Inequalities with absolute value question Solve: $ 5< \left\vert\dfrac{x+10}{x-10}\right\vert<6$
attempt at a solution: 
Dividing into two: 
$5<\left\vert \dfrac{x+10}{x-10}\right\vert  $     And  $\left\vert \dfrac{x+10}{x-10}\right\vert<6  $ 
For first we solve: 
 $ \dfrac{x+10}{x-10}<-5  $  or $ \dfrac{x+10}{x-10} >5  $  which yields $( -∞,15)$
For Second: 
$  -6<\dfrac{x+10}{x-10}  $  and $ \dfrac{x+10}{x-10} <6  $  which yields $(14, ∞)$
intersecting the two we get $(14,15)$
This solution was deemed wrong by the text book. 
Is there any other way of solving this? Is there a mistake in this method of solution? 
 A: Given that $10$ isn't even in the domain of the given function, your solution to the first inequality cannot be correct. Without actually seeing the details, though, it's hard to say what went wrong.
As an alternative, note that $\frac{x+10}{x-10}>0$ for $x>10$ and for $x<-10,$ and $\frac{x+10}{x-10}<0$ for $-10<x<10$. It's far simpler, then, to proceed casewise.
If $x<-10$, the following are equivalent: $$5<\left|\frac{x+10}{x-10}\right|<6\\5<\frac{x+10}{x-10}<6\\5(x-10)>x+10>6(x-10)\\5x-50>x+10>6x-60,$$ but $5x-50>x+10$ yields $x>15,$ so we can rule out all $x<-10$ as solutions to the inequality.
If $x>10,$ the following are equivalent: $$5<\left|\frac{x+10}{x-10}\right|<6\\5<\frac{x+10}{x-10}<6\\5(x-10)<x+10<6(x-10)\\5x-50<x+10<6x-60.$$ $5x-50<x+10$ yields $x<15,$ while $x+10<6x-60$ yields $14<x,$ so all $x$ with $14<x<15$ will be solutions to the inequality.
If $-10<x<10,$ the following are equivalent: $$5<\left|\frac{x+10}{x-10}\right|<6\\-5>\frac{x+10}{x-10}>-6\\-5(x-10)<x+10<-6(x-10)\\-5x+50<x+10<-6x+60.$$ $-5x+50<x+10$ yields $\frac{20}3<x,$ while $x+10<-6x+60$ yields $x<\frac{50}7,$ so all $x$ with $\frac{20}3<x<\frac{50}7$ will be solutions to the inequality.
Hence, the solution set is $$\left(\frac{20}3,\frac{50}7\right)\cup(14,15).$$
A: Distinguish three cases:
(i) $\quad x<-10$:
Here the given equation amounts to
$$5<{-x-10\over10-x}<6,$$
or $50-5x<-x-10<60-6x$. In particular $4x>60$, which is incompatible with $x<-10$. So there are no solutions $x<-10$. Note that $x=-10$ isn't a solution either.
(ii) $\quad-10<x<10$:
Here the given equation amounts to
$$5<{x+10\over10-x}<6,$$
or $50-5x<x+10<60-6x$, which is equivalent with ${20\over3}<x<{50\over7}$. Note that these $x$ indeed lie in the interval $\ ]{-10},10[\ $. For $x=10$ the given condition makes no sense.
(iii) $\quad x>10$:
Here the given equation amounts to
$$5<{x+10\over x-10}<6,$$
or $5x-50<x+10<6x-60$, which is equivalent with $14<x<15$. These $x$ are indeed $>10$.
It follows that within ${\mathbb R}$ the solution set of the given equation is
$$\bigl]{20\over3},{50\over7}\bigr[\ \ \cup \ \ ]14,15[\ \ ,$$
but the equation is meaningless at $x=10$.
