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

1 Answer
Sep 13, 2017

Answer:

#16b^2 - 22b - 22 < 0#

Explanation:

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

Solving the LHS..

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

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

Collecting like terms

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

#(3b + 2)/((2b + 1) (b + 1)) > 8/15#

Expanding the denominator

#(3b + 2)/(2b^2 + 2b + b + 1) > 8/15#

#(3b + 2)/(2b^2 + 3b + 1) > 8/15#

Cross multiplying

#15(3b + 2) > 8(2b^2 + 3b + 1)#

Expanding..

#45b + 30 > 16b^2 + 24b + 8#

Restructuring the equation.. Note #-># 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!

#16b^2 + 24b + 8 < 46b + 30#

Collecting like terms...

#16b^2 + 24b - 46b + 8 - 30 < 0#

#16b^2 - 22b - 22 < 0 -> "Quadratic Equation"#

Should I go further...?!