Solve |3x-7|-|x-8| > 4 (solve algebraically)

Plutonium88

So I broke it into a number line and calculated when

|3x-7|
(3x-7) x>orequAl 7/3
-(3x-7) x < 7/3 (strict)

|x-8|
-(x-8) x<8
(x-8) x>orequal 8


So for x<7/3

-(3x-7) - [-(x-8)] > 4

Simplified to x < -5/2

For the interval 7/3<x<8
(3x-7)-[-(x-8)]
x>19/4

This is part of solution be ause it lies within the interval 7/3<x<8

Now for x>8

(3x-7)-(x-8)
X<3/2

Now I'm not sure how to adress this but I wanna say this is not part of the solution cause it doesn't lie within the interval x>8



And also in terms of writing my Ana after in interval notation how can I do this with the info attained. Do I just plot my newly found restrictions on the number line?

BruceW

you got the first 2 intervals correct, but the last one (for x>8), you didn't quite get right. Once you've got the right answer for the 3rd part, then yes you can draw it out, or just think about what each of the restrictions means, so then you can say what is the final result of all the restrictions.

SammyS

This is not correct. You're still solving|3x-7|-|x-8| > 4, is that right?

So in the region, x≥8, you have (3x-7)-(x-8) > 4.

What does that lead to?

Plutonium88

Ahh my bad, i beleive i wrote the symbol wrong

x > 3/2

But this still isn't part of the interval x≥8?

I just don't know how to consider it

cause 1.5 does not lie in the interval x≥8 (is this correct)**?
(I just dont know how to consider these intervals, with the restrictions attained) - Like for ex: does the restriction have to lie within that interval for it to be part of solution?

So if i look at what i have

X< -5/2

x> 19/4

XE(-∞, -5/2) U (19/4, +∞)


I'm just curious originally with my intervals that i'm taking like x≤7/3, 7/3≤x≤8, x≥8

why don't i consider these on the number line?

BruceW

Think about what it means, you tell it that x>8, then it tells you x>3/2. There is no problem here. Just think, does x>3/2 change the restriction of x>8?

I think you've got the right answer. What does XE mean? And I'm not sure what you mean about considering the original intervals on the number line.. You could look at each of the intervals and write down the restrictions from each, but then your final answer takes them all into account, so its nicer to look at.

Plutonium88

http://s15.postimage.org/g98nz5i2j/line_bmp.png

XE =, or X Belongs To The Interval, i just don't have the special E/ Couldnt find it in my charmap.


So from my number line the intervals that i have, are the same as the answer i stated, above.

And what you`re saying is, that in terms of the restrictions i`ve solved, i only have to take those 3 into account, in order to find the answer (19/4, -5/2 3/2)

and if thats the right answer, it would seem that this is the case


And also, thank you very much for your help. I Really appreciate what you do, learning would be much more difficult without the help of this forum. d:)

BruceW

Oh I get it, like:
[tex]X \in ( - \infty , - \frac{5}{2} ) \cup ( \frac{19}{4} , \infty ) [/tex]
The magic of latex! And yeah, that looks like the right answer to me, since you used the information from each of the 3 important regions to get this answer. I'm glad I've been some help, you had done most of the question in your first post!