MGMAT: Question bank: Is |x| < 1 ?

[Khalid]

Is |x| < 1 ?

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

(2) |x - 3| > 0

What is the best way to solve for statement 1. I have seen the explanation and wondering if there are any easier quick ways! Do I really have to consider 4 scenarios when there is abs. sign on both sides of a equation:

Left and Right >0
Left <0 and Right >0
Left > 0 and Right <0
Left <0 and Right <0

Thanks!

[StaceyKoprince]

If you want to be thorough? Then yes. :) But I think the real test wouldn't give you something this computation intensive unless you were really scoring at the very very top of the range.

Most of the time (and that's usually good enough on this test), you can get away with just two scenarios: both the same and one negated.
both the same: x+1 = 2(x-1) this solves to x=3
one negated: x+1 = -2(x-1) this solves to 3x = 1, or x = 1/3

[kanaks123]

Here is my rule of thumb.

When there is equality between left hand side and right hand side, you always end up with two scenarios.

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

(1) x+1 < 0 x-1 < 0 => -(x+1) = - 2(x-1)
(2) x+1 < 0 x-1 > 0 => -(x+1) = 2(x-1)
(3) x+1 > 0 x-1 < 0 => (x+1) = - 2(x-1)
(4) x+1 > 0 x-1 > 0 => (x+1) = 2(x-1)

As you can see (1) -(x+1) = -2(x-1) is always equal to (x+1) = 2(x+1), which is (4)
Similarly (2) is same as (3)

**Please note this works only for equality. For inequality you must consider all four cases.

[esledge]

Good point, kanaks123. The inequality exception is important.

It's also worth noting that we aren't really ignoring any real solutions. Technically, some of the 4 solutions are invalid for their scenario. These "dummy" solutions are just a by-product of equations with absolute value on both sides.

(1) x+1 < 0, x-1 < 0 => -(x+1) = - 2(x-1) --> x = 3 (INVALID: x = 3 makes x+1>0, i.e. scenario 4)
(2) x+1 < 0, x-1 > 0 => -(x+1) = 2(x-1) --> x = 1/3 (INVALID: x = 1/3 makes x-1<0, i.e. scenario 3)
(3) x+1 > 0, x-1 < 0 => (x+1) = - 2(x-1) --> x = 1/3
(4) x+1 > 0, x-1 > 0 => (x+1) = 2(x-1) --> x = 3