posted by anonymous .
solve: 3+|2y-1|>1 graph the solution set on a number line.
Subtract 3 from each side to get the valid inequality
If 2y>1, then the term inside | | is positive and
2y -1 > -2
2y > -1
Since you already required 2y>1, the allowed value of y are > 1/2 in this case.
IF the term inside || is negative, then
2y < 1 and
1 - 2y > -2
-2y > -3
2y < 3
y < 3/2 are also allowed solutions
The solution is all real numbers. No matter what value of y you select, the inequality is valid. That should have been immediately obvious, since the term in || cannot be less tnan zero.