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Answer Key 2.5

  1. \(x=\pm 8\)
  2. \(n=\pm 7\)
  3. \(b=\pm 1\)
  4. \(x=\pm 2\)
  5. \(\begin{array}{ll}
    \\ \\ \\ \\ \\
    \begin{array}{rrrrr}
    5&+&8a&=&53 \\
    -5&&&&-5 \\
    \hline
    &&\dfrac{8a}{8}&=&\dfrac{48}{8} \\ \\
    &&a&=&6
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrl}
    5&+&8a&=&-53 \\
    -5&&&&-5 \\
    \hline
    &&\dfrac{8a}{8}&=&\dfrac{-58}{8} \\ \\
    &&a&=&-\dfrac{58}{8}\text{ or }-7\dfrac{1}{4}
    \end{array}
    \end{array}\)
  6. \(\begin{array}{ll}
    \\ \\ \\ \\ \\
    \begin{array}{rrrrl}
    9n&+&8&=&46 \\
    &-&8&&-8 \\
    \hline
    &&\dfrac{9n}{9}&=&\dfrac{38}{9} \\ \\
    &&n&=&\dfrac{38}{9}\text{ or }4\dfrac{2}{9}
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    9n&+&8&=&-46 \\
    &-&8&&-8 \\
    \hline
    &&\dfrac{9n}{9}&=&\dfrac{-54}{9} \\ \\
    &&n&=&-6
    \end{array}
    \end{array}\)
  7. \(\begin{array}{ll}
    \\ \\ \\ \\ \\
    \begin{array}{rrrrr}
    3k&+&8&=&2 \\
    &-&8&&-8 \\
    \hline
    &&\dfrac{3k}{3}&=&\dfrac{-6}{3} \\ \\
    &&k&=&-2
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    3k&+&8&=&-2 \\
    &-&8&&-8 \\
    \hline
    &&\dfrac{3k}{3}&=&\dfrac{-10}{3} \\ \\
    &&k&=&-\dfrac{10}{3}
    \end{array}
    \end{array}\)
  8. \(\begin{array}{ll}
    \\ \\ \\
    \begin{array}{rrrrl}
    3&-&x&=&\phantom{-}6 \\
    -3&&&&-3 \\
    \hline
    &&(-x&=&\phantom{-}3)(-1) \\
    &&x&=&-3
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrl}
    3&-&x&=&-6 \\
    -3&&&&-3 \\
    \hline
    &&(-x&=&-9)(-1) \\
    &&x&=&\phantom{-}9
    \end{array}
    \end{array}\)
  9. \(\begin{array}{rrl}
    \\ \\
    \dfrac{-7}{-7}\left| -3-3r \right|&=&\dfrac{-21}{-7} \\
    |-3-3r|&=&3
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    -3&-&3r&=&3 \\
    +3&&&&+3 \\
    \hline
    &&\dfrac{-3r}{-3}&=&\dfrac{6}{-3} \\ \\
    &&r&=&-2
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    -3&-&3r&=&-3 \\
    +3&&&&+3 \\
    \hline
    &&\dfrac{-3r}{-3}&=&\dfrac{0}{-3} \\ \\
    &&r&=&0
    \end{array}
    \end{array}\)
  10. \(\begin{array}{rrrrr}
    \\ \\
    |2+2b|&+&1&=&3 \\
    &-&1&&-1 \\
    \hline
    |2+2b|&&&=&2
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    2&+&2b&=&2 \\
    -2&&&&-2 \\
    \hline
    &&2b&=&0 \\
    &&b&=&0
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    2&+&2b&=&-2 \\
    -2&&&&-2 \\
    \hline
    &&\dfrac{2b}{2}&=&\dfrac{-4}{2} \\ \\
    &&b&=&-2
    \end{array}
    \end{array}\)
  11. \(\begin{array}{rrl}
    \\ \\
    \dfrac{7}{7}|-7x-3|&=&\dfrac{21}{7} \\
    |-7x-3|&=&3
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    \\ \\
    -7x&-&3&=&3 \\
    &+&3&&+3 \\
    \hline
    &&\dfrac{-7x}{-7}&=&\dfrac{6}{-7} \\ \\
    &&x&=&-\dfrac{6}{7}
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    -7x&-&3&=&-3 \\
    &+&3&&+3 \\
    \hline
    &&-7x&=&0 \\
    &&x&=&0
    \end{array}
    \end{array}\)
