earthling Wrote:Guys,
I am confused with Stacy's remark "Scenario 3: One positive, one negative. And now I've already got an idea of the range of numbers I should be testing, because Scenario 1 resulted in x < -1 and Scenario 2 resulted in x > 1. So... let's see what would happen if x is between those two numbers. For |x + 1|, if -1 < x < 1, then the contents will be positive. For |x -1|, if -1 < x < 1, then the contents will be negative. "
Why wouldn't you test two more scenarios i.e. Left Side is negative and right sand is positive or Right Side is -ve and left side is positive. This we can establish the condition that x not equal to 1. I guess my confusion is what condition are we testing in Stacy's remark for scenario 3? How does it help in knowing that the left side is positive and right is -ve?
Any help will be appreciated. Thanks
(x + 1) is (x - 1) plus two. therefore, it's impossible for (x + 1) to be negative while (x - 1) is positive.
since this "fourth case" is impossible, you don't have to consider it.
by the way:
this is a rather complicated way to solve this equation. for an absolute value equation involving other terms, such as |expression| + other stuff = |expression|, you might need an approach like that.
but for this one, it's sufficient just to consider two possibilities: either (x + 1) and (x - 1) have the same sign, or they have opposite signs. in fact, you can just set the equations up in that way; you don't even have to figure out what the actual signs are.
see here. i'm not sure why there are 2 live threads on this problem, but that's the breaks.