another absolute values puzzle...

Hi, it is given in the question that x<-5. If we're able to get this inequality validated in either of the options, we can say the result is correct.

As you're clear with Choice I & III, I'll only take up choice-II.

You said, if we plug the options x<-5 back in eqn |x+3|>2 then it is invalid.
We'll prove this by taking examples -

Let x=-6, Putting this value back in the eqn, we get
|-6+3|
=|-3|= 3 which is greater than 2

Let x=-7, from the eqn -
|-7+3|= |-4| = 4 which is greater than 2

For all valus of x<-5, |x+3| will be greater than 2

Thus proved

sorry, i need to pay more attention to my calculations.

one question though: as long as one of the options is correct in absolute problems, is that sufficient for conclusion? in the II options above, x<-5 or x>-1... we don't need to worry about x>-1 here?

Hi,

Whether you have to check one or all that depends on the question.

Overhere, in 2nd option, we have x<-5, x>-1
Both are valid solutions to the equation.

If I reframe the question and put it in a simple way, then the question is actually asking -
If x<-5, which of the following must be true ?

So, all you need to focus is whether the given options are valid for x<-5. If there is something additional to required, that does not affect our solution.

topspin360 wrote:i have a question specific to answer choice 2...

If 4 < (7-x)/3, which of the following must be true? (converts to: x<-5)

I. 5<x
II. |x+3|>2
III. -(x+5) is positive.

In choice II, i get to solutions: x>-1 or x<-5. If I plug them back into |x+3|>2, first choice is valid but second is invalid. Which means the answer can't be that x<-5.
Correct answer is that both II and III are right. how can II be right?

|x+y| is the DISTANCE between x and -y.
Thus, II asks the following:
Is the distance between x and -3 is greater than 2?
Plot x<-5 on a number line:

<---x---(-5)......(-3)........

As the number line shows, if x<-5, the distance between x and -3 must be greater than 2.