+0

Algebra 2 inequality problem

0
433
1
+564

Feb 9, 2016

#1
+101369
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

Feb 9, 2016