Problem Solving

i've posted a problem similar to this before. Just posting another problem b/c i'm unable to wrap my head around absolute values and inequalities combined.

why do we need to consider three options: x<-1, -1<x<1, x>1? Don't understand the explanation at all. Does anyone have a link or resource I can use to read up on this topic and fully understand the nuances of it? Would really appreciate any resources you guys might have in mind.


thanks.

Problem:

Is |x| < 1 ?

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

(2) |x - 3| > 0


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient


EXPLANATION::

We can rephrase the question by opening up the absolute value sign. In other words, we must solve all possible scenarios for the inequality, remembering that the absolute value is always a positive value. The two scenarios for the inequality are as follows:

If x > 0, the question becomes "Is x < 1?"
If x < 0, the question becomes: "Is x > -1?"
We can also combine the questions: "Is -1 < x < 1?"

Since Statement 2 is less complex than Statement 1, begin with Statement 2 and a BD/ACE grid.

(1) INSUFFICIENT: There are three possible equations here if we open up the absolute value signs:

1. If x < -1, the values inside the absolute value symbols on both sides of the equation are negative, so we must multiply each through by -1 (to find its opposite, or positive, value):

|x + 1| = 2|x -1| -(x + 1) = 2(1 - x) x = 3
(However, this is invalid since in this scenario, x < -1.)

2. If -1 < x < 1, the value inside the absolute value symbols on the left side of the equation is positive, but the value on the right side of the equation is negative. Thus, only the value on the right side of the equation must be multiplied by -1:

|x + 1| = 2|x -1| x + 1 = 2(1 - x) x = 1/3

3. If x > 1, the values inside the absolute value symbols on both sides of the equation are positive. Thus, we can simply remove the absolute value symbols:

|x + 1| = 2|x -1| x + 1 = 2(x - 1) x = 3

Thus x = 1/3 or 3. While 1/3 is between -1 and 1, 3 is not. Thus, we cannot answer the question.

(2) INSUFFICIENT: There are two possible equations here if we open up the absolute value sign:

1. If x > 3, the value inside the absolute value symbols is greater than zero. Thus, we can simply remove the absolute value symbols:

|x - 3| > 0 x - 3 > 0 x > 3

2. If x < 3, the value inside the absolute value symbols is negative, so we must multiply through by -1 (to find its opposite, or positive, value).

|x - 3| > 0 3 - x > 0 x < 3

If x is either greater than 3 or less than 3, then x is anything but 3. This does not answer the question as to whether x is between -1 and 1.

(1) AND (2) SUFFICIENT: According to statement (1), x can be 3 or 1/3. According to statement (2), x cannot be 3. Thus using both statements, we know that x = 1/3 which IS between -1 and 1.

topspin360 wrote:i've posted a problem similar to this before. Just posting another problem b/c i'm unable to wrap my head around absolute values and inequalities combined.

why do we need to consider three options: x<-1, -1<x<1, x>1? Don't understand the explanation at all. Does anyone have a link or resource I can use to read up on this topic and fully understand the nuances of it? Would really appreciate any resources you guys might have in mind.


thanks.

Problem:

Is |x| < 1 ?

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

(2) |x - 3| > 0


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient


EXPLANATION::

We can rephrase the question by opening up the absolute value sign. In other words, we must solve all possible scenarios for the inequality, remembering that the absolute value is always a positive value. The two scenarios for the inequality are as follows:

If x > 0, the question becomes "Is x < 1?"
If x < 0, the question becomes: "Is x > -1?"
We can also combine the questions: "Is -1 < x < 1?"

Since Statement 2 is less complex than Statement 1, begin with Statement 2 and a BD/ACE grid.

(1) INSUFFICIENT: There are three possible equations here if we open up the absolute value signs:

1. If x < -1, the values inside the absolute value symbols on both sides of the equation are negative, so we must multiply each through by -1 (to find its opposite, or positive, value):

|x + 1| = 2|x -1| -(x + 1) = 2(1 - x) x = 3
(However, this is invalid since in this scenario, x < -1.)

2. If -1 < x < 1, the value inside the absolute value symbols on the left side of the equation is positive, but the value on the right side of the equation is negative. Thus, only the value on the right side of the equation must be multiplied by -1:

|x + 1| = 2|x -1| x + 1 = 2(1 - x) x = 1/3

3. If x > 1, the values inside the absolute value symbols on both sides of the equation are positive. Thus, we can simply remove the absolute value symbols:

|x + 1| = 2|x -1| x + 1 = 2(x - 1) x = 3

Thus x = 1/3 or 3. While 1/3 is between -1 and 1, 3 is not. Thus, we cannot answer the question.

