Chapter 4.6, Problem 1CFU

To determine

To describe: The process used to solve $\frac{b+1}{3\left(b-2\right)}=\frac{5b}{6}+\frac{1}{b-2}$

Answer to Problem 1CFU

  $b=2,\frac{2}{5}$

Explanation of Solution

Given information: $\frac{b+1}{3\left(b-2\right)}=\frac{5b}{6}+\frac{1}{b-2}$

Calculation:

Given:

  $\frac{b+1}{3\left(b-2\right)}=\frac{5b}{6}+\frac{1}{b-2}$

Take L.C.M on right side

  $\frac{b+1}{3\left(b-2\right)}=\frac{5b\left(b-2\right)+6}{6\left(b-2\right)}$

Simplify

  $\frac{b+1}{3\left(b-2\right)}=\frac{5b^2-10b+6}{6\left(b-2\right)}$

Multiply both sides $3\left(b-2\right)$

  $b+1=\frac{5b^2-10b+6}{2}$

Multiply both sides 2

  $2b+2=5b^2-10b+6$

Simplify

  $5b^2-12b+4=0$

Use quadratic formula

  $b=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\cdot 5\cdot 4}}{2\cdot 5}$

Simplify

  $b=2,\frac{2}{5}$