No taker for this question ?logitech wrote:Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
Well (0,2) satisfy both equation.
So?
ABS
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
EricLien9122
- Master | Next Rank: 500 Posts
- Posts: 106
- Joined: Thu May 22, 2008 10:50 am
- Thanked: 3 times
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.
|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.
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.
-
logitech
- Legendary Member
- Posts: 2134
- Joined: Mon Oct 20, 2008 11:26 pm
- Thanked: 237 times
- Followed by:25 members
- GMAT Score:730
Nope you are absolutely right! Thank you.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.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
-
jimmiejaz
- Master | Next Rank: 500 Posts
- Posts: 207
- Joined: Sun Mar 11, 2007 6:16 pm
- Location: Mumbai
- Thanked: 11 times
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!!!!!!
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!!!!!!
-
logitech
- Legendary Member
- Posts: 2134
- Joined: Mon Oct 20, 2008 11:26 pm
- Thanked: 237 times
- Followed by:25 members
- GMAT Score:730
I am glad you found that method easy man. I know your test day is coming up soon.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!!!!!!
BEST OF LUCK! BEAT THE SHIT OUT OF THAT GMAT THING!!
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
-
jimmiejaz
- Master | Next Rank: 500 Posts
- Posts: 207
- Joined: Sun Mar 11, 2007 6:16 pm
- Location: Mumbai
- Thanked: 11 times
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.logitech wrote:I am glad you found that method easy man. I know your test day is coming up soon.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!!!!!!
BEST OF LUCK! BEAT THE SHIT OUT OF THAT GMAT THING!!
Best of luck.
Cheers mate!!!!
