Dear friends and moderators,

I didn't even know how to begin the following question:

If y > 0, what is the value of x?
1. |x - 3| > y
2. |x - 3| < - y

OA: B

Kindly explain.
Thanks
KH

Dear friends

After staring at the question for about half hour or so, I have come up with a solution. Kindly correct me if wrong.

If y > 0, what is the value of x?

1. |x - 3| > y
Taking numbers:
x: -2, 1 both can satisfy the above equation. Insufficient.

2. |x - 3| < - y
Since |x-3| is an absolute value, the smallest it can go is 0.
And since y is given to be >0, thus - y will give a negative value which will cause the equation to fall apart unless it is 0.

so |x-3| = 0.
x = 3.

Fundamentally sound?

Thanks
KH

looks good.

incidentally, you wrote the inequality signs as strict inequality (< and >) instead of weak inequality (< and >) this time, a condition that would actually render the problem impossible. if the problem is as written in the first post, then this solution is perfect.

Yes Ron
The inequality signs follow the 1st post.

1. |x - 3| > y
2. |x - 3| < - y

Thanks
KH

Thanks for the explanation!

Hi,

what is the answer to this question? I got D when I solved it. Someone help If I got this wrong

1) if y ≥ 0 then
|x-3| ≥ 0 ,
Case 1
-(x-3) ≥ 0 , -x + 3 ≥0 , -x ≥ -3 , x ≤ 3
or Case 2
(x-3) ≥ 0 , x ≥ 3

Combining the 2 cases,
if x≥ 3 or x ≤ 3 then x = 3 SUFFICIENT

2) |x-3| ≤ -y
Since y ≥ 0 y must 0
hence |x-3| = 0
case 1
-(x-3) = 0
-x+3 = 0 , x = 3
Case 2
x-3 = 0
x = 3

Plugging this answer back in the the equation |x-3| = 0 only case 2 is satisfies the condition and therefore x = 3 SUFFICIENT

Please let us know the answer for this.

your mistake is in red. you mistook "or" for "and".
if it said "and", that would be the correct solution. but it doesn't; it's "or".

the solution to "x≥ 3 or x ≤ 3" is ALL numbers.
when you have "or", you only have to satisfy ONE of the statements. any number satisfies at least one of these, so, there you go.

also, you can solve that statement by inspection.
if you see |quantity| ≥ 0, then this is ALWAYS true NO MATTER what is inside the absolute value!
it's impossible to find any quantity such that |quantity| < 0, so |quantity| ≥ 0 is true for everything, all the time.

Having a bit of trouble with this. Rule that I understand:

|quantity| > A: quantity > A OR quantity <-A
|quantity| < A: -A < quantity < A

Applied to this problem:
(1) |x-3| >= y, where y>=0

case 1:
x-3 >= 0
x >= 3

case 2:
x-3 <= -0
x <= 3. INSUFFICIENT.



(2) |x-3| <= -y, where y>=0

case 1:
|x-3|<= -(0), where y is still positive, just the sign outside the brackets is not.
x<=3

case 2:
|x-3|>=-(-0)
x>=3

Could you explain why |x-3|=0?

What is the OA? Thanks!

OA is "B"

Victor,

Just because y >=0 doesn't mean that | x - 3 | >= 0. You cannot make that substitution.

For example, if y >=0, then one possibility is y = 1. In this case,
| x - 3 | >= 1 meaning x - 3 >= 1 or x - 3 <= -1; we get x >= 4 or x <= 2. Here, x is NOT 3.

_________________
Ben Ku
Instructor
ManhattanGMAT

Ben,

Can you explain why I cannot substitute. The above poster does so as well (Ron replied to that one). Confused here. Thanks.

Given:
y>=0

(1)
|x-3| >=y
|x-3| >= y >=0.

Therefore, |x-3|>=0.

Also, is there a better way to do this problem without plugging in numbers?

i think i see what ben is getting at here. in a sense, you guys are both correct -- you're just talking past each other.

YOU are using the transitive property:
if a > b, and b > c, then a > c.
this is of course true.

BEN, however, is not talking so much about the inequalities themselves as he's talking about their solution sets.
for instance:
let's say that x > y and y > 10.
in this case,
* we do know that x > 10 (in other words, it's impossible that x > 10 is false); in that sense, you are correct.
* where you are going wrong here, though -- and what ben is trying to point out -- is that you are assuming that, given the above two inequalities, that x can automatically assume ANY value > 10. in other words, it seems that you are taking the two inequalities x > y and y > 10, together, as a guarantee that x can actually equal 10, or any greater number.
this isn't necessarily true, depending upon the value of y. for instance, if y is 15, then x is now restricted to numbers that are 15 and higher.

hope this makes more sense