How do you solve #| 2x + 1 | <| 3x - 2 | #?
First find the zeros of each expression in the modules:
Now we must divide the inequation in three parts:
1) When x>=2/3, both expression are larger than 0, therefore:
is equivalent to:
3x-2 will be <=0, therefore
both expressions are negative so we must get the inverse for both ones:
Now the solution is the union of the three solutions: