show your work and solve: 3+[2y-1]>or equal to 1. []= absloute value signs?

Answer 1

With absolute values, you need to actually solve 2 separate inequalities since negative AND positive values will work.

Here, however, you will notice that no matter what the absolute value part is, when you add 3 to it it will always be greater than 1.

So all numbers work.

Answer 2

First, subract 3 from both sides, giving you [2y-1]>=-2.
Now, this tells you y = all real numbers. Why?
Because the absolute value of any number is > or = to zero, or all positive. And since any number like that is >= -2, then y is all real.

Answer 3

3 + [2y -1] >= 1
[2y - 1] >= -2

At this point one might think:

2 >= (2y - 1) >= -2
3 >= 2y >= -1
3/2 >= y >= -1/2

However, an absolute value is always positive, so [2y -1] >= -2 is true for all real numbers.

Answer 4

3+|2y-1|>=1

two problems
3+2y-1>=1
3+(-2y-1)>=1

3+2y-1>=1
2+2y>=1
2y>=-1
y>=-1/2

3+(-2y-1)>=1
3-2y+1>=1
4-2y>=1
-2y>=-3
y<=3

y>=-1/2 or y <=3

Every single number is either greater than -1/2 or less than 3

So the Universal Set is the answer

Answer 5

Trick question.
3 plus anything (that's positiove) is greater than 1, so all real numbers at least.

Are you working with imaginary numbers (i)? If so, that might have a solution.

Answer 6

"| |" are abs. value signs. "[ ]" are brackets.
Assuming that they're brackets:
3 + [2y - 1] >= 1
2y - 1 >= -2
2y >= -1
y >= -.5