SOLUTION: !x!_> 3

Algebra.Com
Question 45227: !x!_> 3
Answer by rapaljer(4671)   (Show Source): You can put this solution on YOUR website!


Do you understand that absolute value of a positive number is positive, the absolute value of 0 is zero, and the absolute value of a negative number is actually the positive number associated with it. Like absolute value of 3 is 3, absolute value of -3 is 3, absolute value of -5 is 5, etc. Be sure you understand this first.

Now, draw a numberline, and by trial and error try to find the numbers on the numberline where the absolute value of the number will result in a number that is greater than or equal to 3.

Trial and error results in all numbers that are from 3 on the numberline, and to the right on the numberline, like 4, 5, 6, etc.

Also, trial and error results in the number -3, and all numbers to the left of -3, like -4, -5, -6, etc.

This means that . In interval notation, this would be
(-inf, -3] U [3, inf).

R^2 at SCC
In interval notation

RELATED QUESTIONS

3+x=x+3 (answered by rfer)
x^3-x=3 (answered by greenestamps)
x/3+x/3+x (answered by jim_thompson5910)
x(x-3)=x (answered by Alan3354)
g(x)=... (answered by Student3354)
3^x*3^3+3^x=84 (answered by Fombitz)
(x-3)^3 -... (answered by jsmallt9)
x>3 (answered by Mathtut)
{{{ 4/x+3/(x-3)=x/(x-3)-1/3... (answered by nyc_function)