Math
posted by Jon .
( / means absolute value)
Solve: 3+/2y1/>=1. Graph the solution set on a number line.
The number line I have is:
5 4 3 2 1 0 1 2 3 4 5 6
First solution:
3+2y1>=1
2y4>=1
2y>=5
y>=5/2
I'm stuck b/c I had to correct the first part. I had a whole different answer. The graph for this would be an open circle between 2 and 3 and the arrow going right.

2y1 = +(2y1) if y >1/2
2y1 = (2y1) if y <1/2
so do the whole problem first for y >1/2
3 + 2y  1 >/= 1
2 + 2y >/= 1
1+y >/= 1/2
y >/= 1/2
so this is true if y >1/2
now do the whole problem for y<1/2
3  2y +1  1 >/= 1
3  2y >/= 1
2y >/= 2
y </= 1
so it is true for any y greater than 1/2 and for any y less than 1 and so the entire real number line
Try some spots
y = 0 yes
y = 1 yes
y = +1 yes
y = 1/2 yes
y = 1/2 yes etc etc etc
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