Chapter 6.6, Problem 43WE

To determine

Simplify the given equation.

Answer to Problem 43WE

Simplified form of $\left(2-\frac{n}{n+1}+\frac{n}{1-n}\right)÷\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$ is $\boxed{-1}$ .

Explanation of Solution

Given information:

  $\left(2-\frac{n}{n+1}+\frac{n}{1-n}\right)÷\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$

Formula used:

  $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

  $\left(2-\frac{n}{n+1}+\frac{n}{1-n}\right)÷\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$

Get common denominator:

  $\left(\frac{2\left(n+1\right)\left(1-n\right)-n\left(n+1\right)+n\left(1-n\right)}{\left(n+1\right)\left(1-n\right)}\right)÷\left(\frac{n+1-\left(n-1\right)}{\left(n+1\right)\left(1-n\right)}\right)$

Now FOIL and distribute:

  $\left(\frac{-2n^2+2-n+n^2+n^2+n}{\left(n+1\right)\left(1-n\right)}\right)÷\left(\frac{n+1-n+1}{\left(n+1\right)\left(1-n\right)}\right)$

Combine like terms:

  $\left(\frac{2}{\left(n+1\right)\left(1-n\right)}\right)÷\left(\frac{2}{\left(n+1\right)\left(1-n\right)}\right)$

Now writing as multiplication by the help of identity $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

  $\left(\frac{2}{\left(n+1\right)\left(1-n\right)}\right)\times \left(\frac{\left(n+1\right)\left(1-n\right)}{2}\right)$

Cancel $2$ and $n+1$ :

  $\frac{-\left(n-1\right)}{n-1}$

Now cancel $n-1$

  $=\boxed{-1}$