what is the interval solution of |(2/3)x - 5| >=3??

Answer 1

Where are you having a problem? I'm not simply handing you the answer -- that's cheating.

You get two cases with absolute value stuff of this form:

|x| >= c
x>=0 and x >= c
x < 0 and -x >= c

You can take it from there.

Answer 2

Look at the limiting condition without absolute signs, that (2/3)x - 5 = 3 => x=12. For positive numbers, any value> 12 is OK. So a restricted zone starts below 12. At x=3, the |(2/3)x-5| is again equal to 3, and the inequality holds for anything less than 3. So the solution is restricted to all values EXCEPT 3=< x <=12

Answer 3

2 - 3x > -a million 2 + a million > 3x 3/3 > x a million > x which ability x is below a million x+ 5 > 0 x > -5 which ability x is larger than -5 because of the fact that x is larger than -5 it cant be equivalent to -5 for this reason answer is D

Answer 4

Absolute value of a difference is the distance between two values.

The distance between 2/3x and 5 must be more than or equal to 3.

Either 2/3x is 3 added to 5 (or greater)

or.. 2/3x is 3 subtracted from 5 (or less)

Knowing this.. write two inequalities and solve

2/3x >= 5 + 3

2/3x >= 8

x>= 12


2/3x <= 5 - 3

2/3x <= 2

x <= 3

Therefore, the values for x must either be less than or equal to 3 or greater than or equal to 12.

Using interval notation:(- ∞,3] U [12,∞)

Answer 5

7

Answer 6

(2/3)x-5 >=3 so 2/3x >= 8 or x>=12
Or, (2/3)x-5<=-3 so (2/3)x <= 2 or x<=3