Chapter 4.7, Problem 43E

To determine

To solve: $\frac{a+2}{2a+1}=\frac{a}{3}+\frac{3}{4a+2}$

Answer to Problem 43E

  $a=\frac{3}{2},-\frac{1}{2}$

Explanation of Solution

Given information: $\frac{a+2}{2a+1}=\frac{a}{3}+\frac{3}{4a+2}$

Calculation:

Given

  $\frac{a+2}{2a+1}=\frac{a}{3}+\frac{3}{4a+2}$

Note the domain exclusion

Domain: $2a+1\ne 0\to a=-\frac{1}{2}$

Multiply both sides by common denominator

  $\begin{array}{l}\left(6\right)\left(2a+1\right)\left(\frac{a+2}{2a+1}=\frac{a}{3}+\frac{3}{2\left(2a+1\right)}\right)\\ 6a+12=4a^2+2a+9\end{array}$

Gather quadratic terms to one side

  $4a^2-4a-3=0$

Factor and set = 0

  $\begin{array}{l}\left(2a-3\right)\left(2a+1\right)=0\\ \to a=\frac{3}{2},-\frac{1}{2}\end{array}$