How do you solve the inequality $\frac{1}{2 b + 1} + \frac{1}{b + 1} > \frac{8}{15}$ $1/(2b+1)+1/(b+1)>8/15$ ?

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How do you solve the inequality $\frac{1}{2 b + 1} + \frac{1}{b + 1} > \frac{8}{15}$ $1/(2b+1)+1/(b+1)>8/15$ ?

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Answer 1 · 2017-09-13T16:43:39

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

Explanation:

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

Solving the LHS..

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

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

Collecting like terms

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

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

Expanding the denominator

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

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

Cross multiplying

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

Expanding..

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

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!

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

Collecting like terms...

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

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

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How do you solve the inequality $\frac{1}{2 b + 1} + \frac{1}{b + 1} > \frac{8}{15}$ $1/(2b+1)+1/(b+1)>8/15$ ?

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Explanation:

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