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    h(x) is defined by

    h(x) = (2x +7)/(5x-2)

    show that hh(x) = x

    I've put (2x +7)/(5x-2) into wherever I see x, but I've no idea how to simplify it because it looks super complex
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    (Original post by ihatePE)
    h(x) is defined by

    h(x) = (2x +7)/(5x-2)

    show that hh(x) = x

    I've put (2x +7)/(5x-2) into wherever I see x, but I've no idea how to simplify it because it looks super complex
    are you allowed to say let x = 1 then plug 1 in (which gives you 3). then plug 3 back in and you get 1
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    Try rewriting ur numerator and denominator as one fraction for each, with the numerator as (5x-2) for both.

    It should cancel nicely from there.

    (Original post by ihatePE)
    h(x) is defined by

    h(x) = (2x +7)/(5x-2)

    show that hh(x) = x

    I've put (2x +7)/(5x-2) into wherever I see x, but I've no idea how to simplify it because it looks super complex
    If u still dont know how to proceed let me know.👊🏾
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    (Original post by Shaanv)
    Try rewriting ur numerator and denominator as one fraction for each, with the numerator as (5x-2) for both.

    It should cancel nicely from there.



    If u still dont know how to proceed let me know.👊🏾
    when separating I got (2x)/(5x-2). and 7/(5x-2)
    so for (2x)/(5x-2)
    I got 4x+14 when simplified (?) I think I done this wrong as well

    for 7/(5x-2)
    I got (35x-14)/(10x+35) which is where I'm stuck again because I can't see any common factors to simplify this fraction
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    Its really hard for me to explain with words, and im no good at latex so i hope this helps.

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    (Original post by ihatePE)
    when separating I got (2x)/(5x-2). and 7/(5x-2)
    so for (2x)/(5x-2)
    I got 4x+14 when simplified (?) I think I done this wrong as well

    for 7/(5x-2)
    I got (35x-14)/(10x+35) which is where I'm stuck again because I can't see any common factors to simplify this fraction
    Sorry forgot to quote u. ^^^^
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    (Original post by Shaanv)
    Its really hard for me to explain with words, and im no good at latex so i hope this helps.

    sorry I can't see the pic
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    top of the fraction: 2(2x+7)/(5x-2) + 7 which simplifies to (4x+14)/(5x-2) + 7 change 7 into 7(5x-2)/(5x-2)
    so you get (4x+14 +35x-14)/5x-2 = 39x/(5x-2)

    bottom of fraction:
    5(2x+7)/(5x-2) -2 which simplifies to (10x+35)/5x-2) -2 change -2 into -2(5x-2)/(5x-2)
    so you get (10x+35-10x+4)/(5x-2) = 39/(5x-2)

    then cancel the 5x-2 because its on the top and bottom so you get 39x/39 which = x
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    Name:  2099A696-6C30-4F2E-A6B9-F34DF8C52EF5.jpg.jpeg
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    (Original post by ihatePE)
    sorry I can't see the pic
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    There’s an easier way. Consider what hh(x) means. It means that if you start with x, run it through function h, then put the result of that through h again you get back to x. That can only happen if h is the same as [inverse h]. So all you need to do is show that [inverse h](x) = h(x) and you’re done.
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    (Original post by Cor395)
    are you allowed to say let x = 1 then plug 1 in (which gives you 3). then plug 3 back in and you get 1
    that is a useful check, but it is not a proof. you need to show that it is true for all values of x in the domain.
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    I mean it's pretty easy to multiply the numerators in your head to get

     \displaystyle \frac{ \frac{4x + 14}{5x - 2} + 7}{\frac{10x + 35}{5x - 2} - 2} . After that you just multiply numerator and denominator by 5x - 2.
 
 
 
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