How do you solve and write the following in interval notation: #|2x-1|+7>=1#?

1 Answer
Jul 14, 2017

Answer:

The solution set for #|2x-1|+7>=1# is #{x in RR}#

Explanation:

Subtract 7 from both sides

#|2x-1| >= -6#

The left hand side is always considered as positive. So it can not be equal to #-6# Thus #|2x-1|+7>=1# is wrong. It should read #|2x-1|+7>1#

The form of #|2x-1|+7>1 # means that all real values of x will make the inequality be true.

The solution set for #|2x-1|+7>=1# is #{x in RR}#

To illustrate this, I shall substitute #1/2# for x:

#|2(1/2)-1|+7 >=1#

#|0|+ 7>=1#

#7 >=1 #

This inequality is true and I have chosen the value for x where any other value will only make the number on the left side become greater. Therefore, the solution set is all real values of x.