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viniterranova

  • 4 years ago

(x +1) / (2-x) < 4

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  1. Dockworker
    • 4 years ago
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    first find when (x+1)/(2-x)=4 or is undefined

  2. Dockworker
    • 4 years ago
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    you will need these values to solve the inequality

  3. viniterranova
    • 4 years ago
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  4. viniterranova
    • 4 years ago
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    May this help you.

  5. Dockworker
    • 4 years ago
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    yes, i know that is the question, but it is important that you solve the equation (x+1)/(2-x)=4 before you solve (x+1)/(2-x)<4

  6. viniterranova
    • 4 years ago
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    Sorry....but may u do it?

  7. Dockworker
    • 4 years ago
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    (x+1)/(2-x)=4 x+1=4(2-x) x+1=8-4x 5x=7 x=7/5 we also must know when (x+1)/(2-x)=4 is undefined it is undefined at x=2, because the denominator 2-x=0

  8. Dockworker
    • 4 years ago
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    now we can split the number line up into 3 intervals using these 2 numbers

  9. Dockworker
    • 4 years ago
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    first we will simplify the rational inequality

  10. Dockworker
    • 4 years ago
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    well i guess that's not necessary in this case, but usually you want to get 0 on one side of the inequality

  11. Dockworker
    • 4 years ago
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    we have the values we will use to form our intervals, x=2, x=7/5 this splits the domain of the function into the following intervals: (-infinity, 7/5), (7/5, 2), (2, infinity)

  12. Dockworker
    • 4 years ago
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    we will pick a value at random from these intervals and see if it is less than 4. if it is, the entire interval is part of the solution

  13. Dockworker
    • 4 years ago
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    from the first interval we can use x=0 (0+1)/(2-0)<4? 1/2<4? TRUE x<7/5 is part of the solution

  14. Dockworker
    • 4 years ago
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    next we will pick the value 8/5 from the 2nd interval (8/5+1)/(2-8/5)<4? (13/5)/(2/5)<4? 13/2<4? FALSE, this is not part of solution

  15. Dockworker
    • 4 years ago
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    from 3rd interval we can use x=3 (3+1)/(2-3)<4? 4/-1<4? -4<4? TRUE x>2 is part of solution so entire solution is x<7/5 or x>2 now you see why we had to solve the related equation

  16. viniterranova
    • 4 years ago
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    Thanks for the answer

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