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anonymous

  • one year ago

Hey! I really need help on this :( I will fan and medal. Using the equation below as a model, fill in numbers in the place of a and b to create a rational equation that has an extraneous solution. (x + a)/ax = b/x Part 1. Show all work to solve for x in the equation and check the solution. Part 2. Explain how to identify the extraneous solution and what it means.

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  1. anonymous
    • one year ago
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    @Vocaloid

  2. anonymous
    • one year ago
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    I don't necessarily want an answer I just want someone to work it out with me :)

  3. netlopes1
    • one year ago
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    \[\frac{x+a}{ax}=\frac{b}{x}\rightarrow Is \quad this?\]

  4. anonymous
    • one year ago
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    Um, I think you're asking me what values I plugged in for a and b. Here: a = 3 and b = 1.

  5. netlopes1
    • one year ago
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    First, remember, "x" must be different of zero, ok?

  6. anonymous
    • one year ago
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    Ok :) so that means if I work it out for x and it comes out to x = 0 then it is not a solution, correct?

  7. netlopes1
    • one year ago
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    If x=0, both denominators of fractions become null and the equation will be indefined, or better, there is no solution.

  8. netlopes1
    • one year ago
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    Item "I.", ok? \[\frac{x+a}{ax}=\frac{b}{x}\rightarrow \frac{x+a=ab}{ax} \rightarrow \]Do you understand so far?

  9. anonymous
    • one year ago
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    No, I'm sorry but I do not understand... Also, my question says that I need to come up with some values for a and b. So, I think I might need to put those values in before I start working it out :) my values are: a = 3 and b = 1

  10. netlopes1
    • one year ago
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    Why your values are a=3 and b=1? These values are in the question.... but if "yes", it's simpler... replace these values in the equation, directly... look

  11. netlopes1
    • one year ago
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    \[\frac{x+a}{ax}=\frac{b}{x}\rightarrow \frac{x+3}{3x}=\frac{1}{x} \rightarrow\]

  12. anonymous
    • one year ago
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    @itsdaniela @raiders88 @Twetapoo @bishop.courtney2014 @Hero @amistre64 @ganeshie8 @saifoo.khan @paki @wio @Preetha

  13. anonymous
    • one year ago
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    Ok :)

  14. netlopes1
    • one year ago
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    And solve the equation for x

  15. anonymous
    • one year ago
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    But I still do not know how to figure that out. I guess cross multiply?

  16. anonymous
    • one year ago
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    @IrishBoy123

  17. netlopes1
    • one year ago
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    Yes, look: \[\frac{x+3}{3x}=\frac{1}{x} \rightarrow x(x+3)=3x.1 \rightarrow ok?\]

  18. anonymous
    • one year ago
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    Ok, so I would come out with x^2 + 3x = 3x?

  19. anonymous
    • one year ago
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    But then it would be x^2 = 0... I don't think that is correct.

  20. netlopes1
    • one year ago
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    This is the cross multiply, ok? So, if x^2 +3x = 3x, or ??? Look

  21. anonymous
    • one year ago
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    I don't understand. I think you subtract the 3x from each side...

  22. anonymous
    • one year ago
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    Which would leave you with x^2 = 0 but then you can't take the square root of 0 so I'm not sure what to do next. But is that what extraneous means?

  23. netlopes1
    • one year ago
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    Remember, when I said that "x" must be different of zero? And now, your solução is exactly zero? This is an inconsistency solution, so, to a=3 and b=1, this equation has no solution.... ok.... but x^2 is a quadratic function, whose graph is a parabola, remember?

  24. anonymous
    • one year ago
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    Umm, I don't understand.

  25. anonymous
    • one year ago
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    You don't have to help me if you don't want to, I know that this is probably very frustrating. I'm sorry that I stink at math lol

  26. anonymous
    • one year ago
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    But thank you for your time :) I will medal :)

  27. anonymous
    • one year ago
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    Ok, thanks @netlopes1

  28. netlopes1
    • one year ago
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    Ok, but I will try again, ok?

  29. netlopes1
    • one year ago
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    It's no excuse, but my English is terrible (I'm Brazilian) and mathematics nomenclature is hard for me .... but I promise I will improve and I never give up ...

  30. anonymous
    • one year ago
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    It's fine :) I'm just not good at math. Yes! It is always good to never give up! Always try your very best :) and if you have done your best, that's all that matters. Thanks again! And I finally figured it out :)

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