Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 05 May 2015, 07:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# x 3| = |2x 3| When you have an equation like this and you

Author Message
TAGS:
Manager
Joined: 11 Nov 2007
Posts: 61
Followers: 1

Kudos [?]: 5 [0], given: 0

x 3| = |2x 3| When you have an equation like this and you [#permalink]  10 Jan 2008, 23:38
|x – 3| = |2x – 3|

When you have an equation like this and you are considering two scenarios +/-. How do you do the negative?

Do you make both sides negative....
-x+3 = -2X+3 ?

Or just one side?
-x+3 = 2x-3?
Director
Joined: 09 Jul 2005
Posts: 595
Followers: 2

Kudos [?]: 26 [0], given: 0

Re: absolute value [#permalink]  11 Jan 2008, 04:11
I would draw both lines and then calculate the values of x for which the lines cross each other. In this case I obtain x=0 and x=2.
Director
Joined: 14 Oct 2007
Posts: 759
Location: Oxford
Schools: Oxford'10
Followers: 13

Kudos [?]: 183 [0], given: 8

Re: absolute value [#permalink]  11 Jan 2008, 09:22
aliensoybean wrote:
|x – 3| = |2x – 3|

When you have an equation like this and you are considering two scenarios +/-. How do you do the negative?

Do you make both sides negative....
-x+3 = -2X+3 ?

Or just one side?
-x+3 = 2x-3?

this is what you do

|x - 3| = |2x - 3|

remove the absolute sign and solve

x-3 = 2x-3 --> x = 0

now minus just one side (as minusing both doesn't make sense since you are doing the same thing to each side. Also which side you minus doesn't matter either)

3 - x = 2x - 3 --> x=2

just to prove the above, minus the other side
x - 3 = 3 - 2x --> x=2
Director
Joined: 09 Jul 2005
Posts: 595
Followers: 2

Kudos [?]: 26 [0], given: 0

Re: absolute value [#permalink]  11 Jan 2008, 11:20
aliensoybean wrote:
|x – 3| = |2x – 3|

When you have an equation like this and you are considering two scenarios +/-. How do you do the negative?

Do you make both sides negative....
-x+3 = -2X+3 ?

Or just one side?
-x+3 = 2x-3?

this is what you do

|x - 3| = |2x - 3|

remove the absolute sign and solve

x-3 = 2x-3 --> x = 0

now minus just one side (as minusing both doesn't make sense since you are doing the same thing to each side. Also which side you minus doesn't matter either)

3 - x = 2x - 3 --> x=2

just to prove the above, minus the other side
x - 3 = 3 - 2x --> x=2

First of all both equations are in fact the same. That is why you obtain the same value for x. Second, in my opinion, the suggested method does not work for all the cases for which both expressions change of sign for different values of x.
VP
Joined: 09 Jul 2007
Posts: 1108
Location: London
Followers: 6

Kudos [?]: 74 [0], given: 0

Re: absolute value [#permalink]  11 Jan 2008, 14:58
automan wrote:
I would draw both lines and then calculate the values of x for which the lines cross each other. In this case I obtain x=0 and x=2.

best method is this i guess. if you draw it is clear
Director
Joined: 08 Jun 2007
Posts: 584
Followers: 2

Kudos [?]: 79 [0], given: 0

Re: absolute value [#permalink]  11 Jan 2008, 21:41
automan wrote:
I would draw both lines and then calculate the values of x for which the lines cross each other. In this case I obtain x=0 and x=2.

how do you draw . can you elaborate?
Director
Joined: 09 Jul 2005
Posts: 595
Followers: 2

Kudos [?]: 26 [0], given: 0

Re: absolute value [#permalink]  12 Jan 2008, 01:54
ashkrs wrote:
automan wrote:
I would draw both lines and then calculate the values of x for which the lines cross each other. In this case I obtain x=0 and x=2.

how do you draw . can you elaborate?

Let's draw |x – 3|.

If (x-3)>0 then |x – 3|=x-3. Therefore, for x>3 you will have f(x)=x-3
If (x-3)<0 then |x – 3|=-x+3. Therefore, for x<3 you will have f(x)=3-x

Both (x-3) and (3-x) can be drawn very quickly.
VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

Re: absolute value [#permalink]  15 Jan 2008, 05:34
automan wrote:
ashkrs wrote:
automan wrote:
I would draw both lines and then calculate the values of x for which the lines cross each other. In this case I obtain x=0 and x=2.

how do you draw . can you elaborate?

Let's draw |x – 3|.

If (x-3)>0 then |x – 3|=x-3. Therefore, for x>3 you will have f(x)=x-3
If (x-3)<0 then |x – 3|=-x+3. Therefore, for x<3 you will have f(x)=3-x

Both (x-3) and (3-x) can be drawn very quickly.

I would make the square of both sides and find that x could be equal both to 0 or to 2
Re: absolute value   [#permalink] 15 Jan 2008, 05:34
Similar topics Replies Last post
Similar
Topics:
1 Do you like 3D tattoo? 5 13 Dec 2013, 01:31
Is SQRT = 3-x? 1) x is not 3 2) -x*lxl > 0 Would you 5 04 Jun 2008, 16:47
The OG has the following equation.... -3y <= -2x - 6 It 4 08 May 2008, 13:13
Suppose you like 1% milk, but you only have 3% whole milk 7 23 Dec 2007, 16:32
How many roots does this equation have? 2^x = x - 1 0 1 2 3 11 05 Nov 2007, 09:53
Display posts from previous: Sort by