[latexpage]

1. \(\dfrac{\cancel{15}3\cancel{m^3}}{4\cancel{n^2}}\cdot \dfrac{\cancel{17}\cancel{n^3}}{\cancel{30}\cancel{10}2\cancel{m^3}}\cdot \dfrac{\cancel{3}1m^4}{\cancel{34}2n^{\cancel{2}}}\Rightarrow \dfrac{3m^4}{16n}\)
2. \(\dfrac{5v^2-25v}{5v+25}\cdot \dfrac{10v}{v^2-11v+30}\Rightarrow \dfrac{\cancel{5}v\cancel{(v-5)}}{\cancel{5}(v+5)}\cdot \dfrac{10v}{\cancel{(v-5)}(v-6)}\Rightarrow \dfrac{10v^2}{(v+5)(v-6)}\)
3. \(\phantom{1}\)
   \(\left(\dfrac{8}{2x}=\dfrac{2}{x}+1\right)(2x) \\ \)
   \(\begin{array}{rrl}
   8&=&2\cdot 2+1(2x) \\
   8&=&\phantom{-}4+2x \\
   -4&&-4 \\
   \midrule
   \dfrac{4}{2}&=&\dfrac{2x}{2} \\ \\
   x&=&2
   \end{array}\)
4. \(\begin{array}{l}
   \dfrac{\left(\dfrac{x^2}{y^2}-16\right)y^3}{\left(\dfrac{x+4y}{y^3}\right)y^3}\Rightarrow \dfrac{x^2y-16y^3}{x+4y}\Rightarrow \dfrac{y(x^2-16y^2)}{x+4y}\Rightarrow\dfrac{y(x-4y)\cancel{(x+4y)}}{\cancel{x+4y}} \\ \\
   \Rightarrow y(x-4y)
   \end{array}\)
5. \(\begin{array}{l}
   \\
   5y+2\cdot 9y+6y\sqrt{y} \\
   23y+6y\sqrt{y}
   \end{array}\)
6. \(\dfrac{(28)(7+3\sqrt{5})}{(7-3\sqrt{5})(7+3\sqrt{5})}\Rightarrow \dfrac{196+84\sqrt{5}}{49-9\cdot 5}\Rightarrow \dfrac{196+84\sqrt{5}}{4}\Rightarrow 49+21\sqrt{5}\)
7. \(\begin{array}{l}
   \\ \\ \\ \\ \\ \\ \\
   (27a^{-\frac{3}{8}})^{\frac{1}{3}} \\ \\
   27^{\frac{1}{3}}a^{-\frac{3}{8}\cdot \frac{1}{3}} \\ \\
   3a^{-\frac{1}{8}} \\ \\
   \dfrac{3}{a^{\frac{1}{8}}}\Rightarrow \dfrac{3}{\sqrt[8]{a}}
   \end{array}\)
8. \(\phantom{1}\)
   \((\sqrt{3x-2})^2=(\sqrt{5x+4})^2 \\ \)
   \(\begin{array}{rrrrrrrr}
   &3x&-&2&=&5x&+&4 \\
   -&3x&-&4&&-3x&-&4 \\
   \midrule
   &&&-6&=&2x&& \\ \\
   &&&x&=&\dfrac{-6}{2}&=&-3
   \end{array}\)
9. \(\phantom{1}\)
   1. \(\begin{array}{rrl}
      \\ \\ \\
      \dfrac{2x^2}{2}&=&\dfrac{72}{2} \\ \\
      x^2&=&36 \\
      x&=&\pm 6
      \end{array}\)
   2. \(\begin{array}{rrl}
      \\ \\
      2x^2-8x&=&0 \\
      2x(x-4)&=&0 \\
      x&=&0, 4
      \end{array}\)
10. \(\phantom{1}\)
    1. \(\begin{array}{rrl}
       \\
       (x+5)(x+1)&=&0 \\
       x&=&-1, -5
       \end{array}\)
    2. \(\begin{array}{l}
       \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
       \text{Quadratic:} \\ \\
       x^2-10x+4=0 \\ \\
       \dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(4)}}{2} \\ \\
       \dfrac{-10\pm \sqrt{100-16}}{2} \\ \\
       \dfrac{10\pm \sqrt{84}}{2} \\ \\
       \dfrac{10\pm 2\sqrt{21}}{2}\Rightarrow 5 \pm \sqrt{21}
       \end{array}\)
11. \(\phantom{1}\)
    \(\left(\dfrac{8}{4x}=\dfrac{2}{x}+3\right)(4x) \\ \)
    \(\begin{array}{rrrrl}
    8&=&8&+&3(4x) \\
    -8&&-8&& \\
    \midrule
    \dfrac{0}{12}&=&\dfrac{12x}{12}&& \\ \\
    x&=&0&&\therefore \text{Undefined. No solution}
    \end{array}\)
12. \(\begin{array}{rrl}
    \\ \\ \\ \\ \\ \\ \\
    \text{Let }u&=&x^2 \\ \\
    u^2-17u+16&=&0 \\
    (u-16)(u-1)&=&0 \\ \\
    (x^2-16)(x^2-1)&=&0 \\
    (x-4)(x+4)(x-1)(x+1)&=&0 \\
    x&=& \pm 1, \pm 4
    \end{array}\)
13. \(\begin{array}{ll}
    \begin{array}{rrl}
    \\ \\ \\ \\
    L&=&W+6 \\
    \text{Area}&=&12+\text{Perimeter} \\ \\
    L\cdot W&=& 12+2L+2W \\
    (W+6)W&=&12+2(W+6)+2W \\
    W^2+6W&=&12+2W+12+2W
    \end{array}
    & \hspace{0.25in}
    \begin{array}{rrrrcrr}
    \\ \\ \\ \\ \\ \\ \\
    0&=&W^2&+&6W&& \\
    &&&-&4W&-&24 \\
    \midrule
    0&=&W^2&+&2W&-&24 \\
    0&=&(W&+&6)(W&-&4) \\
    W&=&\cancel{-6},&4&&& \\ \\
    L&=&W&+&6&& \\
    L&=&4&+&6&& \\
    L&=&10&&&& \\
    \end{array}
    \end{array}\)
14. \(\phantom{1}\)
    \(x, x+2, x+4 \\ \)
    \(\begin{array}{ll}
    \begin{array}{rrrrrrrrrrr}
    &&x(x&+&4)&=&31&+&x&+&2 \\
    x^2&+&4x&&&=&33&+&x&& \\
    &-&x&-&33&&-33&-&x&& \\
    \midrule
    x^2&+&3x&-&33&=&0&&&&
    \end{array}
    & \hspace{0.25in}
    \begin{array}{l}
    \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\
    \dfrac{-3\pm \sqrt{3^2-4(1)(-33)}}{2} \\ \\
    \dfrac{-3\pm \sqrt{141}}{2}
    \end{array}
    \end{array}\)
15. \(\begin{array}{rrrrcl}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    &d&=&r&\cdot &t \\
    \text{To outpost:}&60&=&(B&-&C)5 \\
    \text{Back:}&60&=&(B&+&C)3 \\ \\
    &60&=&5B&-&5C \\
    &60&=&3B&+&3C \\ \\
    &12&=&B&-&C \\
    +&20&=&B&+&C \\
    \midrule
    &32&=&2B&& \\
    &\therefore B&=&16&\text{ km/h}& \\ \\
    &\therefore B&+&C&=&\phantom{-}20 \\
    &16&+&C&=&\phantom{-}20 \\
    -&16&&&&-16 \\
    \midrule
    &&&C&=&4\text{ km/h}
    \end{array}\)