<s>[INST] Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
 [/INST] Three answers:
i) $\frac n0$ is undefined.  If you are ever given an expression $\frac 5{x+3}$ you know from the very beginning that it is impossible for these to be $\frac 50$ so it is impossible for $x +3 =0$.  $x +3 \ne 0$ and $x \ne -3$.
So whatever you do, you know $x = -3$ will not be an acceptable solution.
So to get rid of the fractions you went from 
$A: \frac 5{x+3} + \frac 2{2x +6} = 4$ to
$B:  5(2x+6) + 2(x+3) = 4(x+6)(x+3)$
These two equations are not the same.  If $A$ is true then $B$ is true and you can get $B$ from $A$ but if $B$ is true you don't know that $A$ is true. All the solutions to $A$ are solutions to $B$, but $B$ has (maybe) more solutions than $A$ does. In $A$ it is impossible for $x + 3 = 0$ and for $2x + 6 = 0$.  But in $B$ it is not.  
This is could an "extraneous solution".  You did this the correct way to solve but you need to note that when you multiply by an "unknown" to get rid of a denominator, you must check whether the "unknown" adds some "extraneous" solutions.
So you solve $B$ and get $x = -1.5$ or $x =-3$.  But you already know that $x=-3$ is not an acceptable answer.  It is an acceptable answer to $B$ but not to $A$.  So the answer in $x = -1.5$.
ii) When you have $\frac ab + \frac cd = e$ so $ad +cb = ebd$ notice you only need to multiply by the greatest common multiple of $b$ and $d$.  You don't have to multiply by all of $b$ and $b$.
To get rid of the denominator in 
$\frac 5{x+3} + \frac 2{2x+6} = 4$ 
we don't have to multiply by $(x+3)(2x+6)$.  $2x + 6 = 2(x+3)$ so we only have to multiply by $2x+6$.
$\frac 5{x+3} + \frac 2{2x+6} = 4$
$5*2 + 2 = 4(2x + 3)$.
Indeed we could have only multiplied by $x+3$ to get $5 + \frac 22 = 4(x+3)$.
iii) You could have reduced $\frac 2{2x+6} = \frac 1{x+3}$ from the start.
$\frac 5{x+3} + \frac 2{2x + 6} = 4$
$\frac 5{x+3} + \frac 1{x+3} = 4$
$\frac 6{x+3} =4 $
$6 = 4(x+3)$.
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Note ii) and iii) show that this is not actually a quadratic problem.
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