<s>[INST] 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? 
 [/INST] 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 $$\left(\frac{20}3,\frac{50}7\right)\cup(14,15).$$
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