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Thread: basic algebra

  1. #1
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    basic algebra

    x-p/x+p = p+q/p-q

    sorry about the...inaccuracy.please solve this I tried this like 10 times already and i always get 2(xq+p2) or something and I know its wrong. please help me
    thnx

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  2. #2
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    Re: basic algebra

    Quote Originally Posted by NormalKid View Post
    [SIZE=6]x-p/x+p = p+q/p-q

    First of all, please use grouping symbols.

    Is the problem $\displaystyle \frac{x-p}{x+p}=\frac{p+q}{p-q}~?$

    The way you have posted it reads:
    $\displaystyle x-\frac{x}{p}+p=p+\frac{q}{p}-q$.

    Which is it?
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  3. #3
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    Re: basic algebra

    the first one
    how do u do thaat?
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  4. #4
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    Re: basic algebra

    Hello, NormalKid!

    What is the question?
    I'll take a guess . . .


    $\displaystyle \text{Solve for }x\!:\;\frac{x-p}{x+p} \:=\:\frac{p+q}{p-q}$

    Cross-multiply: .$\displaystyle (x-p)(p-q) \:=\:(x+p)(p+q)$

    .n. . . . . . . . $\displaystyle {\color{red}\rlap{//}}px - qx - p^2 + {\color{green}\rlap{//}}pq \:=\:{\color{red}\rlap{//}}px + qx + p^2 + {\color{green}\rlap{//}}pq$

    n . . . . . . . . . . . . . . . . . . $\displaystyle 2qx \:=\:-2p^2$

    . . . . . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:-\frac{p^2}{q}$
    Thanks from NormalKid
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  5. #5
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    Re: basic algebra

    Quote Originally Posted by Soroban View Post
    Hello, NormalKid!

    What is the question?
    I'll take a guess . . .



    Cross-multiply: .$\displaystyle (x-p)(p-q) \:=\x+p)(p+q)$

    .n. . . . . . . . $\displaystyle {\color{red}\rlap{//}}px - qx - p^2 + {\color{green}\rlap{//}}pq \:=\:{\color{red}\rlap{//}}px + qx + p^2 + {\color{green}\rlap{//}}pq$

    n . . . . . . . . . . . . . . . . . . $\displaystyle 2qx \:=\:-2p^2$

    . . . . . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:-\frac{p^2}{q}$
    good guessing Soroban sir.
    so my guess was correct !
    THANK YOU EVERYBODY. SP.THANKS TO YOU SIR.
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