Solving Absolute Value Inequalities without Changing Directions

[cragar]

Homework Statement

|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

 

[Dunkle]

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

 

[cragar]

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)

 

[statdad]

A shorter idea (for future reference)
<br /> |x| &lt; y<br />

is the same as
<br /> -y &lt; x &lt; y<br />

If your problem begins with \le then replace &lt; with \le in the simplification.

 

[cragar]

would we then have to do a sign chart

 

[symbolipoint]

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?

 

[cragar]

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

 

[symbolipoint]

Reread post #6 and #2. Check your answer.

 

[symbolipoint]

Start with |y|<1, where y = 1/(x-1).

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]

In my opinion, the simplest way to solve most inequalities is to first solve the associated equation. |1/(x-1)|= 1 reduces to 1/(x-1)= 1 or 1/(x-1)= -1. Multiplying both sides of each gives 1= x-1 and 1= 1- x. In the first case, x= 2 and in the second x= 0. The point of that is that continuous functions can change from "<" to ">" on where they are "=". Of course, |1/(x-1)| is NOT continuous at x= 1 so we must add that possibility: the inequality can change at x= 0, x= 1, and x= 2. If we take x= -1< 0, |1/(-1-1)|= 1/2< 1 so the given inequality is true for all x< 0. Check a value of x between 0 and 1, a value of x between 1 and 2, and a value of x larger than 2 to determine which of those intervals also satisfy the inequality.