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Answer Key 8.7

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  1. \(\begin{array}{rrcrrrl}
    \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&2(x)&&&& \\ \\
    3x(2x)&-&x&-&2&=&0 \\
    6x^2&-&x&-&2&=&0 \\
    (3x&-&2)(2x&+&1)&=&0 \\ \\
    &&&&x&=&\dfrac{2}{3}, -\dfrac{1}{2}
    \end{array}\)
  2. \(\begin{array}{rrcrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&x&+&1&& \\ \\
    (x&+&1)(x&+&1)&=&\phantom{-}4 \\
    x^2&+&2x&+&1&=&\phantom{-}4 \\
    &&&-&4&&-4 \\
    \midrule
    x^2&+&2x&-&3&=&0 \\
    (x&-&1)(x&+&3)&=&0 \\ \\
    &&&&x&=&1, -3
    \end{array}\)
  3. \(\begin{array}{rrcrrrllrrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&x&-&4&&&&&& \\ \\
    x(x&-&4)&+&20&=&5x&-&2(x&-&4) \\
    x^2&-&4x&+&20&=&5x&-&2x&+&8 \\
    &-&3x&-&8&&&-&3x&-&8 \\
    \midrule
    x^2&-&7x&+&12&=&0&&&& \\
    (x&-&4)(x&-&3)&=&0&&&& \\ \\
    &&&&x&=&3,&4&&&
    \end{array}\)
  4. \(\begin{array}{rrrrrrrrllr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&x&-&1&&&&&& \\ \\
    x^2&+&6&+&x&-&2&=&\phantom{-}2x(x&-&1) \\
    &&x^2&+&x&+&4&=&\phantom{-}2x^2&-&2x \\
    &-&2x^2&+&2x&&&&-2x^2&+&2x \\
    \midrule
    &&-x^2&+&3x&+&4&=&0&& \\
    &&x^2&-&3x&-&4&=&0&& \\
    &&(x&-&4)(x&+&1)&=&0&& \\ \\
    &&&&&&x&=&4, 1&& \\
    \end{array}\)
  5. \(\begin{array}{rrcrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&x&-&3&& \\ \\
    x(x&-&3)&+&6&=&2x \\
    x^2&-&3x&+&6&=&2x \\
    &-&2x&&&&-2x \\
    \midrule
    x^2&-&5x&+&6&=&0 \\
    (x&-&3)(x&-&2)&=&0 \\ \\
    &&&&x&=&2, 3
    \end{array}\)
  6. \(\begin{array}{rrcrrrlrrrrrcrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&-&1)(3&-&x)&&&&&&&& \\ \\
    (x&-&4)(3&-&x)&=&\phantom{-}12(x&-&1)&+&(x&-&1)(3&-&x) \\
    -x^2&+&7x&-&12&=&\phantom{-}12x&-&12&-&x^2&+&4x&-&3 \\
    +x^2&-&16x&+&15&&-12x&+&12&+&x^2&-&4x&+&3 \\
    \midrule
    &&-9x&+&3&=&0&&&&&&&& \\
    &&&&3&=&9x&&&&&&&& \\ \\
    &&&&x&=&\dfrac{3}{9}\hspace{0.1in}\text{ or}&\dfrac{1}{3}&&&&&&&
    \end{array}\)
  7. \(\begin{array}{rrcrcrrrrrcrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(2m&-&5)(3m&+&1)(2)&&&&&& \\ \\
    3m(3m&+&1)(2)&-&7(2m&-&5)(2)&=&3(2m&-&5)(3m&+&1) \\
    18m^2&+&6m&-&28m&+&70&=&18m^2&-&39m&-&15 \\
    -18m^2&&&+&39m&+&15&&-18m^2&+&39m&+&15 \\
    \midrule
    &&&&17m&+&85&=&0&&&& \\
    &&&&&-&85&&-85&&&& \\
    \midrule
    &&&&&&\dfrac{17m}{17}&=&\dfrac{-85}{17}&&&& \\ \\
    &&&&&&m&=&-5&&&&
    \end{array}\)
  8. \(\begin{array}{rrcrrrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(1&-&x)(3&-&x)&& \\ \\
    (4&-&x)(3&-&x)&=&12(1&-&x) \\
    12&-&7x&+&x^2&=&12&-&12x \\
    -12&+&12x&&&&-12&+&12x \\
    \midrule
    &&x^2&+&5x&=&0&& \\
