Midterm 3: Version E Answer Key

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1. \(\dfrac{\cancel{12}1\cancel{m^3}}{\cancel{5}1\cancel{n^2}}\cdot \dfrac{\cancel{15}\cancel{3}1\cancel{n^3}}{\cancel{36}\cancel{3}1\cancel{m^6}}\cdot \dfrac{\cancel{8}4m^{\cancel{4}}}{\cancel{6}3n^{\cancel{2}}}\Rightarrow \dfrac{4m}{3n}\)
2. \(\dfrac{\cancel{x}1\cancel{(x+2)}}{\cancel{(x+2)}\cancel{(x+7)}}\cdot \dfrac{\cancel{2}1\cancel{(x+7)}}{\cancel{2}1x^{\cancel{3}2}}=\dfrac{1}{x^2}\)
3. \(\phantom{1}\)
   \(\left(\dfrac{x-3}{7}-\dfrac{x-15}{28}=\dfrac{3}{4}\right)(28) \\ \)
   \(\begin{array}{rrrrrrrrl}
   4(x&-&3)&-&(x&-&15)&=&3(7) \\
   4x&-&12&-&x&+&15&=&21 \\
   &+&12&&&-&15&&-15+12 \\
   \midrule
   &&&&&&3x&=&18 \\
   &&&&&&x&=&6
   \end{array}\)
4. \(\begin{array}{l}
   \dfrac{\left(\dfrac{x^2}{y^2}-36\right)y^3}{\left(\dfrac{x+6y}{y^3}\right)y^3}\Rightarrow \dfrac{x^2y-36y^3}{x+6y}\Rightarrow \dfrac{y(x^2-36y^2)}{x+6y}\Rightarrow \dfrac{y(x-6y)\cancel{(x+6y)}}{\cancel{(x+6y)}} \\ \\
   \Rightarrow y(x-6y)
   \end{array}\)
5. \(\begin{array}{l}
   \\ \\ \\ \\ \\ \\
   \sqrt{x^6\cdot x\cdot y^4\cdot y}+2xy\sqrt{36\cdot x\cdot y^4\cdot y}-\sqrt{x\cdot y^2\cdot y} \\ \\
   x^3y^2\sqrt{xy}+2xy\cdot 6y^2\sqrt{xy}-y\sqrt{xy} \\ \\
   x^3y^2\sqrt{xy}+12xy^3\sqrt{xy}-y\sqrt{xy} \\ \\
   \sqrt{xy}(x^3y^2+12xy^3-y)
   \end{array}\)
6. \(\dfrac{\sqrt{7}}{3-\sqrt{7}}\cdot \dfrac{3+\sqrt{7}}{3+\sqrt{7}}\Rightarrow \dfrac{3\sqrt{7}+7}{9-7}\Rightarrow \dfrac{3\sqrt{7}+7}{2}\)
7. \(\left(\dfrac{\cancel{x^0}1y^4}{z^{-12}}\right)^{\frac{1}{4}}\Rightarrow \dfrac{y^{4\cdot \frac{1}{4}}}{z^{-12\cdot \frac{1}{4}}}\Rightarrow \dfrac{y^1}{z^{-3}}\Rightarrow yz^3\)
8. \(\phantom{1}\)
   \((\sqrt{4x-5})^2=(\sqrt{2x+3})^2 \\ \)
   \(\begin{array}{rrrrrrrr}
   &4x&-&5&=&2x&+&3 \\
   -&2x&+&5&&-2x&+&5 \\
   \midrule
   &&&2x&=&8&& \\
   &&&x&=&4&&
   \end{array}\)
9. \(\phantom{1}\)
   1. \(\left(\dfrac{x^2}{3}=27\right)(3)\Rightarrow x^2=81 \Rightarrow x=\pm 9\)
   2. \(\begin{array}{rrl}
      \\ \\ \\
      27x^2+3x&=&0 \\
      3x(9x+1)&=&0 \\
      x&=&0, -\dfrac{1}{9}
      \end{array}\)
10. \(\phantom{1}\)
    1. \(\begin{array}{rrl}
       \\
       (x-12)(x+1)&=&0 \\
       x&=&12, -1
       \end{array}\)
    2. \(\begin{array}{rrl}
       \\ \\
       x^2+13x+12&=&0 \\
       (x+12)(x+1)&=&0 \\
       x&=&-1, -12
       \end{array}\)
11. \(\phantom{1}\)
    \(\left(\dfrac{2}{x}=\dfrac{2x}{3x+8}\right)(x)(3x+8) \\ \)
    \(\begin{array}{ll}
    \begin{array}{rrl}
    2(3x+8)&=&2x^2 \\
    6x+16&=&2x^2 \\
    0&=&2x^2-6x-16 \\
    0&=&2(x^2-3x-8)
    \end{array}
    & \hspace{0.25in}
    \begin{array}{l}
    \\ \\ \\
    \dfrac{-(-3)\pm \sqrt{(-3)^2-4(1)(-8)}}{2(1)} \\ \\
    \dfrac{3\pm \sqrt{9+32}}{2} \\ \\
    \dfrac{3\pm \sqrt{41}}{2}
    \end{array}
    \end{array}\)
12. \(\begin{array}{rrl}
    \\ \\
    (x^2-64)(x^2+1)&=&0 \\
    (x-8)(x+8)(x^2+1)&=&0 \\
    x&=&\pm 8
    \end{array}\)
13. \(\begin{array}{rrcrrrrrrrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    &&&&A&=&20&+&P&&&& \\ \\
    &&L(L&-&5)&=&20&+&2(L&-&5)&+&2L \\
    L^2&-&5L&&&=&20&+&2L&-&10&+&2L \\
    &-&4L&-&10&&-10&-&4L&&&& \\
    \midrule
    L^2&-&9L&-&10&=&0&&&&&& \\
    (L&-&10)(L&+&1)&=&0&&&&&& \\
    &&&&L&=&10,&\cancel{-1}&&&&& \\ \\
    &&&&W&=&L&-&5&&&& \\
    &&&&W&=&10&-&5&&&& \\
    &&&&W&=&5&&&&&&
    \end{array}\)
14. \(\phantom{1}\)
    \(x, x+2, x+4 \\ \)
    \(\begin{array}{rrcrrrrrcrr}
    &&x(x&+&4)&=&35&+&10(x&+&2) \\
    x^2&+&4x&&&=&35&+&10x&+&20 \\
    &-&10x&-&55&&-55&-&10x&& \\
    \midrule
    x^2&-&6x&-&55&=&0&&&& \\
    (x&-&11)(x&+&5)&=&0&&&& \\
    &&&&x&=&11,&-5&&&
    \end{array}\)
    \(\phantom{1}\)
    \(\text{numbers are }11,13,15\text{ or }-5,-3,-1\)
15. \(\begin{array}{rrrrrrcrrr}
    \\ \\ \\ \\ \\ \\
    &&&d_{\text{d}}&=&d_{\text{u}}&+&9&& \\ \\
    &3(5&+&r)&=&4(5&-&r)&+&9 \\
    &15&+&3r&=&20&-&4r&+&9 \\
    -&15&+&4r&&-15&+&4r&& \\
    \midrule
    &&&7r&=&14&&&& \\
    &&&r&=&2&\text{km/h}&&&
    \end{array}\)