"

Answer Key 5.7

  1. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrr}
    &&3H&=&30 \\
    &&H&=&10 \\ \\
    H&+&4S&=&18 \\
    10&+&4S&=&18 \\
    -10&&&&-10 \\
    \hline
    &&\dfrac{4S}{4}&=&\dfrac{8}{4} \\ \\
    &&S&=&2
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrcrrr}
    \\ \\
    &&2S&-&2B&=&2 \\
    &&2(2)&-&2B&=&2 \\
    &&4&-&2B&=&2 \\
    &&-4&&&&-4 \\
    \hline
    &&&&\dfrac{-2B}{-2}&=&\dfrac{-2}{-2} \\ \\
    &&&&B&=&1 \\ \\
    B&+&H&\times &S&=&? \\
    1&+&(10&\times &2)&=&? \\
    &&1&+&20&=&21
    \end{array}
    \end{array}\)
  2. \(\begin{array}{rr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \begin{array}{rrrrr}
    &&3B&=&30 \\
    &&B&=&10 \\ \\
    B&+&2H&=&20 \\
    10&+&2H&=&20 \\
    -10&&&&-10 \\
    \hline
    &&\dfrac{2H}{2}&=&\dfrac{10}{2} \\ \\
    &&H&=&5
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrrrr}
    &&H&+&4M&=&9 \\
    &&5&+&4M&=&9 \\
    &&-5&&&&-5 \\
    \hline
    &&&&4M&=&4 \\
    &&&&M&=&1 \\ \\
    H&+&M&\times &B&=&? \\
    10&+&(5&\times &1)&=&? \\
    &&10&+&5&=&15
    \end{array}
    \end{array}\)
  3. \(\phantom{1}\)
    \(\begin{array}{rrrrrrrrrl}
    \text{Row 2}&-a&-&2b&-&2c&+&2d&=&-8 \\
    \text{Row 3}&2a&-&b&-&c&-&d&=&\phantom{-}5 \\
    \text{Column 1}&a&-&b&+&c&+&2d&=&-1 \\
    \text{Column 3}&a&-&2b&-&c&-&d&=&\phantom{-}3 \\ \\
    &(-a&-&2b&-&2c&+&2d&=&-8)(2) \\
    &-2a&-&4b&-&4c&+&4d&=&-16 \\
    +&2a&-&b&-&c&-&d&=&\phantom{-0}5 \\
    \hline
    &&&-5b&-&5c&+&3d&=&-11 \\ \\
    &-a&-&2b&-&2c&+&2d&=&-8 \\
    +&a&-&b&+&c&+&2d&=&-1 \\
    \hline
    &&&-3b&-&c&+&4d&=&-9 \\ \\
    &-a&-&2b&-&2c&+&2d&=&-8 \\
    +&a&-&2b&-&c&-&d&=&\phantom{-}3 \\
    \hline
    &&&(-4b&-&3c&+&d&=&-5)(-4) \\
    &&&16b&+&12c&-&4d&=&20 \\
    &&+&-3b&-&c&+&4d&=&-9 \\
    \hline
    &&&&&13b&+&11c&=&11 \\ \\
    &&&(-4b&-&3c&+&d&=&-5)(-3) \\
    &&&12b&+&9c&-&3d&=&\phantom{-}15 \\
    &&+&-5b&-&5c&+&3d&=&-11 \\
    \hline
    &&&&&7b&+&4c&=&4
    \end{array}\)
    \(\phantom{1} \\ \\ \\ \\ \)
    \(\begin{array}{rrrrrrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    &&&&(13b&+&11c&=&11)(-4) \\
    &&&&(\phantom{0}7b&+&4c&=&\phantom{0}4)(11) \\ \\
    &&&&-52b&-&44c&=&-44 \\
    &&&+&77b&+&44c&=&\phantom{-}44 \\
    \hline
    &&&&&&25b&=&0 \\
    &&&&&&b&=&0 \\ \\
    &&&&7b&+&4c&=&4 \\
    &&&&7(0)&+&4c&=&4 \\
    &&&&&&\dfrac{4c}{4}&=&\dfrac{4}{4} \\ \\
    &&&&&&c&=&1 \\ \\
    &&-4b&-&3c&+&d&=&-5 \\
    &&-4(0)&-&3(1)&+&d&=&-5 \\
    &&&&-3&+&d&=&-5 \\
    &&&&+3&&&&+3 \\
    \hline
    &&&&&&d&=&-2 \\ \\
    a&-&2b&-&c&-&d&=&\phantom{-}3 \\
    a&-&2(0)&-&1&-&(-2)&=&\phantom{-}3 \\
    &&a&-&1&+&2&=&\phantom{-}3 \\
    &&&&a&+&1&=&\phantom{-}3 \\
    &&&&&-&1&&-1 \\
    \hline
    &&&&&&a&=&2
    \end{array}\)

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.