Midterm 2: Prep Answer Key

Midterm Two Review

  1. x-2y=-4
    x y
    −4 0
    0 2
    −2 1
    x+y=5
    x y
    0 5
    5 0
    2 3

  2. \begin{array}{rrcrlrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ 2x&-&y&=&0&\Rightarrow &y=2x \\ 3x&+&4y&=&-22&& \\ \\ \therefore 3x&+&4(2x)&=&-22&& \\ 3x&+&8x&=&-22&& \\ &&11x&=&-22&& \\ &&x&=&-2&& \\ \\ &&y&=&2x&& \\ &&y&=&2(-2)&=&-4 \\ \end{array}
    (-2,-4)
  3. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(2x&-&5y&=&15)(2) \\ &(3x&+&2y&=&13)(5) \\ \midrule &4x&-&10y&=&30 \\ +&15x&+&10y&=&65 \\ \midrule &&&19x&=&95 \\ &&&x&=&5 \\ \\ &\therefore 3(5)&+&2y&=&\phantom{-}13 \\ &15&+&2y&=&\phantom{-}13 \\ &-15&&&&-15 \\ \midrule &&&2y&=&-2 \\ &&&y&=&-1 \end{array}
    (5,-1)
  4. \begin{array}{rr} \begin{array}{rrrrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &&&(5x&+&6z&=&-4)(-1) \\ \\ &5x&+&y&+&6z&=&-2 \\ +&-5x&&&-&6z&=&\phantom{-}4 \\ \midrule &&&&&y&=&2 \\ \\ &&&\therefore 2y&-&3z&=&\phantom{-}3 \\ &&&2(2)&-&3z&=&\phantom{-}3 \\ &&&-4&&&&-4 \\ \midrule &&&&&-3z&=&-1 \\ &&&&&z&=&\dfrac{1}{3} \\ \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ 5x&+&6z&=&-4 \\ 5x&+&6\left(\dfrac{1}{3}\right)&=&-4 \\ 5x&+&2&=&-4 \\ &-&2&&-2 \\ \midrule &&5x&=&-6 \\ &&x&=&-\dfrac{6}{5} \end{array} \end{array}
    -\dfrac{6}{5}, 2, \dfrac{1}{3}
  5. \begin{array}{rrrrrr} \\ \\ \\ &4a^2&-&9a&+&2 \\ &-a^2&+&4a&+&5 \\ +&3a^2&-&a&+&9 \\ \midrule &6a^2&-&6a&+&16 \end{array}
  6. 8x^4-12x^2y^2-15x^2y^2-3x^4\Rightarrow 5x^4-27x^2y^2
  7. \begin{array}{l} \\ \\ \\ 6-2\left[3x-20x+8-1\right] \\ 6-2\left[-17x+7\right] \\ 6+34x-14 \\ 34x-8 \end{array}
  8. 25a^{-10}b^6\text{ or } \dfrac{25b^6}{a^{10}}
  9. \begin{array}{l} \\ 8a^2(a^2+10a+25) \\ 8a^4+80a^3+200a^2 \end{array}
  10. \begin{array}{l} \\ 4ab^2(a^2-4) \\ 4a^3b^2-16ab^2 \end{array}
  11. \begin{array}{rrrrrrrr} \\ \\ \\ \\ \\ &x^2&-&4x&+&7\phantom{x}&& \\ \times &&&x&-&3\phantom{x}&& \\ \midrule &x^3&-&4x^2&+&7x&& \\ +&&-&3x^2&+&12x&-&21 \\ \midrule &x^3&-&7x^2&+&19x&-&21 \\ \end{array}
  12. \begin{array}{rrrrrrrrrr} \\ \\ \\ \\ \\ \\ &2x^2&+&x&-&3\phantom{x^2}&&&& \\ \times &2x^2&+&x&-&3\phantom{x^2}&&&& \\ \midrule &4x^4&+&2x^3&-&6x^2&&&& \\ &&&2x^3&+&x^2&-&3x&& \\ +&&&&&-6x^2&-&3x&+&9 \\ \midrule &4x^4&+&4x^3&-&11x^2&-&6x&+&9 \end{array}
