# How do you solve the inequality 1/(2b+1)+1/(b+1)>8/15?

Sep 13, 2017

$16 {b}^{2} - 22 b - 22 < 0$

#### Explanation:

$\frac{1}{2 b + 1} + \frac{1}{b + 1} > \frac{8}{15}$

Solving the LHS..

$\frac{1 \left(b + 1\right) + 1 \left(2 b + 1\right)}{\left(2 b + 1\right) \left(b + 1\right)} > \frac{8}{15}$

$\frac{b + 1 + 2 b + 1}{\left(2 b + 1\right) \left(b + 1\right)} > \frac{8}{15}$

Collecting like terms

$\frac{b + 2 b + 1 + 1}{\left(2 b + 1\right) \left(b + 1\right)} > \frac{8}{15}$

$\frac{3 b + 2}{\left(2 b + 1\right) \left(b + 1\right)} > \frac{8}{15}$

Expanding the denominator

$\frac{3 b + 2}{2 {b}^{2} + 2 b + b + 1} > \frac{8}{15}$

$\frac{3 b + 2}{2 {b}^{2} + 3 b + 1} > \frac{8}{15}$

Cross multiplying

$15 \left(3 b + 2\right) > 8 \left(2 {b}^{2} + 3 b + 1\right)$

Expanding..

$45 b + 30 > 16 {b}^{2} + 24 b + 8$

Restructuring the equation.. Note $\to$ Why am doing this because I don't want any form of confusion to come in.. to make it more simpler to understand..

Also when restructuring equations, the inequality sign changes!

$16 {b}^{2} + 24 b + 8 < 46 b + 30$

Collecting like terms...

$16 {b}^{2} + 24 b - 46 b + 8 - 30 < 0$

$16 {b}^{2} - 22 b - 22 < 0 \to \text{Quadratic Equation}$

Should I go further...?!