
  11. \phantom{1}
    \left(-a + \dfrac{40a}{12}-\dfrac{5}{4}= -\dfrac{19}{4}\right)(12) \\
    \begin{array}{rrrrrrr} -12a&+&40a&-&15&=&-57 \\ &&&+&15&&+15 \\ \midrule &&&&\dfrac{28a}{28}&=&\dfrac{-42}{28} \\ \\ &&&&a&=&-\dfrac{3}{2} \end{array}
  12. \phantom{1}
    \left(-\dfrac{7k}{12}+\dfrac{1}{3}-\dfrac{10k}{3}=-\dfrac{13}{8} \right)(24) \\
    \begin{array}{rrrrrrr} -14k&+&8&-&80k&=&-39 \\ &-&8&&&&-8 \\ \midrule &&&&\dfrac{-94k}{-94}&=&\dfrac{-47}{-94} \\ \\ &&&&k&=&\dfrac{1}{2} \end{array}
  13. \phantom{1}
    \left(\dfrac{55}{6} = -\dfrac{15p}{4}+\dfrac{25}{6}\right)(12) \\
    \begin{array}{rrlrr} 110&=&-45p&+&50 \\ -50&&&-&50 \\ \midrule \dfrac{60}{-45}&=&\dfrac{-45p}{-45}&& \\ \\ p&=&-\dfrac{4}{3}&& \end{array}
  14. \phantom{1}
    \left(-\dfrac{2x}{6}+\dfrac{3}{8}-\dfrac{7x}{2}=-\dfrac{83}{24}\right)(24) \\
    \begin{array}{rrrrrrr} -8x&+&9&-&84x&=&-83 \\ &-&9&&&&-9 \\ \midrule &&&&\dfrac{-92x}{-92}&=&\dfrac{-92}{-92} \\ \\ &&&&x&=&1 \end{array}
  15. \phantom{1}
    \left(-\dfrac{5}{8}=\dfrac{5}{4}r-\dfrac{15}{8}\right)(8) \\
    \begin{array}{rrlrr} -5&=&10r&-&15 \\ +15&&&+&15 \\ \midrule \dfrac{10}{10}&=&\dfrac{10r}{10}&& \\ \\ r&=&1&& \end{array}
  16. \phantom{1}
    \left(\dfrac{1}{12}=\dfrac{4x}{3}+\dfrac{5x}{3}-\dfrac{35}{12}\right)(12) \\
    \begin{array}{rrlrrrr} 1&=&16x&+&20x&-&35 \\ +35&&&&&+&35 \\ \midrule \dfrac{36}{36}&=&\dfrac{36x}{36}&&&& \\ \\ x&=&1&&&& \end{array}
  17. \phantom{1}
    \left(-\dfrac{11}{3}+\dfrac{3b}{2}=\dfrac{5b}{2}-\dfrac{25}{6}\right)(6) \\
    \begin{array}{rrrrrrr} -22&+&9b&=&15b&-&25 \\ +22&-&15b&&-15b&+&22 \\ \midrule &&\dfrac{-6b}{-6}&=&\dfrac{-3}{-6}&& \\ \\ &&b&=&\dfrac{1}{2}&& \end{array}
  18. \phantom{1}
    \left(\dfrac{7}{6}-\dfrac{4n}{3}=-\dfrac{3n}{2}+2n+3\right)(6) \\
    \begin{array}{rrrrlrrrr} 7&-&8n&=&-9n&+&12n&+&18 \\ -7&+&9n&&+9n&-&12n&-&7 \\ &-&12n&&&&&& \\ \midrule &&\dfrac{-11n}{-11}&=&\dfrac{11}{-11}&&&& \\ \\ &&n&=&-1&&&& \end{array}

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book