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Math Help - Two difficult solve for x qns.

  1. #1
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    Two difficult solve for x qns.

    Working on this set of questions, and Ive come to two difficult qns, which Im not sure which route I should take at tackling them, its come to a point where Ive written and erased it so many times.. my page has ripped *pressure*
    Any help would be greatly appreciated.
    Solve for x:
    1.
    x/m + n = x/n + m

    2.
    (1/ x + a) + (1 / x + 2a) = 2 /x + 3a

    Thank you.
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  2. #2
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    if i understand the questions correctly:

    \frac{x}{m}+n = \frac{x}{n}+m

    x+nm = x+nm

    0 = 0

    which means this equation is true for every x...

    and the second one is similar:

    (\frac{1}{x}+a)+(\frac{1}{x}+2a) = \frac{2}{x}+3a

    \frac{2}{x}+3a = \frac{2}{x}+3a

    which also means this equation is true for every x, except x=0,
    when \frac{1}{x} is not defined.
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  3. #3
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    How did you simplify the first one, vonflex1?

    I got

    \frac{x}{m}+n=\frac{x}{n}+m

    \frac{x}{m}-\frac{x}{n}=m-n

    \frac{xn-xm}{mn}=m-n

    \frac{x(n-m)}{mn}=m-n

    x=\frac{mn(m-n)}{n-m}

    but I might just be confusing myself somewhere
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  4. #4
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by dkaksl View Post
    How did you simplify the first one, vonflex1?

    I got

    \frac{x}{m}+n=\frac{x}{n}+m

    \frac{x}{m}-\frac{x}{n}=m-n

    \frac{xn-xm}{mn}=m-n

    \frac{x(n-m)}{mn}=m-n

    x=\frac{mn(m-n)}{n-m}

    but I might just be confusing myself somewhere
    You're correct, but you can simplify it further:

    x=\frac{mn(m-n)}{n-m}

    x=\frac{mn(m-n)}{-(m-n)}

    x=-mn
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  5. #5
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    sorry about the first one... my bad.
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  6. #6
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    any ideas about the second question?
    cos the answer for the first one is right thank you
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  7. #7
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    did i write that second question right in latex?

    is this the exact question or is there a

    x+2a, x+a, x+3a in the denominator instead of just the x?
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  8. #8
    Senior Member nikhil's Avatar
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    Lightbulb

    I hope your question is
    [1/(x+a)]+[1/(x+2a)]=2/(x+3a)
    this is equal to
    (x+a+x+2a)/[(x+a)(x+2a)]= 2/(x+3a)
    =(2x+3a)/[(x+a)(x+2a)]= 2/(x+3a)
    now cross multiply and we get
    2x^2+9ax+9a^2=2x^2+6ax+4a^2
    or
    3ax=-5a^2
    finally
    x=-5a/3

    2) or on the second thought if your question is
    [(1/x)+a]+[(1/x)+2a]=[(2/x)+3a]
    then it is an identity. x can have any value except 0
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