Answer Key 5.6

  1. \begin{array}{rr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \begin{array}{rrrrrrrrrl} &t&+&x&+&y&+&z&=&\phantom{-}6 \\ +&-t&+&x&-&y&-&z&=&-2 \\ \midrule &&&&&&&\dfrac{2x}{2}&=&\dfrac{4}{2} \\ \\ &&&&&&&x&=&2 \\ \\ &(-t&+&x&-&y&-&z&=&-2)(-1) \\ &t&-&x&+&y&+&z&=&\phantom{-}2 \\ +&-t&+&3x&+&y&-&z&=&\phantom{-}2 \\ \midrule &&&&&2x&+&2y&=&\phantom{-}4 \\ \\ &&&&&2(2)&+&2y&=&\phantom{-}4 \\ &&&&&-4&&&&-4 \\ \midrule &&&&&&&2y&=&\phantom{-}0 \\ &&&&&&&y&=&\phantom{-}0 \end{array} &\hspace{0.25in} \begin{array}{rrrrrrrrrl} &t&+&2x&+&2y&+&4z&=&17 \\ +&-t&+&3x&+&y&-&z&=&2 \\ \midrule &&&5x&+&3y&+&3z&=&19 \\ &&&5(2)&+&3(0)&+&3z&=&19 \\ &&&&&10&+&3z&=&19 \\ &&&&&-10&&&&-10 \\ \midrule &&&&&&&\dfrac{3z}{3}&=&\dfrac{9}{3} \\ \\ &&&&&&&z&=&3 \\ \\ &t&+&x&+&y&+&z&=&6 \\ &t&+&2&+&0&+&3&=&6 \\ &&&&&t&+&5&=&6 \\ &&&&&&-&5&&-5 \\ \midrule &&&&&&&t&=&1 \\ \end{array} \end{array}
  2. \phantom{1} \\
    \begin{array}{rr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \begin{array}{rrrrrrrrrr} \\ \\ \\ \\ \\ \\ &t&+&x&-&y&+&z&=&-1 \\ +&-t&+&3x&+&y&-&z&=&1 \\ \midrule &&&&&&&4x&=&0 \\ &&&&&&&x&=&0 \\ \\ &&\therefore &t&-&y&+&z&=&-1 \\ &&&-t&+&2y&+&z&=&3 \\ &&&-t&+&y&-&z&=&1 \\ &&&-2t&+&y&-&3z&=&0 \\ \\ &&&-t&+&2y&+&z&=&3 \\ +&&&-t&+&y&-&z&=&1 \\ \midrule &&&&&-2t&+&3y&=&4 \\ \\ &&&(-t&+&y&-&z&=&1)(-3) \\ &&&3t&-&3y&+&3z&=&-3 \\ +&&&-2t&+&y&-&3z&=&0 \\ \midrule &&&&&(t&-&2y&=&-3)(2) \\ &&&&&2t&-&4y&=&-6 \\ +&&&&&-2t&+&3y&=&4 \\ \midrule &&&&&&&-y&=&-2 \\ &&&&&&&y&=&2 \\ \end{array} & \hspace{0.25in} \begin{array}{rrrrrrr} \\ t&-&y&+&z&=&-1 \\ t&-&2&+&z&=&-1 \\ &+&2&&&&+2 \\ \midrule &&t&+&z&=&1 \\ \\ -t&+&2y&+&z&=&3 \\ -t&+&2(2)&+&z&=&3 \\ -t&+&4&+&z&=&3 \\ &-&4&&&&-4 \\ \midrule &&-t&+&z&=&-1 \\ +&&t&+&z&=&1 \\ \midrule &&&&2z&=&0 \\ &&&&z&=&0 \\ \\ &&t&+&z&=&1 \\ &&t&+&0&=&1 \\ &&&&t&=&1 \\ \end{array} \end{array}

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

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