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Math Help - Rational inequality?

  1. #1
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    Rational inequality?

    Can anyone help with this rational inequality
    7/(x+1)>7
    I think its zero but apparently the real answer is -1<x>0.
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  2. #2
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    Quote Originally Posted by homeylova223 View Post
    Can anyone help with this rational inequality 7/(x+1)>7
    You need to be sure that x+1>0~\&~x+1<1.
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  3. #3
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    Quote Originally Posted by homeylova223 View Post
    Can anyone help with this rational inequality
    7/(x+1)>7
    I think its zero but apparently the real answer is -1<x>0.
    \displaystyle\frac{7}{x+1}>7

    if (x+1)<1 and positive

    \Rightarrow\ -1<x<0
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  4. #4
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    The first thing we all want to do is multiply both sides of the inequality by (x+1), but we need to be careful because the inequality changes if we multiply by a negative value, so 2 cases:

    Case 1) x+1 < 0 or x < -1
    Then 7 < 7(x + 1) and 7 < 7x + 7 and we get x > 0
    so taking all the conditions, we have x < -1 and x > 0, but this is not possible, so case 1 gives no solutions

    Case 2) x + 1 > 0 or x > -1
    Then 7 > 7(x+1) and 7 > 7x + 7 and we get x < 0
    so taking all the conditions we have x > -1 and x < 0, this is a range and can be written as -1 < x < 0
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  5. #5
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    [soze=3]Hello, homeylova223![/size]

    \dfrac{7}{x+1}\:>\:7

    \text{I think it's zero} . WHAT is zero?

    We have: . \dfrac{7}{x+1} \:>\:7

    Divide by 7: . \dfrac{1}{x+1} \:> \:1

    Then we have: . \displaystyle \frac{1}{x+1} - 1 \:>\:0 \quad\Rightarrow\quad \frac{-x}{x+1} \:>\:0

    Multiply by -1: . \dfrac{x}{x+1} \:<\:0



    If a fraction is negative: the numerator and denominator have opposite signs.


    [1]\;\begin{array}{cccccccc} x \:>\:0 \\ x+1\:<\:0 & \Rightarrow & x \:<\:-1 \end{array}

    . . . This says: . x > 0 and x < -1 . . . impossible



    [2]\;\begin{array}{ccccccc}x \:<\:0 \\ x+1\:>\:0 & \Rightarrow & x \:>\:-1 \end{array}

    . . . This says: . x < 0 and x > -1

    This is possible: . -1 \:<\:x\:<\:0

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  6. #6
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    Hello, homeylova223!

    \dfrac{7}{x+1}\:>\:7

    \text{I think it's zero} . WHAT is zero?

    We have: . \dfrac{7}{x+1} \:>\:7

    Divide by 7: . \dfrac{1}{x+1} \:> \:1

    Then we have: . \displaystyle \frac{1}{x+1} - 1 \:>\:0 \quad\Rightarrow\quad \frac{-x}{x+1} \:>\:0

    Multiply by -1: . \dfrac{x}{x+1} \:<\:0



    If a fraction is negative: the numerator and denominator have opposite signs.


    [1]\;\begin{array}{cccccccc} x \:>\:0 \\ x+1\:<\:0 & \Rightarrow & x \:<\:-1 \end{array}

    . . . This says: . x > 0 and x < -1 . . . impossible



    [2]\;\begin{array}{ccccccc}x \:<\:0 \\ x+1\:>\:0 & \Rightarrow & x \:>\:-1 \end{array}

    . . . This says: . x < 0 and x > -1

    This is possible: . -1 \:<\:x\:<\:0

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