  12. \(\begin{array}{ll}
    \\ \\ \\ \\ \\
    \begin{array}{rrrrr}
    -4&-&3n&=&2 \\
    +4&&&&+4 \\
    \hline
    &&\dfrac{-3n}{-3}&=&\dfrac{6}{-3} \\ \\
    &&n&=&-2
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    -4&-&3n&=&-2 \\
    +4&&&&+4 \\
    \hline
    &&\dfrac{-3n}{-3}&=&\dfrac{2}{-3} \\ \\
    &&n&=&-\dfrac{2}{3}
    \end{array}
    \end{array}\)
  13. \(\begin{array}{rrrrrrr}
    \\ \\ \\ \\
    8|5p &+&8|&-&5&=&11 \\
    &&&+&5&&+5 \\
    \hline
    &&\dfrac{8}{8}|5p &+&8|&=&\dfrac{16}{8} \\
    &&|5p &+&8|&=&2
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    5p&+&8&=&2 \\
    &-&8&&-8 \\
    \hline
    &&\dfrac{5p}{5}&=&\dfrac{-6}{5} \\ \\
    &&p&=&-\dfrac{6}{5}
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    5p&+&8&=&-2 \\
    &-&8&&-8 \\
    \hline
    &&\dfrac{5p}{5}&=&\dfrac{-10}{5} \\ \\
    &&p&=&-2
    \end{array}
    \end{array}\)
  14. \(\begin{array}{rrrrrrl}
    \\ \\ \\ \\
    3&-&|6n&+&7|&=&-40 \\
    -3&&&&&&-3 \\
    \hline
    &&(-|6n&+&7|&=&-43)(-1) \\
    &&|6n&+&7|&=&43
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    6n&+&7&=&43 \\
    &-&7&&-7 \\
    \hline
    &&\dfrac{6n}{6}&=&\dfrac{36}{6} \\ \\
    &&n&=&6
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    6n&+&7&=&-43 \\
    &-&7&&-7 \\
    \hline
    &&\dfrac{6n}{6}&=&\dfrac{-50}{6} \\ \\
    &&n&=&-\dfrac{25}{3}
    \end{array}
    \end{array}\)
  15. \(\begin{array}{rrrrrrr}
    \\ \\ \\ \\
    5|3&+&7m|&+&1&=&51 \\
    &&&-&1&&-1 \\
    \hline
    &&\dfrac{5}{5}|3&+&7m|&=&\dfrac{50}{5} \\
    &&|3&+&7m|&=&10
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    3&+&7m&=&10 \\
    -3&&&&-3 \\
    \hline
    &&\dfrac{7m}{7}&=&\dfrac{7}{7} \\ \\
    &&m&=&1
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    3&+&7m&=&-10 \\
    -3&&&&-3 \\
    \hline
    &&\dfrac{7m}{7}&=&\dfrac{-13}{7} \\ \\
    &&m&=&-\dfrac{13}{7}
    \end{array}
    \end{array}\)
  16. \(\begin{array}{rrrrrrr}
    \\ \\ \\ \\
    4|r&+&7|&+&3&=&59 \\
    &&&-&3&&-3 \\
    \hline
    &&\dfrac{4}{4}|r&+&7|&=&\dfrac{56}{4} \\
    &&|r&+&7|&=&14
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    r&+&7&=&14 \\
    &-&7&&-7 \\
    \hline
    &&r&=&7
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    r&+&7&=&-14 \\
    &-&7&&-7 \\
    \hline
    &&r&=&-21
    \end{array}
    \end{array}\)
  17. \(\begin{array}{rrrrrrr}
    \\ \\ \\ \\
    -7&+&8|-7x&-&3|&=&73 \\
    +7&&&&&&+7 \\
    \hline
    &&\dfrac{8}{8}|-7x&-&3|&=&\dfrac{80}{8} \\
    &&|-7x&-&3|&=&10
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    -7x&-&3&=&10 \\
    &+&3&&+3 \\
    \hline
    &&\dfrac{-7x}{-7}&=&\dfrac{13}{-7} \\ \\
    &&x&=&-\dfrac{13}{7}
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    -7x&-&3&=&-10 \\
    &+&3&&+3 \\
    \hline
    &&\dfrac{-7x}{-7}&=&\dfrac{-7}{-7} \\ \\
    &&x&=&1
    \end{array}
    \end{array}\)
  18. \(\begin{array}{rrrrrrr}
    \\ \\ \\ \\
    8|3&-&3n|&-&5&=&91 \\
    &&&+&5&&+5 \\
