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Inequalities help

  • Thread starter jvignacio
  • Start date
  • #1
7
0
hey guys just checking if this is correct....
--------------------------------------------------------------------------------

|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
 

Answers and Replies

  • #2
statdad
Homework Helper
1,495
35
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]
 
  • #3
7
0
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 ?
 
  • #4
statdad
Homework Helper
1,495
35
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?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,770
911
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 "<".
 
  • #6
7
0
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 ?
 
  • #7
7
0
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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