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

  • Thread starter Plutonium88
  • Start date
In summary, the conversation discusses solving for the intervals of a complex equation involving absolute values. The intervals for x were determined to be x< -5/2, x> 19/4, and x>3/2. The solution is represented by the interval notation XE(-∞, -5/2) U (19/4, +∞). The conversation also addresses the importance of considering all the restrictions in order to find the final answer.
  • #1
Plutonium88
174
0
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 want to 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?
 
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  • #2
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.
 
  • #3
Plutonium88 said:
So I broke it into a number line and calculated when

|3x-7|
(3x-7) x ≥ 7/3
-(3x-7) x < 7/3 (strict)

|x-8|
-(x-8) x<8
(x-8) x ≥ 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 cause it lies within the interval 7/3<x<8

Now for x>8

(3x-7)-(x-8)
X<3/2
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?
Now I'm not sure how to address this but I want to 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?
 
  • #4
BruceW said:
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.

Ahh my bad, i believe 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 don't 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?
 
  • #5
Plutonium88 said:
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 don't 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?

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?

Plutonium88 said:
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?
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.
 
  • #6
BruceW said:
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.


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 that's 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:)
 
  • #7
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!
 

1. What does the equation |3x-7|-|x-8| > 4 mean?

The equation |3x-7|-|x-8| > 4 represents a mathematical inequality, where the absolute value of 3x-7 minus the absolute value of x-8 is greater than 4.

2. How do I solve this equation algebraically?

To solve this equation algebraically, we need to isolate the variable on one side of the inequality symbol. We can do this by using inverse operations and following the order of operations.

3. What are the steps to solve this equation?

The steps to solve this equation are as follows:

  1. Use the distributive property to remove the absolute value bars.
  2. Combine like terms on each side of the inequality.
  3. Add or subtract the same number to both sides of the inequality to isolate the variable term.
  4. Divide both sides of the inequality by the coefficient of the variable to solve for x.
  5. Check your solution by substituting it back into the original inequality.

4. Can I use the graphing method to solve this equation?

No, the graphing method is not recommended for solving inequalities as it does not accurately represent the solutions. It is best to solve this equation algebraically.

5. What are the possible solutions to this equation?

The possible solutions to this equation depend on the value of x. It is important to check the solution by substituting it back into the original inequality to ensure it satisfies the inequality. The final solution will be in the form of an interval, such as (-∞, 3) or (5, ∞).

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