Answer Key 8.7

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

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