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amorfide Group Title

1/x > 1 how to solve?

  • 2 years ago
  • 2 years ago

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  1. Sir_Rico_of_Eureka Group Title
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    same way as with an equals sign. just leave that alligator mouth there. you need to clear the fractions. try multiplying both sides by the fractions denominator

    • 2 years ago
  2. amorfide Group Title
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    okay so i multiply by x on both sides 1>x so why when i have |dw:1350339428053:dw| why when i have this do i multiply by (x-2)² @Sir_Rico_of_Eureka

    • 2 years ago
  3. Sir_Rico_of_Eureka Group Title
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    You just multiply both sides again to clear the fractions. this time however you have one more step. get x by itself

    • 2 years ago
  4. amorfide Group Title
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    on the question i wrote on the picture, i must multiply by (x-2)² not x-2 i want to know why?

    • 2 years ago
  5. amorfide Group Title
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    @cwrw238

    • 2 years ago
  6. Sir_Rico_of_Eureka Group Title
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    Who said you have to multiply (x-2)^2. Whats the original problem

    • 2 years ago
  7. amorfide Group Title
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    the book because apparently i will lose a solution if i dont

    • 2 years ago
  8. Sir_Rico_of_Eureka Group Title
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    Whats the original problem

    • 2 years ago
  9. amorfide Group Title
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    |dw:1350339676275:dw|

    • 2 years ago
  10. Sir_Rico_of_Eureka Group Title
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    And your book says the solution needs to multiply both sides by (x-2)^2? in the back of the book?

    • 2 years ago
  11. amorfide Group Title
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    yes because if i only multiply by x-2 i lose a solution i dont see how

    • 2 years ago
  12. amorfide Group Title
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    @TuringTest

    • 2 years ago
  13. Sir_Rico_of_Eureka Group Title
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    Hm, If you do the math the same answer comes out. Does it give you an answer to the problem

    • 2 years ago
  14. amorfide Group Title
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    |dw:1350340054622:dw|

    • 2 years ago
  15. TuringTest Group Title
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    you have to keep in mind that x-2 could be negative, so you need to consider cases

    • 2 years ago
  16. Sir_Rico_of_Eureka Group Title
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    I see now, if you look closely simply solving they way we did it at first is saying that x<3 and it can't be a negative number because a negative is never bigger than a positive. Therefore at some number x<3 stops being completly true. How i figured it out is plugged in 2 and saw that it equals 1/0 and you cant divide by zero. Therefore it must be a number between two and three. Thats the logical explanation. But I guess I don't know the specific rules taught in your book.

    • 2 years ago
  17. TuringTest Group Title
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    if x-2<0 the sign of the inequality would change upon multiplication, but (x-2)^2 can never be negative, so that we can use without changing the sign of the inequality

    • 2 years ago
  18. amorfide Group Title
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    thank you very much people!

    • 2 years ago
  19. TuringTest Group Title
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    welcome!

    • 2 years ago
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