Absolute Value EQ w/ Fraction

It doesn't seem especially tough but there's not an example in the book and it was on a quiz today. Please note that I did consult the Inequality sticky at the top of the forum but I didn't see one quite like this. I apologize in advance if I missed it.

$\displaystyle 3 \mid \frac{\2x-1}{3}\mid -1 \geq2$

My first thought was to multiply everything by the denominator 3 so that:

$\displaystyle 3\mid \2x-1\mid -3 \geq6$ Then: add 3 to 6 so that:

$\displaystyle 3\mid \2x-1\mid \geq9$ Then: divide by 3 so that:

$\displaystyle 2x-1\geq3$ Then: add 1 to both sides so that:

$\displaystyle 2x\geq4$ So that...

x is equal to +2 and x is equal to -2

It seemed tidy enough at the time but I have a feeling it's wrong; that maybe I should multiplied 2x-1 by 3 at some point or that I'm remiss in some way or another. Any and all help is appreciated. Thanks.

Re: Absolute Value EQ w/ Fraction

Quote:

Originally Posted by

**Trypanosoma** $\displaystyle 3 \mid \frac{\2x-1}{3}\mid -1 \geq2$

Simply note that

$\displaystyle 3\left|\dfrac{2x-1}{3}\right|=\left|{2x-1}\right|$

Re: Absolute Value EQ w/ Fraction

Quote:

Originally Posted by

**Plato** Simply note that

$\displaystyle 3\left|\dfrac{2x-1}{3}\right|=\left|{2x-1}\right|$

A Picard facepalm moment for me... I've also noticed that not multiplying -1 or 2 that my answer still comes out to same. Will that stand up for precisely similar problems?

But what of something like

A. $\displaystyle 3\mid \frac{\2x-1}{**4**}\mid -1\geq 2$

would I then have to go with the LCD of 12 and then have...

$\displaystyle 9\mid2x-1\mid -3 \geq6$

Again, thank you so much.

Or does that 3 just get left alone, so that:

$\displaystyle 3\mid2x-1\mid -3 \geq6$

And (sorry), would I then divide the right side by the 9 or 3 (depending on which is correct--if correct at all).

Re: Absolute Value EQ w/ Fraction

Quote:

Originally Posted by

**Trypanosoma** A Picard facepalm moment for me... I've also noticed that not multiplying -1 or 2 that my answer still comes out to same. Will that stand up for precisely similar problems?

But what of something like.

I in no way suggested that your solution is wrong.

I just wanted to point that it was awkward.

You should know that $\displaystyle |ab|=|a||b|$.

So your problem quickly simplifies to $\displaystyle |2x-1|\ge 3$.

Thus $\displaystyle 2x-1\ge 3\text{ or }2x-1\le -3~.$