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#1**0 **

Let's do this by process of elimination......

Notice that the graph indicates that 0 is in the solution set.... but...if I put 0 into A, 6/4 >= 4 is not true....so that one is no good

And the graph indicates that a "large" negative value will satisy some inequality.......but.....putting a "large" negative into D will cause it to become false

Look at C.....the graph indicates that -4 can't be part of the solution set.....but.... -3 / [-4 + 2.5] = 2

And this certainly satisfies C

Look at B....we can find the possible intervals that satisfy this as follows:

3/[x + 4] <= 2 get a common denominator on the right

3/[x + 4] <= 2[x + 4] / [x + 4] turn this into an equality

3/ [x + 4] = [2x + 8] / [x + 4] cross-multiply

3[x + 4] = [x + 4][2x + 8]

3x + 12 = 2x^2 + 16x + 32 re-arrange

2x^2 + 13x + 20 = 0 factor

(2x + 5) (x + 4) = 0

Setting each of the factors to 0, we have that x = -5/2 = -2.5 or x = -4

So........the solution will come from these possible intervals

(-inf, -4) , (-4, -2.5] or [-2.5, inf ) [note that -4 can't be included, because it makes the denominator of the inequality = 0 ]

Notice that if we put -3 into 3/[x + 4] we get 3....and this does not satisfy B

So....the solution comes from the other two intervals (-inf, -4) U [-2.5, inf) which is the same as the graphed solution ......so, B is correct

CPhill Feb 9, 2016