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)