Data Sufficiency

Question2 - Is x > 0?

(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

logitech wrote:Question2 - Is x > 0?

(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

No taker for this question ?

Well (0,2) satisfy both equation.

So?

I'll go with E.
X could be 0 or 2, if it's 0 ----> X = 0, if it's 2 ----> X > 0

THANKS.

I don't think 0 satisfy statement 1:

|x+3|=4x-3
|0+3|=4(0)-3
3=-3?

I think it should be A.


Please correct me if I make a mistake.

For (1),
Case a => |x+3| > 0 , when x > -3
Case b => |x+3| < 0 , when x < -3

Now, solve equation (1), for case a, we get
|x+3| = x+3 (case a) => x+3 = 4x-3 => x = 2

Now, solve equation (1), for case b, we get
|x+3| = -(x+3) (case b) => -(x+3) = 4x-3 => x = 0

x=0 is not possible, as per case b, x < -3

so, from (1), we get only one value of x, and that is 2

--------------------------------------------------
Solve, similarily for equation (2)

wherein, possible cases are
(a) x>3 ;
(b) 3/2<x<3;
(c) x<3/2

here, for case (a) x>3
both |x – 3| and |2x – 3| are > 0
On solving, we get x=0, and this is not valid as per case (a), x > 3

for case (b) 3/2<x<3
|x – 3| < 0 and |2x – 3| > 0
On solving, we get x=2

for case (c) x<3
both |x – 3| and |2x – 3| are < 0
On solving, we get x=0

Hence, we neglect case (a), and from case(b) and case (c), we get 2 values of x. i.e 2 and 0
which is insufficient

Hope it helps.

EricLien9122 wrote:I don't think 0 satisfy statement 1:

|x+3|=4x-3
|0+3|=4(0)-3
3=-3?

I think it should be A.


Please correct me if I make a mistake.

Nope you are absolutely right! Thank you.

logitech,

You gave a new method of solving abs values in a previous question.
going by that...
we have |x-3| = |2x-3|
squaring both sides and rearranging we get
3x^2 - 6x = 0
3x(x-2) = 0
x=0 or x=2

hence 2 is insuff....
I must say this is the easiest approach for solving abs equations when both sides have single terms on each side.
Thanks a lot.
Keep rocking!!!!!!

jimmiejaz wrote:logitech,

You gave a new method of solving abs values in a previous question.
going by that...
we have |x-3| = |2x-3|
squaring both sides and rearranging we get
3x^2 - 6x = 0
3x(x-2) = 0
x=0 or x=2

hence 2 is insuff....
I must say this is the easiest approach for solving abs equations when both sides have single terms on each side.
Thanks a lot.
Keep rocking!!!!!!

I am glad you found that method easy man. I know your test day is coming up soon.

BEST OF LUCK! BEAT THE SHIT OUT OF THAT GMAT THING!!

Got it A.

Thanks.

logitech wrote:

jimmiejaz wrote:logitech,

You gave a new method of solving abs values in a previous question.
going by that...
we have |x-3| = |2x-3|
squaring both sides and rearranging we get
3x^2 - 6x = 0
3x(x-2) = 0
x=0 or x=2

hence 2 is insuff....
I must say this is the easiest approach for solving abs equations when both sides have single terms on each side.
Thanks a lot.
Keep rocking!!!!!!

I am glad you found that method easy man. I know your test day is coming up soon.

BEST OF LUCK! BEAT THE SHIT OUT OF THAT GMAT THING!!

Thanks a lot Logitech. I read your story on the I just Beat the GMAT section. I admire your never say die attitude and i have full confidence that this time you will definitely get that magical 700+ score which we all aim for.
Best of luck.
Cheers mate!!!!