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    Ive been given the inequality

    |x-1| > 1/2
    x+1

    find all x ∈ R that satisfy the inequality.

    to get rid of the modulus sign, i seperated the inequality into two, to give

    (x-1) / x+1 > 1/2 and -(x-1) / x+1 > 1/2

    i then rearranged so evything was on the left side and then made the denominator of the -1/2 and (x-1) / x+1 same. i then added/substracted and then rearranged to get x. i did this for the other inequality, combined the two answers and i got the answer:

    -1 < x < 1/3

    i checked the possible x values and they are correct, but i dont no if the answer as a whole is correct. some1 help me please.
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    You've got the answer for the second case only, when l x-1 l = 1 - x.
    What about when l x-1 l = x-1 ?
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    no its for both inequalities... -1 > x > 1/3 is when ive combined the two seperate answers.
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    Then no, it's not.
    x>3 is also a solution.
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    ive tried it a different way...

    i took the denominator x+1 to the other side, and brought the denominator 2 to the left side. ive then just squared both sides, rearranged and factorised to give me the two results..

    (3x-1)(x-3) > 0 so,

    x > 3 and x > 1/3... thanks you were right about the x>3.

    ive got the results, 0 < x < 1/3 and x > 3

    the possible x values satisfy the inequality. is this correct the new method of doin it?
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    It's not actually, you can't just take the denominator to the other side without knowing whether it's positive or negative.
    The method is exactly what you did in your OP, you've must have got your calculations wrong somehow. That's all.
 
 
 
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