Midterm 2: Version D Answer Key

  1. x-2y=-6
    x y
    0 3
    −6 0
    x+y=6
    x y
    0 6
    6 0

    Graph with lines intersecting at (2,4)

  2. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(3x&-&2y&=&0)(5) \\ &(2x&+&5y&=&0)(2) \\ \\ &15x&-&10y&=&0 \\ +&4x&+&10y&=&0 \\ \midrule &&&19x&=&0 \\ &&&x&=&0 \\ \\ &\therefore \cancel{2x}0&+&5y&=&0 \\ &&&5y&=&0 \\ &&&y&=&0 \end{array}
    (0,0)
  3. \begin{array}{rrrrrr} \\ \\ &2x&-&3y&=&8 \\ +&-2x&+&3y&=&4 \\ \midrule &&&0&=&12 \\ \end{array}
    \phantom{1}
    ∴ No solution. Parallel lines.
  4. \begin{array}{ll} \begin{array}{rrrrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(2x&+&y&-&3z&=&-7)(2) \\ \\ &4x&+&2y&-&6z&=&-14 \\ +&&-&2y&+&3z&=&\phantom{-1}9 \\ \midrule &&&4x&-&3z&=&-5 \\ \\ &&&(3x&+&z&=&6)(3) \\ \\ &&&4x&-&3z&=&-5 \\ +&&&9x&+&3z&=&18 \\ \midrule &&&&&13x&=&13 \\ &&&&&x&=&1 \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 3x&+&z&=&6 \\ 3(1)&+&z&=&6 \\ -3&&&&-3 \\ \midrule &&z&=&3 \\ \\ -2y&+&3z&=&9 \\ -2y&+&3(3)&=&9 \\ &&-9&&-9 \\ \midrule &&-2y&=&0 \\ &&y&=&0 \end{array} \end{array}
    (1,0,3)
  5. 36-\cancel{\{-2x-\left[6x-3(5-2x)\right]\}^0}1+3x^2
    36-1+3x^2
    35+3x^2
  6. 3a^2(a^2-4a+4)
    3a^4-12a^3+12a^2
  7. \begin{array}{rrrrrlrrrr} \\ \\ \\ \\ \\ \\ &x^2&+&2x&-&4&&&& \\ \times &x^2&+&2x&-&4&&&& \\ \midrule &x^4&+&2x^3&-&4x^2&&&& \\ &&&2x^3&+&4x^2&-&8x&& \\ +&&&&-&4x^2&-&8x&+&16 \\ \midrule &x^4&+&4x^3&-&4x^2&-&16x&+&16 \end{array}
  8. \polylongdiv{x^4+4x^3+4x^2+10x+20}{x+2}
  9. x^2-3x+6x-18
    x(x-3)+6(x-3)
    (x-3)(x+6)
  10. 3x^2+xy+24xy+8y^2
    x(3x+y)+8y(3x+y)
    (3x+y)(x+8y)
  11. (5x)^3-y^3
    (5x-y)(25x^2+5xy+y^2)
  12. (9y^2-4x^2)(9y^2+4x^2)
    (3y-2x)(3y+2x)(9y^2+4x^2)
  13. \phantom{1}
    B+G=18\Rightarrow G=18-B \\
    \begin{array}{rrrcrrrr} &G&-&4&=&4(B&-&4) \\ &18-B&-&4&=&4B&-&16 \\ +&16+B&&&&+B&+&16 \\ \midrule &&&30&=&5B&& \\ \\ &&&B&=&\dfrac{30}{5}&=&6 \\ \\ &&&\therefore G&=&18&-&B \\ &&&\phantom{\therefore}G&=&18&-&6 \\ &&&\phantom{\therefore}G&=&12&& \end{array}
  14. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(D&+&Q&=&20)(-10) \\ &10D&+&25Q&=&350 \\ \\ &-10D&-&10Q&=&-200 \\ +&10D&+&25Q&=&\phantom{-}350 \\ \midrule &&&15Q&=&150 \\ \\ &&&Q&=&\dfrac{150}{15}\text{ or }10 \\ \\ \therefore &D&+&Q&=&\phantom{-}20 \\ &D&+&10&=&\phantom{-}20 \\ &&-&10&&-10 \\ \midrule &&&D&=&10 \end{array}
  15. \phantom{1}
    A+B=60\Rightarrow B=60-A \\
    \begin{array}{llclrll} \\ \\ \\ \\ \\ \\ \\ &&3.80A&+&3.55B&=&\phantom{-}218.50 \\ 3.80A&+&3.55(60&-&A)&=&\phantom{-}218.50 \\ 3.80A&+&213&-&3.55A&=&\phantom{-}218.50 \\ &-&213&&&=&-213 \\ \midrule &&&&0.25A&=&5.50 \\ \\ &&&&A&=&\dfrac{5.50}{0.25}\text{ or 22 kg} \\ \\ &&&&B&=&60-A \\ &&&&B&=&60-22 \\ &&&&B&=&38\text{ kg} \end{array}

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

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