How do we expand |x + 1| = 2|x – 1|?

[kunaly]

How do we get possible values of 'x' when we have absolute values on both sides of the equation, as in: |x + 1| = 2|x - 1|?

As a matter of background, I got this from the following question:
Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0


Separately, (leaving aside shortcuts tailored to a question) what is the conceptual approach to understand inequalities with absolute value on both sides of the equation?

[atharshiraz]

How do we get possible values of 'x' when we have absolute values on both sides of the equation, as in: |x + 1| = 2|x - 1|?

Treat it like any other absolute value statement. We know that if
|x+a|<b
then it can be "expanded" as:
(x+a)<b
and
-(x+a)<b


The same goes with the statement above:
(x+1)=2(x-1) (equation 1)
and
-(x+1)=2(x-1) (equation 2)

equation 1 says :
x+1=2x-2
and so x = 3

equation 2 says:
-x-1=2x-2
so
x=1/3

[kunaly]

Treat it like any other absolute value statement. We know that if
|x+a|<b
then it can be "expanded" as:
(x+a)<b
and
-(x+a)<b


The same goes with the statement above:
(x+1)=2(x-1) (equation 1)
and
-(x+1)=2(x-1) (equation 2)


thanks for your response. Since the 'right hand side' of the equation also has an absolute value sign, shouldnt we also examine two possible values i.e. for >0 and <0?

[atharshiraz]

thanks for your response. Since the 'right hand side' of the equation also has an absolute value sign, shouldnt we also examine two possible values i.e. for >0 and <0?

I don't know what you mean by "shouldn't we also examine two possible values i.e. for >0 and <0"?

We looked at two possible values so:
(x+1)=2(x-1) (equation 1)
and
-(x+1)=2(x-1) (equation 2)

are two equations you get when you take away the absolute value sign on the LHS and RHS.

The other way to do this is to treat 2|x-1| as the LHS and the alternative is again the same :
2(x-1)=(x+1) (equation 3)
and
-2(x-1)=(x+1) (equation 4)

Equations 1&3 and 2&4 are the same.

Is this what you meant by "shouldnt we also examine two possible values"?

[kunaly]

thanks for your response. Since the 'right hand side' of the equation also has an absolute value sign, shouldnt we also examine two possible values i.e. for >0 and <0?

I don't know what you mean by "shouldn't we also examine two possible values i.e. for >0 and <0"?

We looked at two possible values so:
(x+1)=2(x-1) (equation 1)
and
-(x+1)=2(x-1) (equation 2)

are two equations you get when you take away the absolute value sign on the LHS and RHS.

The other way to do this is to treat 2|x-1| as the LHS and the alternative is again the same :
2(x-1)=(x+1) (equation 3)
and
-2(x-1)=(x+1) (equation 4)

Equations 1&3 and 2&4 are the same.

Is this what you meant by "shouldnt we also examine two possible values"?

Yes, that is what I meant. Many thanks for explaining, makes sense now!

[atharshiraz]

Yes, that is what I meant. Many thanks for explaining, makes sense now!

I just solved a question that created some doubts in me. I FEEL fairly certain that I may be right but any verification would be great.

[tim]

other than the fact that "and" should be "or" in your solution, you've done this correctly. in general, you need to take positive and negative versions of each absolute value in an equation, as kunaly mentioned. however, as athar pointed out, two of the resulting four equations match up with the other two IN THIS CASE, so only looking at two equations is sufficient..

my favorite strategy for absolute value inequalities - regardless of whether the absolute values show up on one or both sides of the inequality - is to treat absolute value as a distance:

|a-b| < |c-d| means
the distance between a and b is less than the distance between c and d

you can often use this insight to tackle the problem intuitively or with a number line rather than solving a bunch of complicated equations or inequalities..