# Absolute Values

• Aug 19th 2009, 03:01 PM
dwat
Absolute Values
Can anyone please hammer out this idea of absolute value for me. Please.

$\displaystyle y = |2x - 3|$

using

$\displaystyle |x| = \begin{cases}x \ldots & x\ge 0 \\\\ -x \ldots & x \le 0\end{cases}$

I know its relatively simple but I just cant seem to get the ideas into my head. I know what it looks like graphically using a graphing utility but once I come to actually solve it by hand, I have no idea what to do

• Aug 19th 2009, 03:10 PM
QM deFuturo
To start, first just substitute 2x - 3 for each occurance of "x" in the definition. Then solve for x. I.e, |2x-3| = 2x-3 when 2x-3 > 0, |2x-3| = -(2x - 3) when 2x - 3 < 0.
• Aug 19th 2009, 03:23 PM
Plato
I love this question. It is at the very base of a great deal of mathematics.

The absolute value of x, $\displaystyle \left| x \right|$ is the distance $\displaystyle x$ is from $\displaystyle 0$.

So we can see $\displaystyle -3$ is three units from zero so $\displaystyle \left| -3 \right|=3$.

So look at this question: Solve $\displaystyle \left| {x + 6} \right| = 2$.
O.K. What number is two units from zero? Well of course that answer is $\displaystyle \pm 2$.
So solve this system $\displaystyle x+6=2\text{ or }x+6=-2$.
• Aug 19th 2009, 03:27 PM
dwat
Ok is this right?

$\displaystyle 2x - 3 \ge 0$

$\displaystyle x \ge \frac{3}{2}$

and

$\displaystyle -2x - 3 \le 0$

$\displaystyle x \ge \frac{-3}{2}$

I've done something wrong haven't I, because that is not the same output my graphing utility is showing me...
• Aug 19th 2009, 03:38 PM
dwat
Quote:

Originally Posted by Plato
I love this question. It is at the very base of a great deal of mathematics.

The absolute value of x, $\displaystyle \left| x \right|$ is the distance $\displaystyle x$ is from $\displaystyle 0$.

So we can see $\displaystyle -3$ is three units from zero so $\displaystyle \left| -3 \right|=3$.

So look at this question: Solve $\displaystyle \left| {x + 6} \right| = 2$.
O.K. What number is two units from zero? Well of course that answer is $\displaystyle \pm 2$.
So solve this system $\displaystyle x+6=2\text{ or }x+6=-2$.

Oh my, that's an explanation of a true teacher hehe, okay so looking at the example you have given me,

does that mean my equation should be

$\displaystyle 2x = 3$

and

$\displaystyle 2x = -3$

so$\displaystyle x = \pm \Big(\frac{3}{2}\Big)$

so how do I interpret that as:

$\displaystyle x \le \frac{3}{2}$

$\displaystyle x\ge \frac{3}{2}$
• Aug 19th 2009, 03:41 PM
QM deFuturo
Quote:

Originally Posted by dwat
Ok is this right?

$\displaystyle 2x - 3 \ge 0$

$\displaystyle x \ge \frac{3}{2}$

and

$\displaystyle -2x - 3 \le 0$

$\displaystyle x \ge \frac{-3}{2}$

I've done something wrong haven't I, because that is not the same output my graphing utility is showing me...

Well, -(2x - 3) is not -2x - 3. You forgot to change the sign of -3.

Also, getting back to your original question, there is nothing to "solve" in this equation. Is this just a question about "how do I graph this equation"?
• Aug 19th 2009, 03:45 PM
QM deFuturo
Quote:

Originally Posted by dwat

so how do I interpret that as:

$\displaystyle x \le \frac{3}{2}$

$\displaystyle x\ge \frac{3}{2}$

2x - 3 will be positive when x > 3/2
2x - 3 will be negative when x < 3/2
• Aug 19th 2009, 04:02 PM
dwat
Quote:

Originally Posted by QM deFuturo
Well, -(2x - 3) is not -2x - 3. You forgot to change the sign of -3.

Also, getting back to your original question, there is nothing to "solve" in this equation. Is this just a question about "how do I graph this equation"?

ohh thanks for correcting my errors.
• Aug 19th 2009, 04:04 PM
dwat
I think I've got the idea of this now....

thank you all for your help

All deserve plus rep from me