resilient wrote:Is |x| <1> 0

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Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0



qa is c

Here's what I have so far:

Question is asking - is -1 < x < 1

Let's look at statement 2:
|x - 3| > 0
This doesn't tell you anything except that x DOES NOT EQUAL 3. For any other value, you can subtract three, take the absolute value and it is positive. 3 does not work because it says 3 - 3 = 0 > 0 not true.
This is insufficient (can be between -1 and 1 or can be higher)

Statement 1:
Let's look at values of x greater than 1 (because the second part of the statement is x-1, so I want to keep this expression positive).
You get:
x + 1 = 2x - 2 rearrange
3 = x This is one answer. And now you know that |x| is not less than 1

You need to find another value of x

This is where I actually failed. I could not figure out how to translate the equations to find a second value of x. (just by trial and error, I know that -3 is NOT the other solution, so I know that at least)

How you would set it up is evaluate each option:
x > 1 (all absolute value sections are positive)
1 > x > -1 (first statement is positive and second statement is negative)
x < -1 (all absolute value sections are negative)

Answer:
At this point, I know the answer is either going to be A or C. If A had given a second answer that is not -1 < x < 1, then that is the answer. If A provides an answer that is in that range, then you need statement B to prove that 3 is not the answer, and the second solution proves that -1 < x < 1.

In the interest of time (and already ruling out -3 as an answer), I guessed C, which is the official answer.


Incidentally - the second solution is 1/3. I figured this out a lot later, but could still use someone's help to set up the absolute value equations properly.

Two solutions can be found by equating the abs equation on the RHS by first considering it + and then -

so I soln

x+1 = 2 (+(x-1))
you get x = 3

II soln

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

so ans is C

Hope this helps

Hey netigen..great explanation but I have just one doubt..did u try changing the expression of stat(1) on both the sides ?

what I mean is, since its absolution on both sides of stat(1) did you try the following:

-(x+1)=2(-(x-1)) This gives the answer for x=-1. Though subsitituting the value of x=-1 in the expression does not hold true for stat(1). But I am just wondering how to approach the problem.