Math

( / 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.

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asked by Jon
  1. |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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    posted by Damon

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