Solving Inequalities: x Range & Answers

[jvignacio]

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

 

[statdad]

If you have
<br /> |x| \le A<br />

then

<br /> -A \le x \le A<br />

but if you have

<br /> |x| \ge A<br />

then

<br /> x \le -A \text{ or } x \ge A<br />

 

[jvignacio]

If you have
<br /> |x| \le A<br />

then

<br /> -A \le x \le A<br />

but if you have

<br /> |x| \ge A<br />

then

<br /> x \le -A \text{ or } x \ge A<br />

so my less than or equal to is correct but other is wrong ?

 

[statdad]

yes - think about the number line. The absolute value of a number shows how far from 0 a number is. If you have (just to make up some numbers)

<br /> |x| \le 9<br />

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

<br /> -9 \le x \le 9 <br />

However, if

<br /> |x| \ge 4 <br />

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

<br /> \text{Either} x \le -4 \text{ or } x \ge 4<br />

By the way, if your inequalities are either &lt; or &gt;, the same type
of steps are used.

Does this help?

 

[HallsofIvy]

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 |x- 2|\ge 4, 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 "<".

 

[jvignacio]

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 |x- 2|\ge 4, 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 ?

 

[jvignacio]

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?