Answer Key 5.6
- \(\begin{array}{rr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrrrrrrrrl}
&t&+&x&+&y&+&z&=&\phantom{-}6 \\
+&-t&+&x&-&y&-&z&=&-2 \\
\hline
&&&&&&&\dfrac{2x}{2}&=&\dfrac{4}{2} \\ \\
&&&&&&&x&=&2 \\ \\
&(-t&+&x&-&y&-&z&=&-2)(-1) \\
&t&-&x&+&y&+&z&=&\phantom{-}2 \\
+&-t&+&3x&+&y&-&z&=&\phantom{-}2 \\
\hline
&&&&&2x&+&2y&=&\phantom{-}4 \\ \\
&&&&&2(2)&+&2y&=&\phantom{-}4 \\
&&&&&-4&&&&-4 \\
\hline
&&&&&&&2y&=&\phantom{-}0 \\
&&&&&&&y&=&\phantom{-}0
\end{array}
&\hspace{0.25in}
\begin{array}{rrrrrrrrrl}
&t&+&2x&+&2y&+&4z&=&17 \\
+&-t&+&3x&+&y&-&z&=&2 \\
\hline
&&&5x&+&3y&+&3z&=&19 \\
&&&5(2)&+&3(0)&+&3z&=&19 \\
&&&&&10&+&3z&=&19 \\
&&&&&-10&&&&-10 \\
\hline
&&&&&&&\dfrac{3z}{3}&=&\dfrac{9}{3} \\ \\
&&&&&&&z&=&3 \\ \\
&t&+&x&+&y&+&z&=&6 \\
&t&+&2&+&0&+&3&=&6 \\
&&&&&t&+&5&=&6 \\
&&&&&&-&5&&-5 \\
\hline
&&&&&&&t&=&1 \\
\end{array}
\end{array}\) - \(\phantom{1} \\ \)
\(\begin{array}{rr}
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\begin{array}{rrrrrrrrrr}
\\ \\ \\ \\ \\ \\
&t&+&x&-&y&+&z&=&-1 \\
+&-t&+&3x&+&y&-&z&=&1 \\
\hline
&&&&&&&4x&=&0 \\
&&&&&&&x&=&0 \\ \\
&&\therefore &t&-&y&+&z&=&-1 \\
&&&-t&+&2y&+&z&=&3 \\
&&&-t&+&y&-&z&=&1 \\
&&&-2t&+&y&-&3z&=&0 \\ \\
&&&-t&+&2y&+&z&=&3 \\
+&&&-t&+&y&-&z&=&1 \\
\hline
&&&&&-2t&+&3y&=&4 \\ \\
&&&(-t&+&y&-&z&=&1)(-3) \\
&&&3t&-&3y&+&3z&=&-3 \\
+&&&-2t&+&y&-&3z&=&0 \\
\hline
&&&&&(t&-&2y&=&-3)(2) \\
&&&&&2t&-&4y&=&-6 \\
+&&&&&-2t&+&3y&=&4 \\
\hline
&&&&&&&-y&=&-2 \\
&&&&&&&y&=&2 \\
\end{array}
& \hspace{0.25in}
\begin{array}{rrrrrrr}
\\
t&-&y&+&z&=&-1 \\
t&-&2&+&z&=&-1 \\
&+&2&&&&+2 \\
\hline
&&t&+&z&=&1 \\ \\
-t&+&2y&+&z&=&3 \\
-t&+&2(2)&+&z&=&3 \\
-t&+&4&+&z&=&3 \\
&-&4&&&&-4 \\
\hline
&&-t&+&z&=&-1 \\
+&&t&+&z&=&1 \\
\hline
&&&&2z&=&0 \\
&&&&z&=&0 \\ \\
&&t&+&z&=&1 \\
&&t&+&0&=&1 \\
&&&&t&=&1 \\
\end{array}
\end{array}\)
