Solving Inequalities: x Range & Answers

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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?