# Math

posted by .

( / means absolute value)

Solve: 3+/2y-1/>=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+2y-1>=1
2y-4>=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.

• Math -

|2y-1| = +(2y-1) if y >1/2
|2y-1| = -(2y-1) 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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