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.
 A: You could solve it this way:
The second fraction, $\frac {2}{2x+6}$ can be simplified into $\frac 1{x+3}$. Thus,
$$\frac 5{x+3}+\frac 2{2x+6}=4\implies\frac 6{x+3}=4\tag1$$
And cross multiplying, we get $4(x+3)=6\implies x=x=-\frac 32$

As to where you went wrong, nothing has gone wrong yet. Use the quadratic formula and see what you get!

 Hint: You will get two solutions. One of them doesn't work (i.e it's extraneous).

A: The quadratic equation you have obtained has two solutions: $-3$ and $-\frac{3}{2}$. The latter one is the correct answer. The former one is incorrect because it would lead to division by $0$ when we plug it into the original equation.
However, the problem need not be solved via quadratic equation. Observe that
$\frac{5}{x + 3} + \color{red}{\frac{2}{2x + 6}} = \frac{5}{x + 3} + \color{red}{\frac{1}{x + 3}} = \frac{6}{x + 3}$. Then the equation could be written as $4(x + 3) = 6$, so $x = \frac{6}{4} - 3 = -\frac{3}{2}$.
A: The solutions of $8x^2+36x+36=0$ are $-1.5$ and $-3$, but the latter is not a solution of the original equation. This new, false solution comes when you multiply the equation by $(x+3)(2x+6)$, which is not the lcm of the denominators.
Note that $2x+6=2(x+3)$.
An example to show what's happening:
Take the equation
$$3x+5=-1$$
The solution is $x=-2$. If you multiply both sides of any equation by a non zero number, you get another equation, whose solutions are the same. But if you multiply both sides by $0$ you get $0=0$.
Now, if you multiply both sides by, say, $x+4$, we don't know if $x+4$ is zero or not. So we obtain another equation with two solutions:
$$3x^2+17x+20=-x-4$$
$$3x^3+18x+24=0$$
$$x=\left\lbrace\begin{align}&-2\\&-4\end{align}\right.$$
One of the solutions is the original one. The other is the value for which the factor $x+4$, that we multiplied by, vanishes.
A: If you reduce your quadratic equation it will become $2x^2+9x+9=0$. Now apply quadratic formula to find the x.
$x=\frac{-9\pm\sqrt{9^2-4*2*9}}{4} $. Solve this one and find x. You will get two answers which are -3  and -1.5
Second one is correct but the first one is not because on the LHS of your question you have 5/x+3 as one of term. If you apply x=-3 here you will be dividing 5 by 0 which is not possible, same in the case of other term(2/2x+6) . I let you conclude now.
A: $\frac{5}{x+3}$+$\frac{2}{2(x+3)}$=4
$\frac{(5*2+2)}{2(x+3)}$=4
12=8(x+3)
x=$\frac{-12}{8}$
x=$\frac{-3}{2}$
A: 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.
A: We have $5\over{x+3}$ + $2\over{2x+6}$ = 4.
This can be rewritten as $10\over{2x+6}$ + $2\over{2x+6}$ = 4.
Multiplying by 2x + 6 on each side, we get: 10 + 2 = 4(2x + 6)
Dividing by 4 on each side: 3 = 2x + 6
$\Rightarrow$ -3 = 2x $\Rightarrow$ x = $-3\over{2}$
