Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11,  x>3,$ which doesn't make sense.
 A: You can't multiply $(x+4) $ both sides as you don't know it is positive or negative.
Correct way: $\dfrac{7}{x+4}>-1$
$ \dfrac{7}{x+4}+1>0 \implies \dfrac{x+11}{x+4}>0$
So

*

*either $x+11$ and $x+4$ both are $>$ than $0$ for which we get $x>-4$


*or both less than $0$ for which we get $x<-11$
A: $$\frac{7}{x+4}>-1 \implies \frac{x+11}{x+4} >0$$
This means
(1): $(x+11)>0 \& x+4>0 \implies x>-4 \& x >-11 \implies x>-4$
(2): $(x+11) <0 \& x+4 <0 \implies x<-11 \& x <-4 \implies x<-11$
So the total answer is union of the two: $x\in (-\infty, -11) \cup (-4, \infty)$ or
$x<-11$ or $x >-4$.
A: Edited for clarity.

Expanding on the comment of sonicsid, when you are faced with the necessity of clearing the denominator of $(x + 4)$, since $x + 4 = 0$ is disallowed, you must break the analysis into two cases:
Case 1: $(x + 4) > 0.$ 
$-1 < \frac{7}{x+4} < 1.$ 
$ (-1) (x+4) < 7 < (x+4).$ 
$(x + 4) > -7$ and $(x + 4) > 7.$
This means that all three of the following constraints must be satisfied for Case 1:

*

*$x > -4.$

*$x > -11.$

*$x > 3.$
Therefore, the Case 1 solutions resolve to 
$x > 3.$

Case 2: $(x + 4) < 0.$ 
$-1 < \frac{7}{x+4} < 1.$ 
$(-1) (x + 4) > 7$ and $7 > (x + 4).$
This means that all three of the following constraints must be satisfied for Case 2:

*

*$x < -4.$

*$x < -11.$

*$x < 3.$
Therefore, the Case 2 solutions resolve to 
$x < -11.$

Putting Case 1 and Case 2 together:
Either $x > 3$ or $x < -11.$
A: I also used to find it difficult so what I did was take $y=x+4$ then the equation transforms to $7/|y|<1$. Now break it into cases.
Case:$1$(y<0)
then $7/-y<1 \implies 7<-y \implies -7>y \implies y \in (-\infty,-7)$
Case:$1$(y>0)
then $7/y<1 \implies 7<y \implies 7<y \implies y \in (7,\infty)$
Now put $y=x+4$ to get the result
A: Alternative
Make sure you only multiply with positive number because otherwise the inequality sign will reverse
$$
\begin{align}
\left|\frac{7}{x+4}\right|\cdot\left|\frac{7}{x+4}\right|&<1\cdot\left|\frac{7}{x+4}\right|\\
\\
&<1\cdot 1\\
\\
\\
7^{2}&<(x+4)^{2}
\end{align}
$$
From here we get the solutions:
$$
x<-11\text{ or }x>3
$$
Notice that I only multiply with $\left|\frac{7}{x+4}\right|$ and $(x+4)^{2}$ which I know are positive numbers.
A: If $k>0$, then $a<b\iff ak<bk$. However, if $k<0$, then $a<b\iff ak\color{red}{>}bk$.
So when multiplying an inequality by a constant, we should be certain whether that constant is positive or negative. In the case of $-1<\frac{7}{x+4}$, multiplying by $x+4$ is not a good idea because we don't know if $x+4<0$ or $x+4>0$. A better idea is to multiply by $(x+4)^2$, which is positive.* Then,
\begin{align}
-1<\frac{7}{x+4}&\iff-(x+4)^2<7(x+4) \\[5pt]
&\iff -x^2-8x-16<7x+28 \\[5pt]
&\iff 0<x^2+15x+44 \\[5pt]
&\iff0<(x+4)(x+11) \\[5pt]
&\iff x<-11\text{ or }x>-4 \, .
\end{align}

*Technically, $(x+4)^2$ could also be zero, if $x=-4$. However, this is a non-issue, as if $x=-4$, then $7/(x+4)$ is undefined, so it is not a solution to $-1<7/(x+4)$; and $x=-4$ is not a solution to $-(x+4)^2<7(x+4)$ either.
A: As noticed you went wrong multiplying for a quantity which can also be negative.
Using that for $A,B>0$ then
$$|A|<B \iff \left|\frac1{A}\right|>\frac1B $$
a very effective way to proceed in this case is the following
$$\left|\frac{7}{x+4}\right|<1 \iff \left|\frac{x+4}{7}\right|>1 \iff |x+4|>7$$
$$\iff x+4<-7 \quad \lor \quad x+4 >7 $$
$$\iff x<-11 \quad \lor \quad x>3 $$
We can also proceed by your way but we need to consider two cases, that is

*

*For $x+4>0 \iff x>-4$
$$\left|\frac{7}{x+4}\right|<1\iff \frac{7}{x+4}<1\iff x+4>7 \iff x>3$$

*

*For $x+4<0 \iff x<-4$
$$\left|\frac{7}{x+4}\right|<1\iff \frac{7}{-(x+4)}<1\iff -(x+4)>7\iff x+4<-7 \iff x<-11$$
A: Just to add to the noise, this is how I would personally attack this problem.  This is a matter of personal style and may not work for you.
$|\frac 7{x+4}| < 1$.
Either $\frac 7{x+4}$ is non-negative (positive or zero) or it is negative.
It is positive if $x+4$ is positive.  It is negative if $x+4$ is negative.  (And it is never $0$ as $\frac ab \ne 0; a=0$ and we can't have $x+4 = 0$ as we can't have $0$ in a denominator).
Case 1:
If $\frac 7{x+4}$ is positive then $x+4$ is positive and then 1) $x > -4$ and 2) $|\frac 7{x+4}| =\frac 7{x+4}$ and $x + 4 > 0$.
So.... $\frac 7{x+4} > 1\implies 7 < x+4 \implies 3<x $ and combining that with $x > -4$ we get.... $x > 3$.
Case 2:
If $\frac 7{x+4}$ is negative then $x+4$ is negative and 1) $x < -4$ and 2) $|\frac 7{x+4}| =\frac 7{-(x+4)}$ and $-(x + 4) > 0$.
So.... $\frac 7{-(x+4)} < 1\implies 7 < -(x+4) = -x -4\implies x< -11$ and combining that with $x< -4$ we get.... $x<-11$.
....
So Case 1: $x > 3$ and Case 2: $x < -11$.
So $x < -11$ or $x > 3$ and the solution set is:
$(-\infty, -11) \cup (3, \infty) =$
$\{$ everything that isn't between $-11$ and $3\}=$
$\mathbb R \setminus [-11,3]$.
