Is x>0?

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greenwich: Master | Next Rank: 500 Posts; Posts: 110; Joined: Sun Jul 18, 2010 4:00 pmFollowed by: 2 members; Thanked: 1 times

by greenwich » Thu Jan 16, 2014 5:23 pm

Is x>0?

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

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Source: Beat The GMAT — Data Sufficiency | Thu Jan 16, 2014 5:23 pm

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Thu Jan 16, 2014 7:11 pm

greenwich wrote:Is x>0?

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

Hi Greenwich,

To find: is X>0 (Yes/No Question)

Statement 1: |x+3|=4x-3
here LHS is Absolute value, which is non-negative
hence 4x-3 >= 0 therefore x >= 3/4
hence x >0 Sufficient.

Statement 2: |x+1|=2x-1
apply the above method,
2x-1>=0
x>=1/2 which is >0
hence sufficient.

Answer is D

Regards,
Uva.

Known is a drop Unknown is an Ocean

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greenwich: Master | Next Rank: 500 Posts; Posts: 110; Joined: Sun Jul 18, 2010 4:00 pmFollowed by: 2 members; Thanked: 1 times

by greenwich » Fri Jan 17, 2014 10:10 am

Why in Statement 1: |x+3|=4x-3
here LHS is Absolute value, which is non-negative?

I think I miss something by solving the problem as follows:

x+3=4x-3
3x=6
x=2

x+3=-4x+3
5x=0
x=0

What did I do wrong in the above problem?

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Jan 17, 2014 10:20 am

greenwich wrote:Why in Statement 1: |x+3|=4x-3
here LHS is Absolute value, which is non-negative?

I think I miss something by solving the problem as follows:

x+3=4x-3
3x=6
x=2

x+3=-4x+3
5x=0
x=0

What did I do wrong in the above problem?

Your approach is solid. However, whenever you deal with absolute value equations, you must check for EXTRANEOUS ROOTS (solutions)
When we plug x = 0 into the original equation, we get:
|0+3| = 4(0)-3
3 = -3 (no good)
So, the solution x = 0 is an extraneous root, which means we must disregard it.
This means that x = 2 is the ONLY solution to the original equation.

I hope that helps.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Jan 17, 2014 10:38 am

greenwich wrote:Is x>0?

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

When solving questions involving ABSOLUTE VALUE, there are 3 steps:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug in the solutions to check for extraneous roots

Now here's one approach:

Target question: Is x > 0?

Statement 1: |x + 3| = 4x - 3
When we apply the rule above, we get two equations
case a: x + 3 = 4x - 3
Rearrange: 6 = 3x
Solve: x = 2
Check for extraneous roots by plugging x = 2 into solution
|2 + 3| = 4(2) - 3
5 = 5
Great, we'll KEEP the solution x = 2

case b: x + 3 = -(4x - 3)
Expand: x + 3 = -4x + 3
Rearrange: 5x = 0
Solve: x = 0
Check for extraneous roots by plugging x = 0 into solution
|0 + 3| = 4(0) - 3
3 = -3
No good. We'll DISREGARD the solution x = 0

So, x MUST equal 2, which means x is definitely greater than 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: |x + 1| = 2x - 1
When we apply the rule above, we get two equations
case a: x + 1 = 2x - 1
Rearrange: 2 = x
Solve: x = 2
Check for extraneous roots by plugging x = 2 into solution
|2 + 1| = 2(2) - 1
3 = 3
Great, we'll KEEP the solution x = 2

case b: x + 1 = -(2x - 1)
Expand: x + 1 = -2x + 1
Rearrange: 3x = 0
Solve: x = 0
Check for extraneous roots by plugging x = 0 into solution
|0 + 1| = 2(0) - 1
1 = -1
No good. We'll DISREGARD the solution x = 0

So, x MUST equal 2, which means x is definitely greater than 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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j_shreyans: Legendary Member; Posts: 510; Joined: Thu Aug 07, 2014 2:24 amFollowed by: 5 members; Thanked: 3 times

by j_shreyans » Thu Jan 29, 2015 10:06 pm

Hi Brent ,

I do not understand that why should we disregard the x=0

if i go through the rule # 1 it says 1. Apply the rule that says: If |x| = k, then x = k and/or x = -k

if i put the x=0 in |x+1|= 2x-1 then it gives

1=-1(without changing the sign) and 1=1 ( with changing the sign)

which goes with above rule.

Please suggest and correct me if i am wrong.

Thanks

Shreyans

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Fri Jan 30, 2015 7:59 am

Brent is illustrating the idea that an absolute value expression cannot be negative. [Note: The variable within the absolute value brackets can be negative, but not the absolute value expression as a whole.]

Take a simple case
|x| = 1. What I'm saying here is that x is 1 unit away from 0 on the number line. Simple enough: x =1 or x = -1.

But what if I tried to tell you that |x| = -1. This equation says that x is -1 units away from 0 on the number line. Well, that doesn't make any sense! A distance can't be negative.

So when we substitute 0 for x in |x+1|= 2x-1, we get |1| = -1. What that equation says is that '1' is -1 units away from 0 on the number line. That makes no sense! We can't say |1| = -1, because it's not true or coherent. Therefore, we have to discard x = 0 is a possibility, because it leads to a logical absurdity.

Veritas Prep | GMAT Instructor

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by [email protected] » Fri Jan 30, 2015 10:47 am

Hi All,

This DS question has a great "concept" shortcut built into it; if you can spot it, then you can avoid doing ALL of the Algebra.

We're asked if X > 0. This is a YES/NO question.

Fact 1: |X+3| = 4X-3

There's a great Number Property rule about absolute values: the result of an absolute value calculation can NEVER be negative.

Here, we're given an equation with an absolute value on "the left" and normal algebra on "the right." Focus on the "right side" of the equation.....

If X = ANY NEGATIVE, then (4X - 3) will be NEGATIVE. Since |X+3| CAN'T be negative, there cannot be a negative answer to this equation.

In that same way, if X = 0, then (4X-3) will also be NEGATIVE. As before, |X+3| CAN'T be negative, so X cannot be 0 either. Once you eliminate the negatives and 0 as possible answers, all that is left are POSITIVE ANSWERS. Since the question asks if X > 0, we don't even need to find the answer(s) to this equation; we KNOW that it/they MUST be positive, so the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT.

Fact 2: |X+1| = 2X - 1

Here, we have the exact same situation as in Fact 2.

If X = ANY NEGATIVE, then (2X-1) will be NEGATIVE
If X = 0, then (2X-1) will be NEGATIVE

Neither of these options is possible when dealing with an absolute value, so X MUST be POSITIVE and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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