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Thread: can you help me with this?

  1. #1
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    can you help me with this?

    am I in the right track??

    $\displaystyle 9\ge\frac{8-x}{4}>0$

    $\displaystyle 9-2\ge\frac{-x}{4}>-2$

    $\displaystyle 7(4)\ge-x>2(4)$

    $\displaystyle 28\ge-x>8$

    i am confused when changing -x to x

    $\displaystyle 28\le x<-8$

    is this correct??

    another example
    -x>-8
    x<-8

    thanks I'm confused
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  2. #2
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    $\displaystyle
    0 < \frac{8-x}{4} \leq 9
    $

    multiply every term by 4 ...

    $\displaystyle
    0 < 8-x \leq 36
    $

    subtract 8 ...

    $\displaystyle
    -8 < -x \leq 28
    $

    one inequality at a time ...

    $\displaystyle -8 < -x$ ... $\displaystyle 8 > x$ ... $\displaystyle x < 8$

    $\displaystyle -x \leq 28$ ... $\displaystyle x \geq -28$

    final solution ...

    $\displaystyle
    -28 \leq x < 8
    $
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  3. #3
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    Quote Originally Posted by skeeter View Post
    $\displaystyle
    0 < \frac{8-x}{4} \leq 9
    $

    multiply every term by 4 ...

    $\displaystyle
    0 < 8-x \leq 36
    $

    subtract 8 ...

    $\displaystyle
    -8 < -x \leq 28
    $

    one inequality at a time ...

    $\displaystyle -8 < -x$ ... $\displaystyle 8 > x$ ... $\displaystyle x < 8$

    $\displaystyle -x \leq 28$ ... $\displaystyle x \geq -28$

    final solution ...

    $\displaystyle
    -28 \leq x < 8
    $

    so this means i will change the signs of both terms?/
    like this -x>8 then x<-8
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  4. #4
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    Quote Originally Posted by princess_21 View Post
    so this means i will change the signs of both terms?/
    like this -x>8 then x<-8
    yes ... what you have done is multiply both sides of the inequality by -1 ... you also must change the direction of the sign anytime you multiply or divide by a negative value.

    for example ...

    5 > 4

    multiply both sides by -1 ...

    -5 < -4
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