Midterm 2: Version C Answer Key

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

    Graph with lines intersecting at (0,-2)

  2. \phantom{1}
    x+y=2\Rightarrow x=2-y \\
    \begin{array}{rrrrrrl} 3(2&-&y)&-&4y&=&13 \\ 6&-&3y&-&4y&=&13 \\ -6&&&&&&-6 \\ \midrule &&&&-7y&=&\phantom{-}7 \\ &&&&y&=&-1 \\ \\ &&&&x&=&2--1 \\ &&&&x&=&3 \end{array}
    (3,-1)
  3. \begin{array}{rrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &4x&-&3y&=&6 \\ +&4x&+&3y&=&2 \\ \midrule &&&8x&=&8 \\ &&&x&=&1 \\ \\ &3y&+&4(1)&=&2 \\ &&-&4&&-4 \\ \midrule &&&3y&=&-2 \\ \\ &&&y&=&-\dfrac{2}{3} \end{array}
    \left(1,-\dfrac{2}{3}\right)
  4. \begin{array}{ll} \begin{array}{rrrrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \left[1\right]&(x&+&y&+&z&=&\phantom{-}6)(-2) \\ \left[2\right]&&&(-2x&+&z&=&-3)(-1) \\ \\ \left[1\right]&-2x&-&2y&-&2z&=&-12 \\ +&&&2y&+&4z&=&\phantom{-}10 \\ \midrule &&&-2x&+&2z&=&-2 \\ +&&\left[2\right]&2x&-&z&=&\phantom{-}3 \\ \midrule &&&&&z&=&1 \\ \end{array} & \hspace{0.25in} \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 2y&+&4z&=&10 \\ 2y&+&4(1)&=&10 \\ &-&4&&-4 \\ \midrule &&2y&=&6 \\ &&y&=&3 \\ \\ -2x&+&z&=&-3 \\ -2x&+&1&=&-3 \\ &-&1&&-1 \\ \midrule &&-2x&=&-4 \\ &&x&=&2 \\ \end{array} \end{array}
    (2,3,1)
  5. 36+\{\cancel{-2x-\left[6x-3(5-2x)\right]\}^0}1+6x^3
    36+1+6x^3\Rightarrow 6x^3+37
  6. 6a^2b(a^2-9)
    6a^4b-54a^2b
  7. \begin{array}{rrrrrlrrrr} \\ \\ \\ \\ \\ \\ &x^2&+&3x&+&5&&&& \\ \times &x^2&+&3x&+&5&&&& \\ \midrule &x^4&+&3x^3&+&5x^2&&&& \\ &&&3x^3&+&9x^2&+&15x&& \\ +&&&&&5x^2&+&15x&+&25 \\ \midrule &x^4&+&6x^3&+&19x^2&+&30x&+&25 \end{array}
  8. \polylongdiv{2x^4+x^3+4x^2-4x-5}{2x+1}
  9. x^2-x+18x-18
    x(x-1)+18(x-1)
    (x-1)(x+18)
  10. 2(a^2-2ab-15b^2)
    2(a^2-5ab+3ab-15b^2)
    2\left[a(a-5b)+3b(a-5b)\right]
    2(a-5b)(a+3b)
  11. (2x)^3-y^3
    (2x-y)(4x^2+2xy+y^2)
  12. (4y^2-x^2)(4y^2+x^2)
    (2y-x)(2y+x)(4y^2+x^2)
  13. \phantom{1}
    B+S=30\Rightarrow B=30-S \\
    \begin{array}{rrrrrrr} B&-&10&=&4(S&-&10) \\ 30-S&-&10&=&4S&-&40 \\ +S&+&40&&+S&+&40 \\ \midrule &&60&=&5S&& \\ \\ &&S&=&\dfrac{60}{5}&=&12 \\ \\ &&\therefore B&=&30&-&S \\ &&B&=&30&-&12 \\ &&B&=&18&& \end{array}
  14. \begin{array}{rrrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &(D&+&N&=&\phantom{1}18)(-1) \\ &(10D&+&5N&=&120)(\div 5) \\ \\ &-D&-&N&=&-18 \\ +&2D&+&N&=&\phantom{-}24 \\ \midrule &&&D&=&6 \\ \\ \therefore &D&+&N&=&18 \\ &6&+&N&=&18 \\ &-6&&&&-6 \\ \midrule &&&N&=&12 \end{array}
  15. \phantom{1}
    \text{if }x=5\%, \text{ then }10-x=30\% \\
    \begin{array}{rrrrcrl} 5x&+&30(10&-&x)&=&25(10) \\ 5x&+&300&-&30x&=&\phantom{-}250 \\ &-&300&&&&-300 \\ \midrule &&&&-25x&=&-50 \\ \\ &&&&x&=&\dfrac{-50}{-25}\text{ or 2 L of 5\%} \\ \\ &&10&-&x&=&\text{8 L of 30\%} \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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