absolute value with three conditions - from MGMAT databank

[guest]

This is from databank Inequalities

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0

for statement 1, there are three conditions that need to be tested what I did is subtracted 2|x - 1| to the left side forming:

|x + 1| - 2|x - 1| = 0

and find the critical positions

-1 < x, -1 < x < 1 and x > 1

I know how to do do case 1 and case 3, but how do I solved case 2, and I don't understand the provided explaination

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

I would greatly appreciate if anyone could provide further insight, what's the logic in determining left side of the equation is position and the right side of the equation is positive.

[rfernandez]

I think your question stems from how to solve absolute value equations in general.

Let's start with the definition of absolute value: |x| = x, if x>0 or -x, if x<0. Let's try it out...

|3| = 3; because the number inside the absolute value was already positive, all we needed to do was drop the absolute value bars.
|-7| = -(-7) = 7; notice here we started with a negative value inside the absolute value so if we want to drop the absolute value bars, we need to multiply the contents by -1, yielding 7.

For more complex absolute value expressions, therefore, our task is to figure out what values make the inside of the absolute value expression positive or negative. If we wish to "open" the absolute values, we'll simply drop the absolute value bars when we know that the expression inside them is positive. Alternately, we'll multiply the expression by -1 when the expression is negative.

This is what the solution leads us through. I'll specifically target the second case, as you requested, when -1 < x < 1:

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

Now, when x is between -1 and 1, the expression inside the absolute value bars on the left side (x + 1) is always positive. That should make sense because no matter what value of x I choose between -1 and 1, the value of (x + 1) will necessarily be a positive number. Therefore, we're allowed to drop the absolute value bars on this left side.

On the right side of the equation, however, the expression inside the absolute value bars (x - 1) is always negative. (Try plugging in a few values of x between -1 and 1 into (x - 1) to confirm that it's always negative.) So if we want to drop the absolute value bars, we'll have to multiply (x - 1) by -1, yielding (1 - x).

Now, the equation can be rewritten as:

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

Rey