Answer Key 8.6

  1. \begin{array}{rrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&5(2)&& \\ \\ 2(m&-&1)&=&5(8) \\ 2m&-&2&=&40 \\ &+&2&&+2 \\ \midrule &&\dfrac{2m}{2}&=&\dfrac{42}{2} \\ \\ &&m&=&21 \end{array}
  2. \begin{array}{rrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&2(x&-&8) \\ \\ 8(x&-&8)&=&\phantom{+}2(8) \\ 8x&-&64&=&\phantom{+}16 \\ &+&64&&+64 \\ \midrule &&\dfrac{8x}{8}&=&\phantom{+}\dfrac{80}{8} \\ \\ &&x&=&\phantom{+}10 \end{array}
  3. \begin{array}{rrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&9(p&-&4) \\ \\ 2(p&-&4)&=&9(10) \\ 2p&-&8&=&90 \\ &+&8&&+8 \\ \midrule &&\dfrac{2p}{2}&=&\dfrac{98}{2} \\ \\ &&p&=&49 \end{array}
  4. \begin{array}{rllll} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&9(n&+&2) \\ \\ 9(9)&=&3(n&+&2) \\ 81&=&3n&+&6 \\ -6&&&-&6 \\ \midrule \dfrac{75}{3}&=&\dfrac{3n}{3}&& \\ \\ n&=&25&& \end{array}
  5. \begin{array}{rrlrl} \\ \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&10(a&+&2) \\ \\ 3(a&+&2)&=&10(a) \\ 3a&+&6&=&10a \\ -3a&&&&-3a \\ \midrule &&\dfrac{6}{7}&=&\dfrac{7a}{7} \\ \\ &&a&=&\dfrac{6}{7} \end{array}
  6. \begin{array}{rrrrrrr} \\ \\ \\ \\ \\ \text{LCD}&=&3(4)&&&& \\ \\ 4(x&+&1)&=&3(x&+&3) \\ 4x&+&4&=&3x&+&9 \\ -3x&-&4&&-3x&-&4 \\ \midrule &&x&=&5&& \end{array}
  7. \begin{array}{rrrrcrr} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&3(p&+&4)&& \\ \\ 2(3)&=&(p&+&4)(p&+&5) \\ 6&=&p^2&+&9p&+&20 \\ -6&&&&&-&6 \\ \midrule 0&=&p^2&+&9p&+&14 \\ 0&=&(p&+&7)(p&+&2) \\ \\ p&=&-2,&-7&&& \\ \end{array}
  8. \begin{array}{rrrrcrr} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&10(n&+&1)&& \\ \\ 5(10)&=&(n&-&4)(n&+&1) \\ 50&=&n^2&-&3n&-&4 \\ -50&&&&&-&50 \\ \midrule 0&=&n^2&-&3n&-&54 \\ 0&=&(n&-&9)(n&+&6) \\ \\ n&=&9,&-6&&& \end{array}
  9. \begin{array}{rrcrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&5(x&-&2)&& \\ \\ (x&+&5)(x&-&2)&=&5(6) \\ x^2&+&3x&-&10&=&\phantom{-}30 \\ &&&-&30&&-30 \\ \midrule x^2&+&3x&-&40&=&0 \\ (x&-&5)(x&+&8)&=&0 \\ \\ &&&&x&=&5, -7 \end{array}
  10. \begin{array}{rrrrcrr} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&5(x&-&3)&& \\ \\ 20&=&(x&-&3)(x&+&5) \\ 20&=&x^2&+&2x&-&15 \\ -20&&&&&-&20 \\ \midrule 0&=&x^2&+&2x&-&35 \\ 0&=&(x&-&5)(x&+&7) \\ \\ x&=&5,&-7&&& \end{array}
  11. \begin{array}{rrcrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&4(m&-&4)&& \\ \\ (m&+&3)(m&-&4)&=&4(11) \\ m^2&-&m&-&12&=&\phantom{-}44 \\ &&&-&44&&-44 \\ \midrule (m^2&-&m&-&56)&=&0 \\ (m&-&8)(m&+&7)&=&0 \\ \\ &&&&m&=&8, -7 \end{array}
  12. \begin{array}{rrcrrrl} \\ \\ \\ \\ \\ \\ \\ \\ \text{LCD}&=&8(x&-&1)&& \\ \\ (x&-&5)(x&-&1)&=&4(8) \\ x^2&-&6x&+&5&=&\phantom{-}32 \\ &&&-&32&&-32 \\ \midrule x^2&-&6x&-&27&=&0 \\ (x&-&9)(x&+&3)&=&0 \\ \\ &&&&x&=&9, -3 \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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