How Do You Solve the Inequality |x-8|<|2x+1|?

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Homework Help Overview

The discussion revolves around solving the inequality |x-8|<|2x+1|, which involves absolute values and requires understanding of their properties. Participants express confusion and seek clarity on the problem-solving process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest sketching graphs of the expressions to gain insights into the inequality. Others express frustration over their attempts and seek clearer explanations of the reasoning involved. There is a focus on identifying critical points and intervals where the expressions change signs.

Discussion Status

The discussion includes various attempts to analyze the inequality, with some participants providing detailed breakdowns of their reasoning. However, there is no explicit consensus on the approach, and multiple interpretations are being explored. Guidance has been offered regarding the use of graphs, but clarity on the original poster's attempts remains lacking.

Contextual Notes

Participants mention the need to show their work as part of the forum's homework help guidelines, indicating that the original poster's previous explanations may not have met this requirement.

Brady
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A pretty simple problem, but I'm confused nonetheless.

|x-8|<|2x+1|

Help would be greatly appreciated.
 
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In addition, write |-5x^5y^2| two ways.
 
You can start by sketching a graph of each side and comparing them. That should give you some major insights!
 
I've pretty much spent hours working on this problem; for some reason, it's still not getting to me. False contradiction and confusion keep popping up...argh
 
please help me
 
Did you make a sketch?
 
..? Didn't I just explain that I tried this problem for a terriby long time, and I would like a clear explanation?
 
Don't we deserve to see your approach?

Brady said:
..? Didn't I just explain that I tried this problem for a terriby long time, and I would like a clear explanation?

Brady said:
A pretty simple problem, but I'm confused nonetheless.

|x-8|<|2x+1|

Help would be greatly appreciated.

|x-8| - |2x+1| < 0

This solution is based on the fact that if a continuous function changes its sign, it might happen only in this function zeroes.

1.
Mark zeroes for each |expression| on the number line: x=8 and x=-1/2.
This two points break the whole number line into three intervals:
I. {-inf, -1/2}: -inf < x < 8
II. [-1/2, 8}: -1/2 <= x < 8
III. [8, +inf}: 8<= x < +inf.

Each expression "keeps" its sign unchanged on each interval.
Draw the number line and put the expressions' signs on the intervals
x-8: ....-...|...-...|...+...
2x+1: ...-...|...+...|...+...

Solve inequality on each region separately following the definition
|a| = -a, if a<0
|a| = a, if a>=0

I.
x-8<0 and 2x+1<0 therefore
-(x-8) - (-(2x+1)) < 0
-x + 8 + 2x + 1 < 0
x < -9 - agrees with -inf < x < 8 condition.

II.
x-8<0 and 2x+1>=0 therefore
-(x-8) - (2x+1) < 0
-x + 8 - 2x - 1 < 0
3x > 7
x > 7/3 - considering -1/2 <= x < 8 restriction
the answer on this interval
7/3 < x < 8

III.
x-8>=0 and 2x+1>=0 therefore
x - 8 - (2x+1) < 0
x - 8 - 2x - 1 < 0
x > - 9 - considering x>=8 restriction
the answer on this interval
x >= 8.

Combininig all the answers,
x < -9 or x > 7/3.

Faster way - just like Tide recommended.
Draw two graphs:
f(x) = |x-8| and g(x) = |2x+1|
(do you need help with plotting them?).

|x-8|<|2x+1| is equvalent to
graph f(x) is below graph g(x).
These two graphs intersect at x=-9 and x=7/3.
Graph f(x) is below graph g(x) left of -9 and right of 7/3.
The same answer.
 
Tide said:
You can start by sketching a graph of each side and comparing them. That should give you some major insights!

Brady said:
please help me
Tide just did!

Tide said:
Did you make a sketch?

Brady said:
..? Didn't I just explain that I tried this problem for a terriby long time, and I would like a clear explanation?

Yes, you did. You did not, however, say whether or not one of the things you tried was graphing the two functions as Tide suggested- that was his question.
In fact, although you say you tried, you have shown us NOTHING of what you tried.
That is, after all, a requirement for homework help on this forum.
 

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