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

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Manager
Joined: 11 Nov 2007
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x 3| = |2x 3| When you have an equation like this and you [#permalink]

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11 Jan 2008, 00:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

|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?

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Director
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11 Jan 2008, 05:11
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.

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Director
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11 Jan 2008, 10:22
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

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Director
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11 Jan 2008, 12:20
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.

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VP
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11 Jan 2008, 15:58
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

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Director
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11 Jan 2008, 22:41
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?

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Director
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12 Jan 2008, 02:54
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.

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VP
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15 Jan 2008, 06:34
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

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Re: absolute value   [#permalink] 15 Jan 2008, 06:34
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