Midterm 2: Version E Answer Key

  1. x-y=-3
    x y
    0 3
    −3 0
    x+2y=3
    x y
    3 0
    0 \dfrac{3}{2}

    Graph with lines intersecting at (-1,2)

  2. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(2x&-&5y&=&-2)(-3) \\ &(3x&-&4y&=&\phantom{-}4)(2) \\ \\ &-6x&+&15y&=&6 \\ +&6x&-&8y&=&8 \\ \midrule &&&7y&=&14 \\ &&&y&=&2 \\ \\ &2x&-&5(2)&=&-2 \\ &2x&-&10&=&-2 \\ &&+&10&&+10 \\ \midrule &&&2x&=&8 \\ &&&x&=&4 \end{array}
    (4,2)
  3. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(4x&+&3y&=&-29)(-2) \\ &(3x&+&2y&=&-21)(3) \\ \\ &-8x&-&6y&=&\phantom{-}58 \\ +&9x&+&6y&=&-63 \\ \midrule &&&x&=&-5 \\ \\ &3(-5)&+&2y&=&-21 \\ &-15&+&2y&=&-21 \\ +&15&&&&+15 \\ \midrule &&&2y&=&-6 \\ &&&y&=&-3 \end{array}
    (-5,-3)
  4. \begin{array}{ll} \begin{array}{rrrrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ &(x&+&y&-&3z&=&0)(-2) \\ \\ &-2x&-&2y&+&6z&=&0 \\ +&2x&-&3y&&&=&16 \\ \midrule &&&-5y&+&6z&=&16 \\ \\ &&&(2y&-&2z&=&-12)(3) \\ \\ &&&6y&-&6z&=&-36 \\ +&&&-5y&+&6z&=&\phantom{-}16 \\ \midrule &&&&&y&=&-20 \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 2x&-&3y&=&16 \\ 2x&-&3(-20)&=&16 \\ 2x&+&60&=&16 \\ &&-60&&-60 \\ \midrule &&2x&=&-44 \\ &&x&=&-22 \\ \\ 2y&-&2z&=&-12 \\ 2(-20)&-&2z&=&-12 \\ -40&-&2z&=&-12 \\ +40&&&&+40 \\ \midrule &&-2z&=&28 \\ &&z&=&-14 \end{array} \end{array}
    (-22,-20,-14)
  5. 5-4\left[2x-2\cancel{(6x-5)^0}1-(7-2x)\right]
    5-4\left[2x-2(1)-7+2x\right]
    5-4\left[4x-9\right]
    5-16x+36
    -16x+41
  6. 3ab^4(a^2-25)
    3a^3b^4-75ab^4
  7. \begin{array}{rrrrrlrrrr} \\ \\ \\ \\ \\ \\ &x^2&+&3x&-&6&&&& \\ \times &x^2&+&3x&-&6&&&& \\ \midrule &x^4&+&3x^3&-&6x^2&&&& \\ &&&3x^3&+&9x^2&-&18x&& \\ +&&&&-&6x^2&-&18x&+&36 \\ \midrule &x^4&+&6x^3&-&3x^2&-&36x&+&36 \end{array}
  8. \polylongdiv{3x^3+18+7x^2}{x+3}
  9. x^2-3z+7x-21
    x(x-3)+7(x-3)
    (x-3)(x+7)
  10. 4x^2(x+1)-9(x+1)
    (x+1)(4x^2-9)
    (x+1)(2x-3)(2x+3)
  11. (2x)^3-(3y)^3
    (2x-3y)(4x^2+6xy+9y^2)
  12. x^4-625x^2+x^2-625
    x^2(x^2-625)+1(x^2-625)
    (x^2+1)(x^2-625)
    (x^2+1)(x-25)(x+25)
  13. \phantom{1}
    B+G=20\Rightarrow B=20-G \\
    \begin{array}{rrrrrrrrrl} &G&-&4&=&2(B&-&4)&& \\ &G&-&4&=&2B&-&8&& \\ \\ &G&-&4&=&2(20&-&G)&-&8 \\ &G&-&4&=&40&-&2G&-&8 \\ &G&-&4&=&32&-&2G&& \\ +&2G&+&4&&4&+&2G&& \\ \midrule &&&3G&=&36&&&& \\ &&&G&=&12&&&& \\ \\ &&&B&=&20&-&G&& \\ &&&B&=&20&-&12&=&8 \\ \end{array}
  14. \phantom{1}
    x=16\%\text{ solution} \\
    \begin{array}{rrrrrrrr} &16x&+&6(20)&=&12(x&+&20) \\ &16x&+&120&=&12x&+&240 \\ -&12x&-&120&&-12x&-&120 \\ \midrule &&&4x&=&120&& \\ \\ &&&x&=&\dfrac{120}{4}&=&30\text{ ml} \\ \end{array}
  15. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(C&+&R&=&60)(-3.40) \\ \\ &-3.40C&-&3.40R&=&-204 \\ +&3.40C&+&3.90R&=&\phantom{-}213\\ \midrule &&&0.50R&=&\phantom{-}9 \\ \\ &&&R&=&\dfrac{9}{0.50}\text{ or 18 kg} \\ \\ &C&+&R&=&\phantom{-}60 \\ &C&+&18&=&\phantom{-}60 \\ &&-&18&&-18 \\ \midrule &&&C&=&\phantom{-}42\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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