EIVs Question Bank - Q7 (Absolute Range): Is |x| < 1

[Guest]

Is |x| < 1 ?

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

(2) |x - 3| > 0

The Q-stem can be simplified to Is -1 < x < 1 ?

Easy statement here is 2. So lets start with that and the BD/ACE grid.

Statement 2: |x - 3| > 0
So with absolute expressions, you have to consider the positive and negative values.
POSITIVE: x > 3
NEGATIVE: x < 3
So x is greater than or less than 3. Which means x can be any number, but 3. This is INSUFF in answering our question "Is -1 < x < 1 ?". So we can cross out BD from our BD/ACE grid.

Now we move to statement 1.
This is where I struggle. Can someone please explain how best to tackle this statement. I am struggling to follow the approach outlined in the Q-bank solution.

Thanks in advance.

[rfernandez]

Is |x| < 1 ?

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

(2) |x - 3| > 0

I disagree with Sudhan's treatment of (1). Here's how to figure this one out:

Statement (2):
Yes, it tells us that x does not equal 3. INSUFFICIENT.

Statement (1):
We'd like to be able to solve for x, but those pesky absolute value bars are getting in the way. We obviously can't just remove the bars, however. One way to approach it is to consider specific intervals for x and alter the equation accordingly.

The intervals to consider are given by the absolute value expressions themselves. Namely, |x + 1| has a "critical value" at x=-1, since this is where the (x+1) is zero. Less than -1, (x+1) is negative, more than -1, (x+1) is positive. Similarly, |x - 1| has a critical value at x=1. So we can "open up" this absolute value equation by expressing it as three different equations over the three intervals x<-1, -1<x<1, and x>1. Let's consider each case:

Case x<-1
|x + 1| = 2|x - 1|
In this interval, (x+1) yields a negative number, so we can replace |x+1| with its opposite, or -(x+1). (x+1) also yields a negative number, so we can replace |x+1| with its opposite, or -(x-1). This gives the following equation:
-(x+1) = -2(x-1)
-x-1 = -2x+2
x=3
Looks like we have a solution, but in fact we don't. We started with the assumption that x<-1, so ending with x=3 means there is no solution here. On to the next case.

Case -1<x<1
|x + 1| = 2|x - 1|
Here, (x+1) is positive, so |x+1| is simply (x+1). However, (x-1) is negative, so |x-1| is equivalent to -(x-1). Resulting equation:
x+1 = -2(x-1)
x+1 = -2x+2
x = 1/3
In this case, the solution lies in the interval -1<x<1, so we have a viable solution. But there's still another case to consider...

Case x>1
|x + 1| = 2|x - 1|
Here, both expressions inside the absolute values are going to yield positive values, so we can simply drop the absolute value bars.
x+1 = 2(x-1)
x+1 = 2x-2
x=3
This solution lies in the interval x>1, so it's also a viable solution.

In summary, the solution to |x + 1| = 2|x - 1| is x=1/3 or x=3. This leads to an INSUFFICIENT result.

Statements (1) and (2):
Combining these results, we have x=1/3. SUFFICIENT.

The correct answer is C.

Rey

[Captain]

Is |x| < 1 ?

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

An alternate way to solve equation 1 is to square both sides. In this case.
x^2 + 2x + 1 = 4x^2 - 8x + 4.
ie. 3x^2 - 10x + 3 = 0
ie. 3x^2 - 9x - x + 3 = 0
ie (3x-1)(x-3) = 0
ie x = 3, 1/3.
In this case both 3 and 1/3 are solutions to |x+1| = 2|x-1|. I have a feeling that this will be true in all cases but I would suggest that you plug in the values into the original equation and check that both the values actually do satisfy the modular equation.

[RonPurewal]

Is |x| < 1 ?

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

An alternate way to solve equation 1 is to square both sides. In this case.
x^2 + 2x + 1 = 4x^2 - 8x + 4.
ie. 3x^2 - 10x + 3 = 0
ie. 3x^2 - 9x - x + 3 = 0
ie (3x-1)(x-3) = 0
ie x = 3, 1/3.
In this case both 3 and 1/3 are solutions to |x+1| = 2|x-1|. I have a feeling that this will be true in all cases but I would suggest that you plug in the values into the original equation and check that both the values actually do satisfy the modular equation.

yeah, that method will work, but it deserves two caveats.
1) you still have to check the solutions to make sure that they work in the original equation
2) it's much more work than the method that i'll outline in the next post.

...but it does work, and, if you're reallyreallyreally good at factoring, it may be the best method for you.

[RonPurewal]

as far as the equation
|x + 1| = 2|x - 1|
goes,
here's what is almost certainly the most efficient way to solve it: just realize that there are only two possible cases.

namely:
1)
(x + 1) and (x - 1) have opposite signs:
in this case, the equation becomes -(x + 1) = 2(x - 1) (or you could put the negative sign on the other one; it doesn't really matter)
-x - 1 = 2x - 2
1 = 3x
1/3 = x

2)
(x + 1) and (x - 1) have the same sign:
in this case, the equation is x + 1 = 2(x - 1)
x + 1 = 2x - 2
3 = x

then check these solutions in the original equation, and make sure that both of them work. (they do)
done.

--

the above method is definitely the best way to proceed when all that's in the equation is 2 absolute values (perhaps preceded by coefficients). if anything is added to, or subtracted from, these absolute values, then you'll have to try all 4 possible combinations of signs: ++, +-, -+, and --. in that case, the method presented above by rey is probably the better way to solve.
but, as usual,the most important thing is to figure out which method is most comfortable, and fastest, for you.

[guest]

The advantage of Rey's approach is that each new quantity inside absolute value signs only adds one more range to check - with the approach above in the general case, each new quantity means 2 more possibilities:

e.g. |x+1|=2|x-3|-5|x+2|-7

using Rey's method you only need to check x<-2,-2<x<-1,-1<x<3,x>3 (Four ranges). If you considered each absolute value quantity as positive or negative, that would give you eight possibilities. Of course, you're HIGHLY unlikely to see an example as complicated as the one above on the GMAT, so any of the approaches should get you though these problems in a reasonable amount of time.

[rfernandez]

Agreed; there are usually multiple paths to get there. Better to have a number of different techniques available to you, and this discussion is a good example of that.