Answer Key 2.4
- \(\phantom{1}\)
\(\left(\dfrac{3}{5}\left(1 + p\right) = \dfrac{21}{20}\right)(20) \\ \)
\(\dfrac{3}{\cancel{5}1}\cdot \cancel{20 }4(1 + p) = \dfrac{21}{\cancel{20}1}\cdot \cancel{20} \\ \)
\(\begin{array}{rrrrr}
12&+&12p&=&\phantom{-}21 \\
-12&&&&-12 \\
\hline
&&\dfrac{12p}{12}&=&\dfrac{9}{12} \\ \\
&&p&=&\dfrac{3}{4}
\end{array}\) - \(\phantom{1}\)
\(\left(-\dfrac{1}{2} = \dfrac{3k}{2} + \dfrac{3}{2}\right)(2) \\ \)
\(\begin{array}{rrrrr}
-1&=&3k&+&3 \\
-3&&&-&3 \\
\hline
\dfrac{-4}{3}&=&\dfrac{3k}{3}&& \\ \\
k&=&-\dfrac{4}{3}&&
\end{array}\) - \(\phantom{1}\)
\(\left(0 = -\dfrac{5}{4}x+\dfrac{6}{4}\right)(4) \\ \)
\(\begin{array}{rrlrr}
0&=&-5x&+&6 \\
+5x&&+5x&& \\
\hline
\dfrac{5x}{5}&=&\dfrac{6}{5}&& \\ \\
x&=&\dfrac{6}{5}&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{3n}{2} - 8 = -\dfrac{29}{12}\right)(12) \\ \)
\(\begin{array}{rrrrr}
18n&-&96&=&-29 \\
&+&96&&+96 \\
\hline
&&\dfrac{18n}{18}&=&\dfrac{67}{18} \\ \\
&&n&=&\dfrac{67}{18}
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{3}{4} - \dfrac{5}{4}m = \dfrac{108}{24}\right)(24) \\ \)
\(\begin{array}{rrrrr}
18&-&30m&=&108 \\
-18&&&&-18 \\
\hline
&&\dfrac{-30m}{-30}&=&\dfrac{90}{-30} \\ \\
&&m&=&-3
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{11}{4} + \dfrac{3}{4}r = \dfrac{160}{32}\right)(32) \\ \)
\(\begin{array}{crcrr}
11\cdot 8&+&3r\cdot 8&=&160 \\
\phantom{-}88&+&24r&=&160 \\
-88&&&&-88 \\
\hline
&&\dfrac{24r}{24}&=&\dfrac{72}{24} \\ \\
&&r&=&3
\end{array}\) - \(\phantom{1}\)
\(\left(2b + \dfrac{9}{5} = -\dfrac{11}{5}\right)(5) \\ \)
\(\begin{array}{rrrrr}
10b&+&9&=&-11 \\
&-&9&&-9 \\
\hline
&&\dfrac{10b}{10}&=&\dfrac{-20}{10} \\ \\
&&b&=&-2
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{3}{2} - \dfrac{7v}{4} = -\dfrac{9}{8}\right)(8) \\ \)
\(\begin{array}{rrrrr}
3\cdot 4&-&7r\cdot 2&=&-9 \\
\phantom{-}12&-&14r&=&-9 \\
-12&&&&-12 \\
\hline
&&\dfrac{-14r}{-14}&=&\dfrac{-21}{-14} \\ \\
&&r&=&\dfrac{3}{2}
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{21n}{6}+\dfrac{3}{2} = \dfrac{3}{2}\right)(6) \\ \)
\(\begin{array}{rrrrr}
\dfrac{21n}{6}&+&3\cdot 3&=&3\cdot 3 \\
&-&9&&-9 \\
\hline
&&\dfrac{21n}{6}&=&0 \\ \\
&&n&=&0