  13. \begin{array}{rrrrrrrrrr} \\ \\ \\ \\ \\ \\ &x^2&+&5x&-&2\phantom{x^2}&&&& \\ \times &2x^2&-&x&+&3\phantom{x^2}&&&& \\ \midrule &2x^4&+&10x^3&-&4x^2&&&& \\ &&&-x^3&-&5x^2&+&2x&& \\ +&&&&&3x^2&+&15x&-&6 \\ \midrule &2x^4&+&9x^3&-&6x^2&+&17x&-&6 \end{array}
  14. \begin{array}{rrrrrrrrrr} \\ \\ \\ \\ \\ (x+4)(x+4)&\Rightarrow &&x^2&+&8x&+&16&& \\ &&\times&&&x&+&4&& \\ \midrule &&&x^3&+&8x^2&+&16x&& \\ &&+&&&4x^2&+&32x&+&64 \\ \midrule &&&x^3&+&12x^2&+&48x&+&64 \end{array}
  15. r^{-4-3}s^{9+9}\Rightarrow r^{-7}s^{18}\Rightarrow \dfrac{s^{18}}{r^7}
    \dfrac{s^{18}}{r^7}
  16. \begin{array}{l} \\ \\ (x^{-2--2}y^{-3-4})^{-1} \\ (1\cancel{x^0}y^{-7})^{-1} \\ y^7 y^7 \end{array}
  17. \polylongdiv{2x^3-7x^2+15}{x-2}
  18. 2^3\cdot 11
  19. 2^5\cdot 3\cdot 7 \left\{ \begin{array}{l} 84=2^2\cdot 3\cdot 7 \\ 96=2^5\cdot 3 \end{array}\right.
  20. x(5y+6z)-3(5y+6z)
    (5y+6z)(x-3)
  21. -12=4\times -3
    1=4+-3 \\
    x^2+4x-3x-12
    x(x+4)-3(x+4)
    (x+4)(x-3)
  22. x^2(x+1)-4(x+1)
    (x+1)(x^2-4)
    x+1)(x-2)(x+2)</li> <li>\(x^3-(3y)^3
    (x-3y)(x^2+3xy+9y^2)
  23. (x^2-36)(x^2+1)
    (x-6)(x+6)(x^2+1)
  24. \begin{array}{lll} \begin{array}{rrrrl} (A&+&B&=&70)(-4) \\ 4A&+&7B&=&430 \end{array} & \Rightarrow \hspace{0.25in} \begin{array}{rrrrrl} \\ \\ \\ \\ &-4A&-&4B&=&-280 \\ +&4A&+&7B&=&\phantom{-}430 \\ \midrule &&&3B&=&\phantom{-}150 \\ \\ &&&B&=&\dfrac{150}{3}\text{ or }50 \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \therefore A&+&B&=&70 \\ \\ A&+&50&=&70 \\ &&-50&&-50 \\ \midrule &&A&=&20 \end{array} \end{array}
  25. \begin{array}{rrcrrrl} \\ \\ \\ \\ \\ \\ \\ 5x&+&21(2)&=&11(x&+&2) \\ \\ 5x&+&42&=&11x&+&22 \\ -5x&-&22&&-5x&-&22 \\ \midrule &&20&=&6x&& \\ \\ &&x&=&\dfrac{20}{6}&=&3\dfrac{1}{3}\text{ litres} \\ \end{array}
  26. \phantom{1}
    B+G=16\Rightarrow B=16-G\text{ or }G=16-B \\
    \begin{array}{ll} \begin{array}{rrrrrrr} \\ \\ \\ \\ \\ G&-&4&=&3(B&-&4) \\ 16-B&-&4&=&3B&-&12 \\ +B&+&12&&+B&+&12 \\ \midrule &&24&=&4B&& \\ \\ &&B&=&\dfrac{24}{4}&=&6 \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \therefore G&=&16&-&B \\ G&=&16&-&6 \\ G&=&10&& \end{array} \end{array}

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book