# Absolute value inequality

|1/(x-1)|<1

## The Attempt at a Solution

is that the same as this -1<1/(x-1)<1

can i do each side by it self then take the values at which they intersect

so i subtracted the 1 then got (-x+2)/(x-1) then made a sign chart with 2 and 1 on it
then took the less than terms so i got (1,2) for the first suoltions
then for the second i got x/(x-1) then made a sign chart with 0 and 1 on it
then i took the greater than terms geting (-inf,0) U (1,inf)
is this right

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is that the same as this -1<1/(x-1)<1
Well, let's check. Suppose x = 0. Does this inequality hold? No because -1 = -1.

Maybe this will help:

|x| < y
=> x < y or -x > -y

ok i see so you took x<y then divided it by a -1 and then flipped the sign
so then would our soultions to my original inequality be
(-inf,1) U (2,inf)

Homework Helper
A shorter idea (for future reference)
$$|x| < y$$

is the same as
$$-y < x < y$$

If your problem begins with $$\le$$ then replace $$<$$ with $$\le$$ in the simplification.

would we then have to do a sign chart

Last edited:
symbolipoint
Homework Helper
Gold Member
crager,
Look again at Dunkle's first response. Graph the relation on a number line and you may more clearly find a path to a solution of his example and to your exercise problem.

In your original expression on the left, the expression inside of the absolute value is either positive, or negative; examine each of these conditions separately. Do you yet need more detailed descriptions?

(-inf,-2) U (2,inf) is this the answer

symbolipoint
Homework Helper
Gold Member

symbolipoint
Homework Helper
Gold Member

Use a number line to help understand this:
Either y<1 OR -y<1.

Replace y with the original expression and solve, and check or first solve for y and then replace y with the original expression and finish solving.
...but be careful. I'm not absolutely sure about my own answer, so I might have made a conceptual error. Still, not that for the particular exercise, x canNOT be 1.

HallsofIvy