Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

One of the explainations from MGMAT is that

"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"

Why is the left side of the equation is positive but the value on the right is negative?

if you take any number x between -1 and 1, x+1 is positive and x-1 is negative. remember, this is only saying that the values inside the absolute value signs are this way. once we apply the absolute value signs, they both become positive. let us know if this helps or if you have further questions..

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Tim Sanders
Manhattan GMAT Instructor

what is OA?
In my opinion it should be C.

from statement 1 x =3 or x=1/3
from statement 2, x>3 or x<3
from both only possible ans is x=1/3.

Looks good!

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Jamie Nelson
ManhattanGMAT Instructor

In this question, what I do not understand from the explanation is the following:

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:

The above is part of the explanation for evaluating statement 1. My question is- what is the reasoning behind multiplying each part of the equations that are negative by -1? Why do we do this?

if you take the absolute value of a quantity that's negative, the sign of that quantity is reversed.

if you take |-5|, you get 5.
if you take |x| when x is negative, you get -x (which would be positive in that case).
if you take |x - 3| when x - 3 is negative, you get -x + 3, or 3 - x (which is then positive).
etc.

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