"

Midterm 2: Version D Answer Key

[latexpage]

  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. \(\phantom{1}\)
    \(A+B=60\Rightarrow B=60-A \\ \)
    \(\begin{array}{llclrll}
    \\ \\ \\ \\ \\ \\ \\
    &&3.80A&+&3.55B&=&\phantom{-}218.50 \\
    3.80A&+&3.55(60&-&A)&=&\phantom{-}218.50 \\
    3.80A&+&213&-&3.55A&=&\phantom{-}218.50 \\
    &-&213&&&=&-213 \\
    \midrule
    &&&&0.25A&=&5.50 \\ \\
    &&&&A&=&\dfrac{5.50}{0.25}\text{ or 22 kg} \\ \\
    &&&&B&=&60-A \\
    &&&&B&=&60-22 \\
    &&&&B&=&38\text{ kg}
    \end{array}\)

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.