Solving Inequalities: x Range & Answers

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Homework Help Overview

The discussion revolves around solving inequalities involving absolute values, specifically the expressions |x-2| ≥ 4 and 1 ≤ |x + 2| ≤ 4. Participants are exploring the correct interpretation and solution methods for these types of inequalities.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the correct application of absolute value properties in inequalities, questioning the validity of initial interpretations and solutions. There is a focus on understanding how to derive ranges from absolute value expressions and the implications of different inequality signs.

Discussion Status

Some participants have provided guidance on the general approach to solving absolute value inequalities, emphasizing the importance of identifying critical points and testing intervals. There is ongoing exploration of whether the original poster's interpretations are correct, with some suggestions for re-evaluating their solutions.

Contextual Notes

Participants are navigating the complexities of inequalities with multiple conditions and are encouraged to clarify their reasoning and assumptions. There is mention of potential confusion regarding the use of "less than or equal to" in the context of absolute values.

jvignacio
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hey guys just checking if this is correct...
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|x-2|(greater than and equal to) 4


= -2 (greater than and equal to) x (greater than and equal to) 6

is this correct? thanks u

and

1 (less than or equal to) | x + 2 | (less than or equal to) 4


= -1 (less than or equal to) x (less than or equal to) 2

correct? thank u
 
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If you have
[tex] |x| \le A[/tex]

then

[tex] -A \le x \le A[/tex]

but if you have

[tex] |x| \ge A[/tex]

then

[tex] x \le -A \text{ or } x \ge A[/tex]
 
statdad said:
If you have
[tex] |x| \le A[/tex]

then

[tex] -A \le x \le A[/tex]

but if you have

[tex] |x| \ge A[/tex]

then

[tex] x \le -A \text{ or } x \ge A[/tex]

so my less than or equal to is correct but other is wrong ?
 
yes - think about the number line. The absolute value of a number shows how far from [tex]0[/tex] a number is. If you have (just to make up some numbers)

[tex] |x| \le 9[/tex]

the number [tex]x[/tex] is at most a distance of nine from zero. Looking at the number line, that means that it must be true that

[tex] -9 \le x \le 9 [/tex]

However, if

[tex] |x| \ge 4 [/tex]

then [tex]x[/tex] is at least four units from zero. Again, looking at the number line, this means that

[tex] \text{Either} x \le -4 \text{ or } x \ge 4[/tex]

By the way, if your inequalities are either [tex]<[/tex] or [tex]>[/tex], the same type
of steps are used.

Does this help?
 
jvignacio said:
so my less than or equal to is correct but other is wrong ?
You originally said "= -2 (less than or equal to) x (less than or equal to) 6" which has two "less than or equal to"s. Both are wrong.

The best way to solve a complicated inequality is to solve the equation first. To solve [itex]|x- 2|\ge 4[/itex], first solve |x- 2|= 4 which reduces to x- 2= 4 or x- 2= -4 and has solutions x= 6 and x= -2. The point is that, since |x-2| is a continuous function, it can only change from "< 4" to "> 4" and vice-versa where it is equal to 4. The two points, x= -2 and x= 6, divide the real number line into 3 intervals and |x-2| must be either greater than or less than 4 throughout each interval. Checking a single value in each of x< -2, -2< x< 6, and x> 6 will tell you which is ">" and which is "<".
 
HallsofIvy said:
You originally said "= -2 (less than or equal to) x (less than or equal to) 6" which has two "less than or equal to"s. Both are wrong.

The best way to solve a complicated inequality is to solve the equation first. To solve [itex]|x- 2|\ge 4[/itex], first solve |x- 2|= 4 which reduces to x- 2= 4 or x- 2= -4 and has solutions x= 6 and x= -2. The point is that, since |x-2| is a continuous function, it can only change from "< 4" to "> 4" and vice-versa where it is equal to 4. The two points, x= -2 and x= 6, divide the real number line into 3 intervals and |x-2| must be either greater than or less than 4 throughout each interval. Checking a single value in each of x< -2, -2< x< 6, and x> 6 will tell you which is ">" and which is "<".

ahh so the solution is all 3 intervals?

should i do it the same way to solve

1 (less than or equal to) | x + 2 | (less than or equal to) 4 ?
 
i got the answer for |x-2|(greater than or equal to) 4

= x (less than or equal to) 2 or x (greater than or equal to) 6?
 

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