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Answer Key 5.4

  1. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    \\
    &a&-&b&+&2c&=&2& \\
    +&a&+&b&+&c&=&3& \\
    \hline
    &&&2a&+&3c&=&5& \\ \\
    &a&-&b&+&2c&=&2& \\
    +&2a&+&b&-&c&=&2& \\
    \hline
    &&&(3a&+&c&=&4)&(-3) \\
    &&&-9a&-&3c&=&-12& \\ \\
    &&&-9a&-&3c&=&-12& \\
    +&&&2a&+&3c&=&5& \\
    \hline
    &&&&&\dfrac{-7a}{-7}&=&\dfrac{-7}{-7}& \\ \\
    &&&&&a&=&1&
    \end{array}
    &\hspace{0.25in}
    \begin{array}{rrrrrrrl}
    &&3a&+&c&=&4& \\
    &&3(1)&+&c&=&4& \\
    &&3&+&c&=&4& \\
    &&-3&&&&-3& \\
    \hline
    &&&&c&=&1& \\ \\
    a&+&b&+&c&=&3& \\
    (1)&+&b&+&(1)&=&3& \\
    &&b&+&2&=&3& \\
    &&&-&2&&-2& \\
    \hline
    &&&&b&=&1&
    \end{array}
    \end{array}\)
  2. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    \\ \\
    &2a&+&3b&-&c&=&12& \\
    +&3a&+&4b&+&c&=&19& \\
    \hline
    &&&5a&+&7b&=&31& \\ \\
    &2a&+&3b&-&c&=&12& \\
    +&a&-&2b&+&c&=&-3& \\
    \hline
    &&&(3a&+&b&=&9)&(-7) \\
    &&&-21a&-&7b&=&-63& \\ \\
    &&&5a&+&7b&=&31& \\
    +&&&-21a&-&7b&=&-63& \\
    \hline
    &&&&&\dfrac{-16a}{-16}&=&\dfrac{-32}{-16}& \\ \\
    &&&&&a&=&2&
    \end{array}
    &\hspace{0.25in}
    \begin{array}{rrrrrrr}
    &&3a&+&b&=&9 \\
    &&3(2)&+&b&=&9 \\
    &&6&+&b&=&9 \\
    &&-6&&&&-6 \\
    \hline
    &&&&b&=&3 \\ \\
    a&-&2b&+&c&=&-3 \\
    (2)&-&2(3)&+&c&=&-3 \\
    2&-&6&+&c&=&-3 \\
    &&-4&+&c&=&-3 \\
    &&+4&&&&+4 \\
    \hline
    &&&&c&=&1
    \end{array}
    \end{array}\)
  3. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &(3x&+&y&-&z&=&7)&(-1) \\
    &-3x&-&y&+&z&=&-7& \\
    +&x&+&3y&-&z&=&5& \\
    \hline
    &&&(-2x&+&2y&=&-2)&(\div 2) \\
    &&&(-x&+&y&=&-1)&(7) \\
    &&&-7x&+&7y&=&-7& \\ \\
    &(3x&+&y&-&z&=&7)&(2) \\
    &6x&+&2y&-&2z&=&14& \\
    +&x&+&y&+&2z&=&3& \\
    \hline
    &&&7x&+&3y&=&17& \\
    +&&&-7x&+&7y&=&-7& \\
    \hline
    &&&&&\dfrac{10y}{10}&=&\dfrac{10}{10}& \\ \\
    &&&&&y&=&1& \\
    \end{array}
    &\hspace{0.25in}
    \begin{array}{rrrrrrr}
    \\ \\
    &&-x&+&y&=&-1 \\
    &&-x&+&(1)&=&-1 \\
    &&-x&+&1&=&-1 \\
    &&&-&1&&-1 \\
    \hline
    &&&&-x&=&-2 \\
    &&&&x&=&2 \\ \\
    x&+&y&+&2z&=&3 \\
    (2)&+&(1)&+&2z&=&3 \\
    &&2z&+&3&=&3 \\
    &&&-&3&&-3 \\
    \hline
    &&&&2z&=&0 \\
    &&&&z&=&0 \\
    \end{array}
    \end{array}\)
  4. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &x&+&y&+&z&=&4&(-1) \\
    &-x&-&y&-&z&=&-4& \\ \\
    &-x&-&y&-&z&=&-4& \\
    +&x&+&2y&+&3z&=&10& \\
    \hline
    &&&(y&+&2z&=&6)&(2) \\
    &&&2y&+&4z&=&12& \\ \\
    &-x&-&y&-&z&=&-4& \\
    +&x&-&y&+&4z&=&20& \\
    \hline
    &&&-2y&+&3z&=&16& \\
    +&&&2y&+&4z&=&12& \\
    \hline
    &&&&&\dfrac{7z}{7}&=&\dfrac{28}{7}& \\ \\
