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If |X-2| = |2X-3| , what are the possible values of x?

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by missrochelle » Sun Aug 22, 2010 8:16 am
If |X-2| = |2X-3| , what are the possible values of x?

Because there are two absolute value expressions, each of which yields two algebraic cases, it seems that we need to test four cases overall.

1. Positive/Positive: (x-2) = (2x-3)
2. Positive/Negative: (x-2) = -(2x-3)
3. Negative/Positive: -(x-2) = (2x-3)
4. Negative/Negative -(x-2) = -(2x-3)

The book states that case 1 and 4 yield the same equation. And 2 and 3 yield the same equation....therefore we only need to test the positive/positive case, and a case where the sign of one equation changes. (1 and 3 for example)

Also, the two cases tested are if 1) neither expression changes signs or 2) one expression changes signs. I am just trying to figure out the step by step approach for the "complex" absolute value equations. Are you supposed to test all four cases first, then see which are different equations?

The example in the problem set (#6) states that we test same same, and both have opposite sign.

In terms of a general strategy: what are we looking for when we decide to take this approach? (is it simply, an absolute value equation with one variable and several constraints?). And lastly, is the only way to figure out which cases to test -- to write out all four and see which are distinct?
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Source: — Problem Solving |

by DarkKnight » Sun Aug 22, 2010 12:26 pm
I am getting, x=1 or x=5/3.
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by missrochelle » Sun Aug 22, 2010 2:58 pm
Thats correct., what methodology did you use?
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by Rahul@gurome » Sun Aug 22, 2010 6:37 pm
There is an easier way.
Square both sides.
We have (x - 2)^2 = (2x - 3)^2
Or x^2 + 4 - 4x = 4x^2 + 9 - 12x
Or 3x^2 - 8x + 5 = 0.
Or (3x - 5)(x-1).
Or x = 5/3 or 1.
These values can be replaced and verified.
Rahul Lakhani
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