Chapter 11.5, Problem 38PPE

To determine

To solve:

  $\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{u^2-u-6}$

And check the solution

Answer to Problem 38PPE

  $u=0$

Explanation of Solution

Given:

  $\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{u^2-u-6}$

Concept Used:

To add two rational expressions:

1. Find Least Common Denominator.
2. To rewrite each rational expression with LCD in above step as denominator.
3. Perform addition.

Calculation:

To solve

  $\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{u^2-u-6}$

First rewrite it as

  $\begin{array}{c}\frac{u+1}{u+2}-\frac{-1}{u-3}=\frac{u-1}{u^2-u-6}\\ \Rightarrow \frac{u+1}{u+2}+\frac{1}{u-3}=\frac{u-1}{u^2-u-6}\end{array}$

Now, to solve above equation first simplify it by adding the two rational expressions in left side, as

  $\begin{array}{c}\frac{u+1}{u+2}+\frac{1}{u-3}=\frac{u-1}{u^2-u-6}\\ \frac{\left(u+1\right)\left(u-3\right)}{\left(u+2\right)\left(u-3\right)}+\frac{1\left(u+2\right)}{\left(u-3\right)\left(u+2\right)}=\frac{u-1}{u^2-u-6}\\ \frac{\left(u+1\right)\left(u-3\right)+\left(u+2\right)}{\left(u+2\right)\left(u-3\right)}=\frac{u-1}{u^2-u-6}\\ \frac{u^2+u-3u-3+u+2}{u^2+2u-3u-6}=\frac{u-1}{u^2-u-6}\\ \frac{u^2-u-1}{u^2-u-6}=\frac{u-1}{u^2-u-6}\\ \frac{u^2-u-1}{u^2-u-6}\text{.}\left(\frac{u^2-u-6}{u-1}\right)=1\end{array}$

Divide the common factor to solve further, as

  $\begin{array}{c}\frac{u^2-u-1}{u^2-u-6}\text{.}\left(\frac{u^2-u-6}{u-1}\right)=1\\ u^2-u-1=u-1\\ \Rightarrow u^2-u-1-u+1=0\\ \Rightarrow u^2-0=0\\ \Rightarrow u^2=0\\ \Rightarrow u=0\end{array}$

Now to check the solution put $u=0$ in left side and right side of given equation and check whether they are equal, as

  $\begin{array}{c}\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{u^2-u-6}\\ \text{Putting }u=0\\ \Rightarrow \frac{0+1}{0+2}=\frac{-1}{0-3}+\frac{0-1}{0^2-0-6}\\ \Rightarrow \frac{1}{2}=\frac{1}{3}+\frac{1}{6}\\ \Rightarrow \frac{1}{2}=\frac{2+1}{6}\\ \Rightarrow \frac{1}{2}=\frac{3}{6}\\ \Rightarrow \frac{1}{2}=\frac{1}{2}\end{array}$

Thus, solution of given equation is:

  $u=0$ .