    \hline
    &&\dfrac{8}{8}|3&-&3n|&=&\dfrac{96}{8} \\
    &&|3&-&3n|&=&12
    \end{array}\)
    \(\phantom{1}\)
    \(\begin{array}{ll}
    \begin{array}{rrrrr}
    3&-&3n&=&12 \\
    -3&&&&-3 \\
    \hline
    &&\dfrac{-3n}{-3}&=&\dfrac{9}{-3} \\ \\
    &&n&=&-3
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrr}
    3&-&3n&=&-12 \\
    -3&&&&-3 \\
    \hline
    &&\dfrac{-3n}{-3}&=&\dfrac{-15}{-3} \\ \\
    &&n&=&5
    \end{array}
    \end{array}\)
  19. \(\begin{array}{ll}
    \\ \\ \\ \\ \\
    \begin{array}{rrrrrrr}
    5x&+&3&=&2x&-&1 \\
    -2x&-&3&&-2x&-&3 \\
    \hline
    &&\dfrac{3x}{3}&=&\dfrac{-4}{3}&& \\ \\
    &&x&=&-\dfrac{4}{3}&&
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrrrr}
    5x&+&3&=&-2x&+&1 \\
    +2x&-&3&&+2x&-&3 \\
    \hline
    &&\dfrac{7x}{7}&=&\dfrac{-2}{7}&& \\ \\
    &&x&=&-\dfrac{2}{7}&&
    \end{array}
    \end{array}\)
  20. \(\begin{array}{ll}
    \\ \\ \\
    \begin{array}{rrrrrrr}
    \\ \\ \\
    2&+&3x&=&4&-&2x \\
    -2&+&2x&&-2&+&2x \\
    \hline
    &&\dfrac{5x}{5}&=&\dfrac{2}{5}&& \\ \\
    &&x&=&\dfrac{2}{5}&&
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrrrr}
    2&+&3x&=&-4&+&2x \\
    -2&-&2x&&-2&-&2x \\
    \hline
    &&x&=&-6&&
    \end{array}
    \end{array}\)
  21. \(\begin{array}{ll}
    \\ \\ \\
    \begin{array}{rrrrrrr}
    3x&-&4&=&2x&+&3 \\
    -2x&+&4&&-2x&+&4 \\
    \hline
    &&x&=&7&&
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrlrr}
    \\ \\ \\
    3x&-&4&=&-2x&-&3 \\
    +2x&+&4&&+2x&+&4 \\
    \hline
    &&\dfrac{5x}{5}&=&\dfrac{1}{5}&& \\ \\
    &&x&=&\dfrac{1}{5}&&
    \end{array}
    \end{array}\)
  22. \(\begin{array}{ll}
    \\ \\ \\
    \begin{array}{rrrrrrr}
    2x&-&5&=&3x&+&4 \\
    -3x&+&5&&-3x&+&5 \\
    \hline
    &&-x&=&9&& \\
    &&x&=&-9&&
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrlrr}
    \\ \\
    2x&-&5&=&-3x&-&4 \\
    +3x&+&5&&+3x&+&5 \\
    \hline
    &&\dfrac{5x}{5}&=&\dfrac{1}{5}&& \\ \\
    &&x&=&\dfrac{1}{5}&&
    \end{array}
    \end{array}\)
  23. \(\begin{array}{ll}
    \\ \\ \\ \\
    \begin{array}{rrrrrrr}
    4x&-&2&=&6x&+&3 \\
    -6x&+&2&&-6x&+&2 \\
    \hline
    &&\dfrac{-2x}{-2}&=&\dfrac{5}{-2}&& \\ \\
    &&x&=&-\dfrac{5}{2}&&
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrrrr}
    4x&-&2&=&-6x&-&3 \\
    +6x&+&2&&+6x&+&2 \\
    \hline
    &&\dfrac{10x}{10}&=&\dfrac{-1}{10}&& \\ \\
    &&x&=&-\dfrac{1}{10}&&
    \end{array}
    \end{array}\)
  24. \(\begin{array}{ll}
    \\ \\
    \begin{array}{rrrrrrr}
    3x&+&2&=&2x&-&3 \\
    -2x&-&2&&-2x&-&2 \\
    \hline
    &&x&=&-5&&
    \end{array}
    & \hspace{0.5in}
    \begin{array}{rrrrlrr}
    \\ \\ \\
    3x&+&2&=&-2x&+&3 \\
    +2x&-&2&&+2x&-&2 \\
    \hline
    &&\dfrac{5x}{5}&=&\dfrac{1}{5}&& \\ \\
    &&x&=&\dfrac{1}{5}&&
    \end{array}
    \end{array}\)

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Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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