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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- Aug 23rd 2009, 10:48 AMjadi929solving 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) - Aug 23rd 2009, 11:04 AMe^(i*pi)
- Aug 23rd 2009, 11:28 AMSoroban
He4llo, jadi929!

Quote:

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

. . $\displaystyle \frac{1}{x+5} + \frac{1}{x+3} \;=\;\frac{3x + 11}{x^2 + 8x +15}$

had an equation with fractions??*ever*

If that's true, this is anintroduction of the subject.**awful**

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__