Is x > 0?

I think u did a calculation mistake in calculating first case of the stmt 1 , x should be 2. So both statements give the same result which is x=0 or 2 so answer should be E

Yes!I did!Thasnk you for pointing it out!
How is the answer D in tht case??

oh got it..you have to put back the values and only 2 will hold:)

I think we are supposed to put the answers back into the equation to see which one works and in this case only x=2 works for both statements so it gives us a definte answer....
x =0 gives 3 = -3 in first stmt and 1=-1 in second stmt...

even I didn't think of it when I tried this the first time but I guess that's how we should be approaching this problem...

so is the OA C or E? If we combine statement A and B we get a definite 2, so shouldnt the answer be C

OA IS D

sorry, my mistake. we still cannot arrive at a conclusion by combining A and B. So, the answer should be E, it cannot be D. Which OG is this from?

uptowngirl92 wrote:Is x > 0?

(1) |x + 3| = 4x - 3

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

Let's think this question through instead of doing it purely algebraically.

In both statements, we know that the the left side is equal to or greater than 0 (since it's absolute value). Therefore, the right side must also be greater than or equal to 0.

So, for (1) we know that:

4x - 3 >= 0
4x >= 3
x >= 3/4

If x is greater than or equal to 3/4, is it always going to be greater than 0? YES!

We can reason through statement (2) the same way to get:

2x - 1 >= 0
2x >= 1
x >= 1/2

If x is greater than or equal to 1/2, is it always going to be greater than 0? YES!

So, each statement confirms on its own that x must be greater than 0: choose (D).

thats a great approach..thanks again Stuart!