fresinha12 wrote:

I got statement 2 to be insuff by just eye balling the equation and realizing that 2 and 0 were solutions..however I am having a hard time putting statemnt 2 in a equation and finding the critical points...can someone pls expand and show how to get the solutions to the following equation

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

1. we have to find key points, in which modules are zero:

1.1. x-3=0 --> x=3

1.2. 2x-3=0 --> x=1.5

2. Now, we carefully eliminate modulas:

2.1. for x e (-∞,1.5)&(3,+∞): x-3=2x-3 --> x=0 (x is in the range)

2.2. for x e [1.5,3]: x-3=-2x+3 --> 3x=6 --> x=2 (x is in the range)

So, we have two solutions: 0 and 2

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