# algebra 2

posted by
**anonymous** on
.

solve: 3+|2y-1|>1 graph the solution set on a number line.

Subtract 3 from each side to get the valid inequality

|2y-1|> -2

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.