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  1. #1
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    solving equations

    i'm supposed to solve the following equation for x, but dont really know where to start

    (1 / (x+5) ) + ( 1 / (x+3) ) = (3x + 11) / (x^2 + 8x +15)
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  2. #2
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    Quote Originally Posted by jadi929 View Post
    i'm supposed to solve the following equation for x, but dont really know where to start

    (1 / (x+5) ) + ( 1 / (x+3) ) = (3x + 11) / (x^2 + 8x +15)
    LHS

    $\displaystyle \frac{1}{x+5} + \frac{1}{x+3} = \frac{(x+3) + (x+5)}{(x+3)(x+5)}$

    on the rhs $\displaystyle (x+3)(x+5) = x^2+8x+15$

    this means the denominators cancel
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  3. #3
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    He4llo, jadi929!

    i'm supposed to solve the following equation for $\displaystyle x$, but dont really know where to start

    . . $\displaystyle \frac{1}{x+5} + \frac{1}{x+3} \;=\;\frac{3x + 11}{x^2 + 8x +15}$
    You've never ever had an equation with fractions??
    If that's true, this is an awful introduction of the subject.

    We have: .$\displaystyle \frac{1}{x+5} + \frac{1}{x+3} \;=\;\frac{3x+11}{(x+5)(x+3)} $


    Multiply through by the LCD, $\displaystyle (x+5)(x+3)$

    . . $\displaystyle {\color{red}\rlap{//////}}(x+5)(x+3)\cdot\frac{1}{{\color{red}\rlap{/////}}x+5} \;+\; (x+5){\color{red}\rlap{//////}}(x+3)\cdot\frac{1}{{\color{red}\rlap{/////}}x+3} $ . $\displaystyle =\;\;{\color{red}\rlap{//////}}(x+5) {\color{red}\rlap{//////}}(x+3)\cdot\frac{3x+11}{{\color{red}\rlap{//////}}(x+5) {\color{red}\rlap{//////}}(x+3)} $


    We have: .$\displaystyle (x + 3) + (x+5) \;=\;3x + 11 \quad\Rightarrow\quad 2x + 8 \;=\;3x+11 \quad\Rightarrow\quad\boxed{ x \;=\;-3}$


    But when we check our answer, we get:

    . . $\displaystyle \frac{1}{\text{-}3+5} + \frac{1}{\text{-}3+3} \;=\;\frac{3(\text{-}3) + 11}{(\text{-}3)^2 + 8(\text{-}3) + 15} \quad\Rightarrow\quad \frac{1}{2} + \frac{1}{0} \;=\;\frac{2}{0} $ .??


    Therefore, the equation has no solution.

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