"

Answer Key 3.7

  1. \(2x+5=25\)
  2. \(4x+12=36\)
  3. \(3x-8=22\)
  4. \(6x-8=22\)
  5. \(x-8=\dfrac{x}{2}\)
  6. \(x-4=\dfrac{x}{2}\)
  7. \(x+x+1+x+2=21\)
  8. \(x+x+2-(x+4)=5\)
  9. \(\begin{array}{rrrrrr}
    \\ \\ \\ \\ \\
    5&+&3x&=&17& \\
    -5&&&&-5& \\
    \hline
    &&3x&=&12& \\ \\
    &&x&=&\dfrac{12}{3}&\text{or } 4
    \end{array}\)
  10. \(\begin{array}{rrrrrr}
    \\ \\ \\ \\ \\
    3x&-&5&=&10& \\
    &+&5&&+5& \\
    \hline
    &&3x&=&15& \\ \\
    &&x&=&\dfrac{15}{3}&\text{or } 5
    \end{array}\)
  11. \(\begin{array}{rrrrrrr}
    \\ \\
    60&+&9x&=&10x&-&2 \\
    +2&-&9x&&-9x&+&2 \\
    \hline
    &&62&=&x&&
    \end{array}\)
  12. \(\begin{array}{rrrrrrr}
    \\ \\
    7x&-&11&=&6x&+&5 \\
    -6x&+&11&&-6x&+&11 \\
    \hline
    &&x&=&16&&
    \end{array}\)
  13. \(\begin{array}{rrrrrrrrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\
    x&+&x&+&1&+&x&+&2&=&108 \\
    &&&&&&3x&+&3&=&108 \\
    &&&&&&&-&3&&-3 \\
    \hline
    &&&&&&&&3x&=&105 \\ \\
    &&&&&&&&x&=&\dfrac{105}{3}\text{ or }35 \\ \\
    &&&&&&&&&\therefore &35, 36, 37
    \end{array}\)
  14. \(\begin{array}{rrrrrrrrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\
    x&+&x&+&1&+&x&+&2&=&-126 \\
    &&&&&&3x&+&3&=&-126 \\
    &&&&&&&-&3&&-\phantom{00}3 \\
    \hline
    &&&&&&&&3x&=&-129 \\ \\
    &&&&&&&&x&=&\dfrac{-129}{3}\text{ or }-43 \\ \\
    &&&&&&&&&\therefore &-43, -42, -41
    \end{array}\)
  15. \(\begin{array}{rrrrrrrrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    x&+&2(x&+&1)&+&3(x&+&2)&=&-76 \\
    x&+&2x&+&2&+&3x&+&6&=&-76 \\
    &&&&&&6x&+&8&=&-76 \\
    &&&&&&&-&8&&-\phantom{0}8 \\
    \hline
    &&&&&&&&6x&=&-84 \\ \\
    &&&&&&&&x&=&\dfrac{-84}{6}\text{ or }-14 \\ \\
    &&&&&&&&&\therefore &-14, -13, -12
    \end{array}\)
  16. \(\begin{array}{rrrrrrrrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    x&+&2(x&+&2)&+&3(x&+&4)&=&\phantom{0}70 \\
    x&+&2x&+&4&+&3x&+&12&=&\phantom{0}70 \\
    &&&&&&6x&+&16&=&\phantom{0}70 \\
    &&&&&&&-&16&&-16 \\
    \hline
    &&&&&&&&6x&=&54 \\ \\
    &&&&&&&&x&=&\dfrac{54}{6}\text{ or }9 \\ \\
    &&&&&&&&&\therefore &9,11,13
    \end{array}\)

License

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

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