Solving inequality with absolute value

[Marc_1]

Hi all, I need your help to solve an inequality involving absolute values.

I would like to know if (i) what I did is correct or not and (ii) if not do you have any intuitions to simplify or solve this inequality ?

[MATH] (C_H - C_L) (\lvert \bar x - v + P_{L} \rvert - \bar x + P_{L}) > \gamma [/MATH]
Here is what I did so far :
[MATH] (C_H - C_L) (\lvert \bar x - v + P_{L} \rvert) +(C_H - C_L)(- \bar x + P_{L}) > \gamma \\ (\lvert \bar x - v + P_{L} \rvert) - (\bar x + P_{L}) > \frac{\gamma}{(C_H - C_L)} \\ (\lvert \bar x - v + P_{L} \rvert) > \frac{\gamma}{(C_H - C_L)} + \bar x - P_{L}[/MATH]
So we have to solve:

[MATH](1): ( \bar x - v + P_{L} ) > \frac{\gamma}{(C_H - C_L)} + \bar x - P_{L}[/MATH]
and

[MATH](2): ( \bar x - v + P_{L} ) > -\frac{ \gamma}{(C_H - C_L)} - \bar x + P_{L}[/MATH]
This would mean that :

[MATH](1): \frac{-v+2P_L}{C_H-C_L} > \gamma\\ (2): - \frac{2\bar{x}-v}{C_H-C_L} > \gamma[/MATH]
Is it correct ?
Thanks in advance, Marc

 

[Dr.Peterson]

You forgot to change > to < in inequality (2). An inequality of the form |u|>b is equivalent to "u>b or u<-b".

You also have not made it clear what you are solving for.

 

[Steven G]

If ab < c it does NOT follow that a<c/b ! This is what you did when you divided by sides by (C_H−C_L)

Consider this example (2)(-3) < 5. No problem here as -6<5.

Divide both sides by (-3) and get 2< -5/3 which is not true. After all no positive number is less than any negative number.

 

[Marc_1]

Thank you for your replies, it really helps me !
@Dr.Peterson, I have to solve this for several equations and find if any value of gamma is similar among them. But if I reach to understand how to do this for one, it will be fine for the others
@Jomo, Ok I see, you are true. I forgot to mention that [MATH](C_H - C_L)[/MATH] is necessarily positive since [MATH]C_H > C_L [/MATH]. It might work with this hypothesis, right ?

 

[Dr.Peterson]

Please try redoing your work, correcting the errors I pointed out. Then we can discuss whether you understand.

If you need examples or instructions, see here:

Solving Absolute-Value Inequalities -- Explained!

To solve absolute-value inequalities, first we must drop the absolute-value bars and restate the inequality in one of two ways, depending on the case.

www.purplemath.com

Absolute Value in Algebra

www.mathsisfun.com

 

[Marc_1]

Ok (I spotted an error of mine)

[MATH] (\lvert \bar x - v + P_{L} \rvert) > \frac{\gamma}{(C_H - C_L)} + \bar x - P_{L}[/MATH]
So we have to solve:

[MATH](1): ( \bar x - v + P_{L} ) > \frac{\gamma}{(C_H - C_L)} + \bar x - P_{L}[/MATH]
and

[MATH](2): ( \bar x - v + P_{L} ) < -\frac{ \gamma}{(C_H - C_L)} - \bar x + P_{L}[/MATH]
This would mean that :

[MATH](1): ( \bar x - v + P_{L} ) > \frac{\gamma}{(C_H - C_L)} + \bar x - P_{L}[/MATH]
[MATH](1): ( - v + P_{L} ) > \frac{\gamma}{(C_H - C_L)} - P_{L}[/MATH]
[MATH](1): ( - v + 2P_{L} ) > \frac{\gamma}{(C_H - C_L)} [/MATH]
[MATH](1): ( - v + 2P_{L} ) (C_H - C_L) > \gamma [/MATH]
[MATH](2): ( \bar x - v + P_{L} ) < -\frac{ \gamma}{(C_H - C_L)} - \bar x + P_{L}[/MATH]
[MATH](2): ( 2\bar x - v ) < -\frac{ \gamma}{(C_H - C_L)} [/MATH]
[MATH](2): ( 2\bar x - v ) (C_H - C_L) < - \gamma [/MATH]
[MATH](2): ( v - 2\bar x ) (C_L - C_H) > \gamma [/MATH]

 

[Dr.Peterson]

It's still unclear what your goal is. You start with an inequality, "something > gamma", and end up with two inequalities, each of which says "something > gamma". Clearly you aren't solving for gamma! All you've done is to split the inequality into two cases (without stating when each case applies). You said you want to "find if any value of gamma is similar among" several inequalities; but what does that mean?

As for your work, the two cases have to be joined not by "and", but by "or". That is, the original inequality will be true if either (1) or (2) is true. You just need to remain aware of this to the end.

Given that, and the assumption that [MATH]C_H>C_L[/MATH] (high and low, I suppose?), your work is all valid. However, at the end what you really have, given my comment about "or", is "gamma < this or gamma < that". That can be restated as "gamma < max(this, that)". Can you see why? If I say "x < 2, or x < 3", then as long as x < 3, it will be satisfied, because whenever x < 2, you know that also x < 3. So the result is really

[MATH]\gamma < \max[(- v + 2P_L) (C_H - C_L), (v - 2\bar x) (C_L - C_H)][/MATH]​

or, if you like,

[MATH]\gamma < \max[2P_L - v, 2\bar x - v] (C_H - C_L)[/MATH]​

But since you end up with gamma right where it started, you could have done it in a different way that makes it clear that you are really just rewriting the absolute value. This approach would start with the fact that |u| = {u, if u>=0; -u, if u<0}. So you'd rewrite [MATH]\lvert \bar x - v + P_{L} \rvert[/MATH] in terms of two cases, [MATH]\bar x \ge v - P_{L}[/MATH] and [MATH]\bar x \lt v - P_{L}[/MATH]. You'd get the same result.