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# (1/(2b+1) + (1/(b+1)) > 8/15

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(1/(2b+1) + (1/(b+1)) > 8/15

how do i solve the inequality

Guest Nov 18, 2017
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1 / (2b + 1)   + 1/ ( b + 1)   > 8 / 15

Let's make this an equality

Get a common denominator on the left

[  ( b + 1 )  +  (2b + 1) ] / [ (2b + 1) ( b + 1) ]  =   8 / 15   simplify

[ 3b + 2]  / [  2b^2 + b + 2b + 1 ]  =  8 / 15       cross multiply

15 [ 3b + 2 ]   =  8 [ 2b^2 + 3b + 1 ]

45b + 30  =  16b^2 + 24b + 8          rearrange as

16b^2 - 21b -22  =  0      factor

(16b + 11) ( b - 2)  = 0

Setting each factor to 0  and solving for b   produces    b = -11/16   and b  = 2

This gives us three intervals to test  in the original inequality

(-inf, -11/16), ( -11/16, 2) and (2, inf)

Choosing b = 0 as a test point in the middle interval...we have that

1 / (2(0) + 1)   + 1/ ( (0) + 1)   > 8 / 15   ???

1/1   +  1 / 1   > 8/15   ???

2 > 8/15    and this is true

Picking test points in the other two intervals will make the inequality false  [ test - 1 and 3 for yourself ]

So.... it appears that the  solution is     -11/16 < b < 2

But...we need to refine this....in the original inequality, b cannot = -1/2  or  -1 because these two values make a denominator  = 0

So.....the correct answer comes from either

( -1, -11/16), ( -11/16 -1/2) or ( -1/2, 2 )

Picking   a test point in the middle interval  [I'll use -10/16 = -5/8 ]  produces

1 / (2(-5/8) + 1)   + 1/ ( (-5/8) + 1)   > 8 / 15  ???

1 / [-2/8]  +  1 / [3/8]  > 8/15   ???

-4   + 8/3  > 8/15     ???    and this is false

So.....the real solutions come from

(-1, -11/16) U ( -1/2, 2)

CPhill  Nov 18, 2017

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