1. ## 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)

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

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

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

this means the denominators cancel

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

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

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

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

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

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

But when we check our answer, we get:

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

Therefore, the equation has no solution.