Here's the question you clicked on:
(a) Explain your strategy in solving inequalities of the form \( \frac{f(x)}{g(x)} < c. \) (b) Solve \(\left| \frac{x-1}{x+1} \right| <1\).
Your answer is not making any sense.
Because whatever is in absolute value can be positive and negative (you wouldn't know), you have to think about it in both cases. For example, |x| = 4 can mean x=4 or x=-4 because the abs. value turns the number positive. so ((x-1)/(x+1)) < 1 and ((x-1)/(x+1)) > -1 -1 < ((x-1)/(x+1)) < 1
\[-1<\frac{ x-1}{ x+1 } <1\]
@ghazi still don't make sense. What is the answer? @Denebel Yes I know. But how do you solved it? Both the denominator and dividend is a function.
@twitter :) multiply both sides by x+1 and solve it
@ghazi yeah? how do you solved that? I am not impressed with your half work. I bet you are plain wrong though. X on both sides is just going to cancel each other and besides you deleted your first response because it was plain stupid. You are not helping.
\[\left| \frac{x-1}{x+1} \right| <1\] \[ \frac{x-1}{x+1} <1\] or \[ \frac{x-1}{x+1} >-1\] you cannot multiply by \(x+1\) because you do not know if it is positive or negative start with \[\frac{x-1}{x+1}-1<0\] solve that one, then \[\frac{x-1}{x+1}+1>0\] solve that one, then take the intersection
\[\frac{x-1}{x+1}-1<0\] \[\frac{-2}{x+1}<0\] \[x>-1\] for the first one
oh this is ancient history nvm
\[\frac{x-1}{x+1}+1>0 \]\[\frac{2x}{x+1}>0\]\[x>0\] Therefore \( x \) is greater than zero. Thanks @satellite73 ! I am your fan now! ;))