    &&&&&z&=&4&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrrr}
    &&y&+&2z&=&6 \\
    &&y&+&2(4)&=&6 \\
    &&y&+&8&=&6 \\
    &&&-&8&&-8 \\
    \hline
    &&&&y&=&-2 \\ \\
    x&+&y&+&z&=&4 \\
    x&+&(-2)&+&(4)&=&4 \\
    &&x&+&2&=&4 \\
    &&&-&2&&-2 \\
    \hline
    &&&&x&=&2
    \end{array}
    \end{array}\)
  5. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &x&+&2y&-&z&=&0& \\
    +&3x&-&2y&-&4z&=&-5& \\
    \hline
    &&&4x&-&5z&=&-5& \\ \\
    &(2x&-&y&+&z&=&15)&(2) \\
    &4x&-&2y&+&2z&=&30& \\
    +&x&+&2y&-&z&=&0& \\
    \hline
    &&&(5x&+&z&=&30)&(5) \\
    &&&25x&+&5z&=&150& \\ \\
    &&&4x&-&5z&=&-5& \\
    +&&&25x&+&5z&=&150& \\
    \hline
    &&&&&\dfrac{29x}{29}&=&\dfrac{145}{29}& \\ \\
    &&&&&x&=&5& \\
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrrr}
    &&5x&+&z&=&30 \\
    &&5(5)&+&z&=&30 \\
    &&25&+&z&=&30 \\
    &&-25&&&&-25 \\
    \hline
    &&&&z&=&5 \\ \\
    x&+&2y&-&z&=&0 \\
    (5)&+&2y&-&(5)&=&0 \\
    5&+&2y&-&5&=&0 \\
    &&&&2y&=&0 \\
    &&&&y&=&0 \\
    \end{array}
    \end{array}\)
  6. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &(x&-&y&+&2z&=&-3)&(2) \\
    &2x&-&2y&+&4z&=&-6& \\
    +&x&+&2y&+&3z&=&4& \\
    \hline
    &&&(3x&+&7z&=&-2)&(-1) \\
    &&&-3x&-&7z&=&2& \\ \\
    &2x&+&y&+&z&=&-3& \\
    +&x&-&y&+&2z&=&-3& \\
    \hline
    &&&3x&+&3z&=&-6& \\
    +&&&-3x&-&7z&=&2& \\
    \hline
    &&&&&\dfrac{-4z}{-4}&=&\dfrac{-4}{-4}& \\ \\
    &&&&&z&=&1& \\
    \end{array}
    &\hspace{0.25in}
    \begin{array}{rrrrrrr}
    &&3x&+&3z&=&-6 \\
    &&3x&+&3(1)&=&-6 \\
    &&3x&+&3&=&-6 \\
    &&&-&3&&-3 \\
    \hline
    &&&&\dfrac{3x}{3}&=&\dfrac{-9}{3} \\ \\
    &&&&x&=&-3 \\ \\
    x&-&y&+&2z&=&-3 \\
    (-3)&-&y&+&2(1)&=&-3 \\
    -3&-&y&+&2&=&-3 \\
    &&-y&-&1&=&-3 \\
    &&&+&1&&+1 \\
    \hline
    &&&&-y&=&-2 \\
    &&&&y&=&2
    \end{array}
    \end{array}\)
  7. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    \\
    &x&+&y&+&z&=&6& \\
    +&2x&-&y&-&z&=&-3& \\
    \hline
    &&&&&\dfrac{3x}{3}&=&\dfrac{3}{3}& \\ \\
    &&&&&x&=&1& \\ \\
    &x&-&2y&+&3z&=&6& \\
    &(1)&-&2y&+&3z&=&6& \\
    &1&-&2y&+&3z&=&6& \\
    &-1&&&&&&-1& \\
    \hline
    &&&-2y&+&3z&=&5& \\ \\
    &x&+&y&+&z&=&6& \\
    &(1)&+&y&+&z&=&6& \\
    &1&+&y&+&z&=&6& \\
    &-1&&&&&&-1& \\
    \hline
    &&&(y&+&z&=&5)&(2) \\
    &&&2y&+&2z&=&10&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrrr}
    \\ \\ \\ \\ \\
    &&-2y&+&3z&=&5 \\
    +&&2y&+&2z&=&10 \\
    \hline
    &&&&\dfrac{5z}{5}&=&\dfrac{15}{5} \\ \\
    &&&&z&=&3 \\ \\
    x&+&y&+&z&=&6 \\
    (1)&+&y&+&(3)&=&6 \\
    1&+&y&+&3&=&6 \\
    &&y&+&4&=&6 \\
    &&&-&4&&-4 \\
    \hline
    &&&&y&=&2 \\
    \end{array}
    \end{array}\)
