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deverhardt

  • 2 years ago

Solve the equation for x. -1+((2x)/(x+3))=((-2)/(x+2))

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  1. deverhardt
    • 2 years ago
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    Multiply the equation by the LCD.

  2. deverhardt
    • 2 years ago
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    Sorry I should have stated to solve the equation for x.

  3. estudier
    • 2 years ago
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    K, let's start again.....

  4. deverhardt
    • 2 years ago
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    okay

  5. estudier
    • 2 years ago
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    What is it u want to do?

  6. deverhardt
    • 2 years ago
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    |dw:1350674953315:dw|

  7. estudier
    • 2 years ago
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    Yes, so what is it u want to do?

  8. deverhardt
    • 2 years ago
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    Multiply (-1), (2x/x+3), and (-2x/x+2) by the LCD of (x+3) and (x+2).

  9. estudier
    • 2 years ago
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    OK, that is the long way around of solving for x. You will get to the same place at the end. I guess your teacher has told u to do it that way, right?

  10. deverhardt
    • 2 years ago
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    yes

  11. estudier
    • 2 years ago
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    LCD of (x+3) and (x+2) is just (x+3)(x+2) I will leave you to it, it is not too hard to do.

  12. deverhardt
    • 2 years ago
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    Okay. Thank you. I know the answer is (-1,0). I just wanted to make sure I was on the right path.

  13. estudier
    • 2 years ago
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    U should also note that x=-3 and x = -2 are not defined for this equation.

  14. deverhardt
    • 2 years ago
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    correct

  15. deverhardt
    • 2 years ago
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    Could you please explain the shorter way of doing this type of equation.

  16. estudier
    • 2 years ago
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    If you add the terms on the left side -> (x-3)/(x+3) = -2/(x+2) Cross multiply -> (x-3)(x+2) = -2(x+3) -> x^2 -x -6 = -2x - 6 -> x^2 + x = 0 -> x( x +1) = 0 -> x = 0 or -1

  17. deverhardt
    • 2 years ago
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    Wow that is much quicker! Thank you so much

  18. estudier
    • 2 years ago
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    yw:)

  19. estudier
    • 2 years ago
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    The other way reminds you that the original equation is not defined for x=-3 and x = -2 But with some practice you will not need reminding....

  20. deverhardt
    • 2 years ago
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    Could I use this shorter way for this equation as well? (see drawing)

  21. deverhardt
    • 2 years ago
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    |dw:1350677688912:dw|

  22. estudier
    • 2 years ago
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    Yes, u can.

  23. deverhardt
    • 2 years ago
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    Thanks. When I add the terms on the LHS I get (z+1) / (z-6). Is this correct?

  24. estudier
    • 2 years ago
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    No, (z+1) + 1*(z-6) / (z-6) = (2z-5) / (z-6)

  25. deverhardt
    • 2 years ago
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    Gotcha. Thanks

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