Inequality with absolute value confusion: |x +1| = 2|x -1|

One way:

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

Square both sides: \((x+1)^2=4(x-1)^2\);

\(x^2+2x+1=4x^2-8x+4\);

\(3x^2-10x+3=0\);

\(x=\frac{1}{3}\) or \(x=3\).

Another way:

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

Two key points: \(x=-1\) and \(x=1\) (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------


A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));

B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));

C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).

So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\).

Hope it helps.

Hi Bunuel thank you for your response. I understood the first process. I got a bit confused with the second way that you mentioned. why did we take -x when considering the blue range? In the green range we took x =1 on one side of the equation x= -1 on the other side? I am not sure I understood clearly. I have never been very strong with inequality; sorry if this is a minor thing that I'm missing somehow.

Thanks again for your help.

In the blue range, the term x+1 becomes negative (x<-1). However, modulus of any term should be positive. So we multiply the term within the modulus by a -1 so that the modulus becomes positive (negative times negative = positive).

For example, if x is negative, |x| = -x. (-x is positive)
If x=-5, then |-5| = -(-5) = 5

Similar logic applies for the green range. Hope this helps. Let me know if you need further clarification.

Hi deeuce
This is a direct example to solve an Absolute value equation of the form |a| = |b|

Before we get into solving it, why don't we have a look at some GMAT standard models tested through Absolute Value Equations?

One of the general models tested on GMAT is |a| = k where k is a constant (Ex- |a-3| = 6))

A second model of testing would be that of using a double mod. ||a|-p| =q where p and q are constants.

And a third model would be |a| =|b|

Each of these cases have a separate way of solving but the fundamental properties of Absolute Values are always used to solve.

Now, that we know we are going to use the third model here, lets deep dive into solving it.

|x +1| = 2|x -1|

Since both LHS & RHS contain modulus, lets try to break it and reach to the first simple model of |a| = k

Let us substitute 2|x+1| as k and hence we have with us

x+1 = k when x >= -1

and

x+1 = (-k) when x < -1

This is using the Fundamental properties of Modulus and their basic operations.

Substitute k as 2 |x-1| in both the equations

So how have we broken this by now?

x + 1 = 2 |x-1| for x >= - 1
and
x + 1 = -2 |x - 1| for x < -1

1st case-
x + 1 = 2 |x-1| for x >=-1
Using the Mod operations, lets have the reference ranges for x and they are
x + 1 = 2 |x-1| ; for x >= -1 and now lets break the second mod
We have now
x + 1 = 2 (x-1) for x >=-1 and x > =1 which gives us a combined range of x > = 1
Solving this, x + 1 = 2x - 2 or x = 3
Is this value in the reference range of x >= 1 ?YES.
3 is >=1 and hence its a valid solution point. Accepted value of x=3.

Now lets see if there are any other values too!
x + 1 = -2 (x-1) for x >= -1 and x < 1 which gives us a combined range of -1 <= x < 1
Solving this x + 1 = -2x + 2 or 3x =1 =>x =1/3
Is 1/3 between -1 and 1 ?
YES.
Hence its a valid solution point. Accepted value of x= 1/3 also


2nd Case
x + 1 = -2 |x - 1| for x < -1
Using the Mod operations, lets have the reference ranges for x and they are,
x + 1 = -2(x-1) for x < -1 and x >=1 . Is it possible for x to be greater than or equal to 1 and also less than -1?NO.
Discard this scenario.

Also,we have one more step to go!
x + 1 = -( -2(x-1) ) for x < -1 and x <1 which gives us a combined range of x <-1
Solving we have, x + 1 = 2x - 2
or x =3 but is this value appropriate for the range of x <-1? Is 3<-1? NO
So discard x=3 for this range.

However, according to the solution above, you have x =3 and x =1/3 as two solution points for x.
So we hope by now we hope you have gained the insights on how to solve Absolute value equations of this form.

GMAT Tip This approach is explained so that you can build up on the fundamentals of Absolute Value and the methods of opening the mod equations and arriving at values using the reference ranges.

Alternately you can use the method of squaring.

This is a fairly GMAT needed approach to tackle questions of this sort.

GMAT Learning The alternate definition of |x| is the square root of \(x^2\)
or
|x| = √\(x^2\)

So, |x+1| = √ \((x+1)^2\) and |x−1| = √\((x−1)^2\) and hence we have the converted equation as

√ \((x+1)^2 \) = 2 √\((x−1)^2\)

So, Squaring both sides,

=>\((x+1)^2\) = 4\((x−1)^2\)

\(x^2\)+2x+1=4\(x^2\)−8x+4

=> 3\(x^2\)−10x+3=0

=> 3\(x^2\) -9x –x -3 = 0

or

=>3x(x-3) -1(x-3) =0

=> (3x-1) (x-3) = 0

=> x= 1/3 , x=3

Attaching the PDF of this solution here and a screenshot(for the first approach) to make it simple and more illustrative for those who are already comfortable with the Mod operations.

You can also have a look at this video on Absolute Values to build on your concepts and strengthen the fundamental on Absolute Values
https://www.youtube.com/watch?v=yGBAIcBhQdM

Hope this post helps you all who are seeking some GMAT guidance on Absolute Values ! :)

Please feel free to reach out to us for any help or GMAT related guidance.

Devmitra Sen
GMAT Quant SME