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    \frac{x+k}{x+4k}>\frac{k}{x}

    Now if i multiply both sides by \frac{1}{k} and then subtract \frac{1}{x} from both sides , I get :
    \frac{x(x+k)-k(x+4k)}{kx(x+4k)}>0


    But if I directly subtract \frac{k}{x} from both sides, I end up with :
    \frac{x(x+k)-k(x+4k)}{x(x+4k)}>0

    Shouldn't i get the same fraction from both methods?
    Where am I going wrong?
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    No, you shouldn't get the same fraction. It's an inequality, not an equation/identity.
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    Woops.
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    (Original post by princejan7)
    Shouldn't i get the same fraction from both methods?
    Where am I going wrong?
    You're going wrong in assuming you should have the same result by both methods.

    Lets take a more obvious example.

    4 > 2

    Divide by 2,

    2 > 1

    and subtract 1 to get:

    1 > 0

    Or the other way, from

    4 > 2

    subtract 2 to get

    2 > 0.

    Both are correct, as in your example.

    If you want to get to your first result, from the second, you need to divide by 2 here, or by k in your example.

    Edit: I am assuming k > 0, otherwise the first method would be incorrect.
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    Anyway, just to make clear, your first attempt is not correct because the value of k is indetermined. Second attempt is better.
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    (Original post by Dekota-XS)
    Anyway, just to make clear, your first attempt is not correct because the value of k is indetermined. Second attempt is better.
    what if it was given that k >0?
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    (Original post by princejan7)
    what if it was given that k >0?
    If k > 0 the first attempt is fine, AND does lead to different results; see my previous post.
 
 
 
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