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Math Help - simplify expressions

  1. #1
    Newbie sabyasachi's Avatar
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    simplify expressions

    WOULD SOMEONE HELP ME WITH THESE SUMS ? HERE ARE THEY :
    1. SOLVE FOR x :

    1/(X) - 1/(X+6 )= 1/(A) - 1/(A+B)

    2. SIMPLIFY:

    A/(A-X) + B/(B-X) + C/( C-X)
    _____________________________
    3/(X) - 1/(X-A) - 1/(X-B) - 1/(X-C)

    3. SIMPLIFY :

    A2/(X-A) + B2/(X-B) + C2/(X-C) + A + B + C
    ___________________________________________
    A/(X-A) + B/(X-B) + C/(X-C)

    YOUR HELP WILL BE APPRECIATED.
    Last edited by sabyasachi; May 24th 2008 at 10:49 PM.
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  2. #2
    A riddle wrapped in an enigma
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    Could you use (parentheses) to clarify what the denominators are?
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by sabyasachi View Post
    WOULD SOMEONE HELP ME WITH THESE SUMS ? HERE ARE THEY :
    1. SOLVE FOR x :

    1/(X) - 1/(X+6 )= 1/(A) - 1/(A+B)
    \frac 1X-\frac{1}{X+6}=\frac 1A-\frac{1}{A+B}

    \frac 1X-\frac{1}{X+6}=\frac{X+6}{X(X+6)}-\frac{X}{X(X+6)}=\boxed{\frac{6}{X(X+6)}}
    And :
    \frac 1A-\frac{1}{A+B}=\frac{A+B}{A(A+B)}-\frac{A}{A(A+B)}=\boxed{\frac{B}{A(A+B)}}

    The equation is now :

    \frac{6}{X(X+6)}=\frac{B}{A(A+B)}

    Taking the inverse :

    \frac 16 \cdot X(X+6)=\frac 1B \cdot A(A+B)

    X(X+6)=\frac 6B \cdot A(A+B)

    Can you continue ?
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  4. #4
    Newbie sabyasachi's Avatar
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    please!

    please can you help me further with that sum?
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  5. #5
    Moo
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    Quote Originally Posted by sabyasachi View Post
    please can you help me further with that sum?
    Develop and solve... But I don't know your level and whether you know about discriminants or not ~
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  6. #6
    Newbie sabyasachi's Avatar
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    got it

    well i m sorry.i just can't sit idle without solving those sums. moo did try to help me but he/she left a certain portion of a sum to be solved by me.but i was still confused a it.
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  7. #7
    Newbie sabyasachi's Avatar
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    Well I Don't

    Well! I Do Not Know About Discriminants Yet. Please Solve The Sum Step-by-step And In Simple Language/method. Much Obliged.
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  8. #8
    Moo
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    Actually, without any further information about A and B, the equation can have no solution... So please post everything you have ~


    There is a direct solution : X=A and B=6.
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  9. #9
    Newbie sabyasachi's Avatar
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    I Know

    Well I Do Know A Bit About Solving Linear, Quadratic Equations Etc. Now I Realize That I Do Know About Discriminants, But I Did Not Know The Exact Term Prior To This.
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  10. #10
    Moo
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    Ok.

    Then, we have :

    Quote Originally Posted by Moo View Post

    X(X+6)=\frac 6B \cdot A(A+B)
    X^2+6X=\underbrace{\frac 6B \cdot A(A+B)}_{M}

    X^2+6X-M=0

    \Delta=36+4M

    --> \sqrt{\Delta}=\sqrt{36+4M}=\sqrt{4(9+M)}=2\sqrt{9+  M}

    X=\frac{-6 \pm 2\sqrt{9+M}}{2}=-3 \pm \sqrt{9+M}

    And 9+M has to be \ge 0, that is to say M \ge -9 \Longleftrightarrow \frac 6B \cdot A(A+B) \ge -9 \Longleftrightarrow 2A(A+B) \ge -3B



    And I don't think we can go further.
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  11. #11
    Newbie sabyasachi's Avatar
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    Please

    This Problem Is Meant For Students Of Class 8 ( In India). Please Simplify Your Method. I Could Understand Till The Inversion,but Faltered While Solving The Remaining Portion.
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  12. #12
    Moo
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    Quote Originally Posted by sabyasachi View Post
    This Problem Is Meant For Students Of Class 8 ( In India). Please Simplify Your Method. I Could Understand Till The Inversion,but Faltered While Solving The Remaining Portion.
    \Delta is the discriminant..

    When you have an equation : ax^2+bx+c=0, \Delta=b^2-4ac

    Thus the solutions are x=\frac{-b \pm \sqrt{\Delta}}{2a}

    If you have never dealt with discriminants, you can't solve generally this equation.



    A particular solution to this is : X=A and B=6, because the two sides of the equation are similar.
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  13. #13
    Newbie sabyasachi's Avatar
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    At Last

    Thanks A Lot. If We Can Arrive At The Solution According To Mathematical Methods, Please Solve It.
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  14. #14
    Moo
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    Quote Originally Posted by sabyasachi View Post
    Thanks A Lot. If We Can Arrive At The Solution According To Mathematical Methods, Please Solve It.
    This is a mathematical method...

    There are so many unknown things that we can't provide a decent solution apart from the one I wrote just above..
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  15. #15
    Newbie sabyasachi's Avatar
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    Something More

    Thank You For Helping Me.i Troubled You The Whole Day.but Now Can You Help Me With The Other Sums (2 Of Them)? I Know You Are Getting Irritated.
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