    &&x(x&+&5)&=&0&& \\ \\
    &&&&x&=&0,&-5&
    \end{array}\)
  9. \(\begin{array}{crrrrrcrrrrrcrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&2(y&-&3)(y&-&4)&&&&&&&& \\ \\
    7(2)(y&-&4)&-&1(y&-&3)(y&-&4)&=&(y&-&2)(2)(y&-&3) \\
    14y&-&56&-&y^2&+&7y&-&12&=&2y^2&-&10y&+&12 \\ \\
    &&&&-\phantom{0}y^2&+&21y&-&68&=&2y^2&-&10y&+&12 \\
    &&&&-2y^2&+&10y&-&12&&-2y^2&+&10y&-&12 \\
    \midrule
    &&&&-3y^2&+&31y&-&80&=&0&&&& \\
    &&&&3y^2&-&31y&+&80&=&0&&&& \\
    &&&&(y&-&5)(3y&-&16)&=&0&&&& \\ \\
    &&&&&&&&y&=&5, &\dfrac{16}{3}&&&
    \end{array}\)
  10. \(\begin{array}{rrrrrrrrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&+&2)(x&-&2)&&&& \\ \\
    1(x&-&2)&+&1(x&+&2)&=&3x&+&8 \\
    x&-&2&+&x&+&2&=&3x&+&8 \\
    &&&&-2x&&&&-2x&& \\
    \midrule
    &&&&&&0&=&x&+&8 \\
    &&&&&&-8&&&-&8 \\
    \midrule
    &&&&&&x&=&-8&&
    \end{array}\)
  11. \(\begin{array}{rrcrcrrrcrcrrrcrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&+&1)(x&-&1)(6)&&&&&&&&&& \\ \\
    (x&+&1)(x&+&1)(6)&-&(x&-&1)(x&-&1)(6)&=&5(x&+&1)(x&-&1) \\
    6(x^2&+&2x&+&1)&-&6(x^2&-&2x&+&1)&=&5(x^2&&-&&1) \\
    6x^2&+&12x&+&6&-&6x^2&+&12x&-&6&=&5x^2&&&-&5 \\
    &&&&&&&&&&24x&=&5x^2&&&-&5 \\
    &&&&&&&&&&-24x&&&-&24x&& \\
    \midrule
    &&&&&&&&&&0&=&5x^2&-&24x&-&5 \\
    &&&&&&&&&&0&=&(5x&+&1)(x&-&5) \\ \\
    &&&&&&&&&&x&=&5, &-\dfrac{1}{5}&&&
    \end{array}\)
  12. \(\begin{array}{rrcrcrrrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&+&3)(x&-&2)&&&& \\ \\
    (x&-&2)(x&-&2)&-&1(x&+&3)&=&1 \\
    x^2&-&4x&+&4&-&x&-&3&=&1 \\
    &&&&&&&-&1&&-1 \\
    \midrule
    &&&&&&x^2&-&5x&=&0 \\
    &&&&&&x(x&-&5)&=&0 \\ \\
    &&&&&&&&x&=&0, 5
    \end{array}\)
  13. \(\begin{array}{rrrrcrrrrrcrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&-&1)(x&+&1)&&&&&& \\ \\
    x(x&+&1)&-&2(x&-&1)&=&4x^2&&&& \\
    x^2&+&x&-&2x&+&2&=&4x^2&&&& \\
    -x^2&&&+&x&-&2&&-x^2&+&x&-&2 \\
    \midrule
    &&&&&&0&=&3x^2&+&x&-&2 \\
    &&&&&&0&=&(3x&-&2)(x&+&1) \\ \\
    &&&&&&0&=&\dfrac{2}{3},&-1&&&
    \end{array}\)
  14. \(\begin{array}{rrrrcrrrr}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&+&2)(x&-&4)&& \\ \\
    2x(x&-&4)&+&2(x&+&2)&=&3x \\
    2x^2&-&8x&+&2x&+&4&=&3x \\
    &&&-&3x&&&&-3x \\
    \midrule
    &&2x^2&-&9x&+&4&=&0 \\
    &&(2x&-&1)(x&-&4)&=&0 \\ \\
    &&&&&&x&=&\dfrac{1}{2}, 4
    \end{array}\)
  15. \(\begin{array}{rrrrcrrrl}
    \\ \\ \\ \\ \\ \\ \\ \\ \\
    \text{LCD}&=&(x&+&1)(x&+&5)&& \\ \\
    2x(x&+&5)&-&3(x&+&1)&=&-8x^2 \\
    2x^2&+&10x&-&3x&-&3&=&-8x^2 \\
    +8x^2&&&&&&&&+8x^2 \\
    \midrule
    &&10x^2&+&7x&-&3&=&0 \\
    &&(10x&-&3)(x&+&1)&=&0 \\ \\
    &&&&&&x&=&\dfrac{3}{10}, -1
    \end{array}\)

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