  8. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &x&+&y&-&z&=&0& \\
    +&2x&+&y&+&z&=&0& \\
    \hline
    &&&3x&+&2y&=&0& \\ \\
    &(x&+&y&-&z&=&0)&(-4) \\
    &-4x&-&4y&+&4z&=&0& \\
    +&x&+&2y&-&4z&=&0& \\
    \hline
    &&&-3x&-&2y&=&0& \\
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrr}
    &-3x&-&2y&=&0 \\
    +&3x&+&2y&=&0 \\
    \hline
    &&&0&=&0 \\ \\
    &&\therefore &x&=&0 \\
    &&&y&=&0 \\
    &&&z&=&0 \\
    \end{array}
    \end{array}\)
  9. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &x&+&y&+&z&=&2& \\
    +&2x&-&y&+&3z&=&9& \\
    \hline
    &&&3x&+&4z&=&11& \\ \\
    &2x&-&y&+&3z&=&9& \\
    +&&&y&-&z&=&-3& \\
    \hline
    &&&(2x&+&2z&=&6)&(-2) \\
    &&&-4x&-&4z&=&-12& \\ \\
    &&&3x&+&4z&=&11& \\
    +&&&-4x&-&4z&=&-12& \\
    \hline
    &&&&&-x&=&-1& \\
    &&&&&x&=&1&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrrr}
    &&2x&+&2z&=&6 \\
    &&2(1)&+&2z&=&6 \\
    &&2&+&2z&=&6 \\
    &&-2&&&&-2 \\
    \hline
    &&&&\dfrac{2z}{2}&=&\dfrac{4}{2} \\ \\
    &&&&z&=&2 \\ \\
    x&+&y&+&z&=&2 \\
    (1)&+&y&+&(2)&=&2 \\
    &&y&+&3&=&2 \\
    &&&-&3&&-3 \\
    \hline
    &&&&y&=&-1 \\
    \end{array}
    \end{array}\)
  10. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &(4x&&&+&z&=&3)&(2) \\
    &8x&&&+&2z&=&6& \\
    +&6x&-&y&-&2z&=&-1& \\
    \hline
    &&&(14x&-&y&=&5)&(3) \\
    &&&42x&-&3y&=&15& \\
    +&&&-2x&+&3y&=&5& \\
    \hline
    &&&&&\dfrac{40x}{40}&=&\dfrac{20}{40}& \\ \\
    &&&&&x&=&\dfrac{1}{2}&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrrr}
    \\ \\ \\ \\ \\
    &&-2x&+&3y&=&5 \\
    &&-2\left(\dfrac{1}{2}\right)&+&3y&=&5 \\
    &&-1&+&3y&=&5 \\
    &&+1&&&&+1 \\
    \hline
    &&&&\dfrac{3y}{3}&=&\dfrac{6}{3} \\ \\
    &&&&y&=&2 \\ \\
    &&4x&+&z&=&3 \\
    &&4\left(\dfrac{1}{2}\right)&+&z&=&3 \\
    &&2&+&z&=&3 \\
    &&-2&&&&-2 \\
    \hline
    &&&&z&=&1 \\
    \end{array}
    \end{array}\)
  11. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &&&x&-&z&=&-2& \\
    +&&&y&+&z&=&5& \\
    \hline
    &&&x&+&y&=&3& \\ \\
    &2x&-&3y&+&z&=&-1& \\
    +&x&&&-&z&=&-2& \\
    \hline
    &&&(3x&-&3y&=&-3)&(\div 3) \\
    &&&x&-&y&=&-1& \\
    +&&&x&+&y&=&3& \\
    \hline
    &&&&&\dfrac{2x}{2}&=&\dfrac{2}{2}& \\ \\
    &&&&&x&=&1&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrr}
    x&-&z&=&-2 \\
    (1)&-&z&=&-2 \\
    1&-&z&=&-2 \\
    -1&&&&-1 \\
    \hline
    &&-z&=&-3 \\
    &&z&=&3 \\ \\
    y&+&z&=&5 \\
    y&+&(3)&=&5 \\
    y&+&3&=&5 \\
    &-&3&&-3 \\
    \hline
    &&y&=&2 \\
    \end{array}
    \end{array}\)
  12. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &(3x&+&4y&-&z&=&11)&(2) \\
    &6x&+&8y&-&2z&=&22& \\
    +&&&y&+&2z&=&-4& \\
    \hline
    &&&(6x&+&9y&=&18)&(\div 3) \\
    &&&2x&+&3y&=&6& \\