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{41}{9} = \dfrac{5}{2}x+\dfrac{10}{6} - \dfrac{1}{3}x\right)(18) \\ \)
\(\begin{array}{rrlrrrr}
41\cdot 2&=&5x\cdot 9&+&10\cdot 3&-&x\cdot 6 \\
82&=&45x&+&30&-&6x \\
-30&&&-&30&& \\
\hline
\dfrac{52}{39}&=&\dfrac{39x}{39}&&&& \\ \\
x&=&\dfrac{4}{3}&&&&
\end{array}\) - \(\phantom{1}\)
\(\left(-a + \dfrac{40a}{12}-\dfrac{5}{4}= -\dfrac{19}{4}\right)(12) \\ \)
\(\begin{array}{rrrrrrr}
-12a&+&40a&-&15&=&-57 \\
&&&+&15&&+15 \\
\hline
&&&&\dfrac{28a}{28}&=&\dfrac{-42}{28} \\ \\
&&&&a&=&-\dfrac{3}{2}
\end{array}\) - \(\phantom{1}\)
\(\left(-\dfrac{7k}{12}+\dfrac{1}{3}-\dfrac{10k}{3}=-\dfrac{13}{8} \right)(24) \\ \)
\(\begin{array}{rrrrrrr}
-14k&+&8&-&80k&=&-39 \\
&-&8&&&&-8 \\
\hline
&&&&\dfrac{-94k}{-94}&=&\dfrac{-47}{-94} \\ \\
&&&&k&=&\dfrac{1}{2}
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{55}{6} = -\dfrac{15p}{4}+\dfrac{25}{6}\right)(12) \\ \)
\(\begin{array}{rrlrr}
110&=&-45p&+&50 \\
-50&&&-&50 \\
\hline
\dfrac{60}{-45}&=&\dfrac{-45p}{-45}&& \\ \\
p&=&-\dfrac{4}{3}&&
\end{array}\) - \(\phantom{1}\)
\(\left(-\dfrac{2x}{6}+\dfrac{3}{8}-\dfrac{7x}{2}=-\dfrac{83}{24}\right)(24) \\ \)
\(\begin{array}{rrrrrrr}
-8x&+&9&-&84x&=&-83 \\
&-&9&&&&-9 \\
\hline
&&&&\dfrac{-92x}{-92}&=&\dfrac{-92}{-92} \\ \\
&&&&x&=&1
\end{array}\) - \(\phantom{1}\)
\(\left(-\dfrac{5}{8}=\dfrac{5}{4}r-\dfrac{15}{8}\right)(8) \\ \)
\(\begin{array}{rrlrr}
-5&=&10r&-&15 \\
+15&&&+&15 \\
\hline
\dfrac{10}{10}&=&\dfrac{10r}{10}&& \\ \\
r&=&1&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{1}{12}=\dfrac{4x}{3}+\dfrac{5x}{3}-\dfrac{35}{12}\right)(12) \\ \)
\(\begin{array}{rrlrrrr}
1&=&16x&+&20x&-&35 \\
+35&&&&&+&35 \\
\hline
\dfrac{36}{36}&=&\dfrac{36x}{36}&&&& \\ \\
x&=&1&&&&
\end{array}\) - \(\phantom{1}\)
\(\left(-\dfrac{11}{3}+\dfrac{3b}{2}=\dfrac{5b}{2}-\dfrac{25}{6}\right)(6) \\ \)
\(\begin{array}{rrrrrrr}
-22&+&9b&=&15b&-&25 \\
+22&-&15b&&-15b&+&22 \\
\hline
&&\dfrac{-6b}{-6}&=&\dfrac{-3}{-6}&& \\ \\
&&b&=&\dfrac{1}{2}&&
\end{array}\) - \(\phantom{1}\)
\(\left(\dfrac{7}{6}-\dfrac{4n}{3}=-\dfrac{3n}{2}+2n+3\right)(6) \\ \)
\(\begin{array}{rrrrlrrrr}
7&-&8n&=&-9n&+&12n&+&18 \\
-7&+&9n&&+9n&-&12n&-&7 \\
&-&12n&&&&&& \\
\hline
&&\dfrac{-11n}{-11}&=&\dfrac{11}{-11}&&&& \\ \\
&&n&=&-1&&&&
\end{array}\)