(2) INSUFFICIENT: There are two possible equations here if we open up the absolute value sign:

1. If x > 3, the value inside the absolute value symbols is greater than zero. Thus, we can simply remove the absolute value symbols:

|x - 3| > 0 x - 3 > 0 x > 3

2. If x < 3, the value inside the absolute value symbols is negative, so we must multiply through by -1 (to find its opposite, or positive, value).

|x - 3| > 0 3 - x > 0 x < 3

If x is either greater than 3 or less than 3, then x is anything but 3. This does not answer the question as to whether x is between -1 and 1.

(1) AND (2) SUFFICIENT: According to statement (1), x can be 3 or 1/3. According to statement (2), x cannot be 3. Thus using both statements, we know that x = 1/3 which IS between -1 and 1.

One of the better ways to do an inequality + modulus question is drawing out the number line. |x-5| is the distance of x from number 5.

We need to find whether |x| < 1

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

First statement tells us that distance of x from -1 is twice the distance of x from +1. (2:1 ratio)

Lets draw the number line

-2 ---------- -1-----------0-----------1---------------2------------3------

Now distance of x from -1 is twice the distance between x and 1.
This means two things.

Either x is between -1 and 1 or x is greater than 1 (in which case x=3 because distance between -1 and 3 is 4 and that between 1 and 3 is 2)
If x is between -1 and 1, |x| < 1, otherwise |x| = 3. Insufficient.

(2) |x - 3| > 0
We know that | any number | >=0 . Since |x - 3| > 0 it means that |x - 3| is not equal to 0 => x is not 3.
This statement is insufficient by itself but we can clearly see that 1) and 2) combined lead to |x| < 1. Sufficient.

I am not sure whether you've read this one before, but I wrote something regarding inequalities and moduli at this link. It might help
Cheers !
https://www.beatthegmat.com/squares-squa ... tml#476551

Let me know if you got more questions

eagleeye,

What values beside 3 satisfies the equation?
|x + 1| = 2|x - 1|

I tried values between -1 and 1 and didn't get a result? What am I missing?

theCEO wrote:eagleeye,

What values beside 3 satisfies the equation?
|x + 1| = 2|x - 1|

I tried values between -1 and 1 and didn't get a result? What am I missing?

Try x=1/3

Thanks eagleye.

theCEO wrote:What values beside 3 satisfies the equation?
|x + 1| = 2|x - 1|

As usual, Eagleeye provides a great explanation - using the number line is a great way to solve absolute value problems.

Another way to solve, especially when you have absolute value expressions on both sides of the equal sign, is to square both sides. Doing so generally leaves you with a quadratic.

In this case:

(|x+1|)^2 = (2|x-1|)^2
(x+1)(x+1) = 4(x-1)(x-1)
x^2 + 2x + 1 = 4x^2 -8x + 4
0 = 3x^2 - 10x + 3
0 = x^2 - (10/3)x + 1

The fraction in the x term makes this a bit more annoying, but we need 2 numbers that multiply to 1 and add up to 3 1/3. Since our numbers multiply to 1, they're inverses (e.g. 1/2 and 2, 1/3 and 3, 1/4 and 4, ...). It shouldn't take too much trial and error to see that we get:

0 = (x - 3)(x - 1/3)
x = 3 or 1/3

Once you have those two solutions, finishing the problem should be very quick.

How should you decide which approach to take? As always, when doing untimed practice try every method you know so that, when you get to timed practice and test day, you know which method works best for you.

topspin360 wrote:
Is |x| < 1 ?

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

(2) |x - 3| > 0

Statement 1: |x+1| = 2|x-1|
Case 1:
x+1 = 2x-2
3 = x

Case 2:
x+1 = -(2x-2)
3x = 1
x = 1/3.

Confirm that both solutions are valid:
|3+1| = 2|3-1|
4=4.

|1/3 + 1| = 2|1/3 - 1|
4/3 = 4/3.

Since x=3 and x=1/3 are both valid solutions, we cannot determine whether |x|<1.
INSUFFICIENT.

Statement 2: |x-3|>0
It's possible that x=0 or that x=10.
INSUFFICIENT.

Statements 1 and 2 combined:
Since x=3 does not satisfy statement 2, x=1/3.
Thus, |x|<1.
SUFFICIENT.

The correct answer is C.

thanks all.