    +&&&-2x&+&y&=&-6& \\
    \hline
    &&&&&4y&=&0& \\
    &&&&&y&=&0& \\ \\
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrr}
    \\ \\ \\
    -2x&+&y&=&-6 \\
    -2x&+&0&=&-6 \\
    &&\dfrac{-2x}{-2}&=&\dfrac{-6}{-2} \\ \\
    &&x&=&3 \\ \\
    y&+&2z&=&-4 \\
    0&+&2z&=&-4 \\
    &&\dfrac{2z}{2}&=&\dfrac{-4}{2} \\ \\
    &&z&=&-2
    \end{array}
    \end{array}\)
  13. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &&&(-2y&+&z&=&-6)&(3) \\
    &&&-6y&+&3z&=&-18& \\
    +&x&+&6y&+&3z&=&30& \\
    \hline
    &&&(x&+&6z&=&12)&(-2) \\
    &&&-2x&-&12z&=&-24& \\
    +&&&2x&+&2z&=&4& \\
    \hline
    &&&&&\dfrac{-10z}{-10}&=&\dfrac{-20}{-10}& \\ \\
    &&&&&z&=&2&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrr}
    \\ \\ \\
    2x&+&2z&=&4 \\
    2x&+&2(2)&=&4 \\
    2x&+&4&=&4 \\
    &-&4&&-4 \\
    \hline
    &&2x&=&0 \\
    &&x&=&0 \\ \\
    -2y&+&z&=&-6 \\
    -2y&+&2&=&-6 \\
    &-&2&&-2 \\
    \hline
    &&\dfrac{-2y}{-2}&=&\dfrac{-8}{-2} \\ \\
    &&y&=&4
    \end{array}
    \end{array}\)
  14. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &(x&-&y&+&2z&=&0)&(2) \\
    &2x&-&2y&+&4z&=&0& \\
    +&x&+&2y&&&=&1& \\
    \hline
    &&&3x&+&4z&=&1& \\ \\
    &&&(2x&+&z&=&4)&(-4) \\
    &&&-8x&-&4z&=&-16& \\
    +&&&3x&+&4z&=&1& \\
    \hline
    &&&&&\dfrac{-5x}{-5}&=&\dfrac{-15}{-5}& \\ \\
    &&&&&x&=&3&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrr}
    x&+&2y&=&1 \\
    3&+&2y&=&1 \\
    -3&&&&-3 \\
    \hline
    &&\dfrac{2y}{2}&=&\dfrac{-2}{2} \\ \\
    &&y&=&-1 \\ \\
    2x&+&z&=&4 \\
    2(3)&+&z&=&4 \\
    6&+&z&=&4 \\
    -6&&&&-6 \\
    \hline
    &&z&=&-2
    \end{array}
    \end{array}\)
  15. \(\begin{array}{rr}
    \\ \\
    \begin{array}{rrrrrrrr}
    &x&+&y&+&z&=&4 \\
    +&&-&y&-&z&=&-4 \\
    \hline
    &&&&&x&=&0
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrr}
    \\ \\ \\ \\ \\ \\
    x&-&2y&=&0 \\
    0&-&2y&=&0 \\
    &&-2y&=&0 \\
    &&y&=&0 \\ \\
    -y&-&z&=&-4 \\
    0&-&z&=&-4 \\
    &&-z&=&-4 \\
    &&z&=&4
    \end{array}
    \end{array}\)
  16. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrrrrrl}
    &(x&+&y&-&z&=&2)&(-2) \\
    &-2x&-&2y&+&2z&=&-4& \\
    +&2x&&&+&z&=&6& \\
    \hline
    &&&-2y&+&3z&=&2& \\
    +&&&2y&-&4z&=&-4& \\
    \hline
    &&&&&-z&=&-2& \\
    &&&&&z&=&2&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\
    2x&+&z&=&6 \\
    2x&+&2&=&6 \\
    &-&2&&-2 \\
    \hline
    &&\dfrac{2x}{2}&=&\dfrac{4}{2} \\ \\
    &&x&=&2 \\ \\
    2y&-&4z&=&-4 \\
    2y&-&4(2)&=&-4 \\
    2y&-&8&=&-4 \\
    &+&8&&+8 \\
    \hline
    &&\dfrac{2y}{2}&=&\dfrac{4}{2} \\ \\
    &&y&=&2
    \end{array}
    \end{array}\)

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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.