# How do you solve  [a / (a+1)] - [1 / (a-1)]?

Nov 6, 2015

$\frac{{a}^{2} - 2 a - 1}{{a}^{2} - 1}$

Look at the method. It is another technique for you to consider.

#### Explanation:

$\textcolor{b l u e}{\text{Things to think about for the method}}$
These are fraction $\to \frac{c o u n t}{s i z e}$

$\textcolor{b r o w n}{\text{To be able to directly add or subtract counts}}$ what you are counting $\textcolor{g r e e n}{\text{must be of the same type or size.}}$

$\frac{c o u n t}{s i z e} \to \left(\text{numerator")/("denominator}\right)$

Neither of $\left(a + 1\right)$ nor $\left(a - 1\right)$ will easily divide into each other. So you have to devise something they will. The easiest way to do this is multiply them together.

Write $\left(a + 1\right) \times \left(a - 1\right)$ as $\left(a + 1\right) \left(a - 1\right)$

Consider $\frac{a}{a + 1}$. How do we change $\left(a + 1\right)$ into $\left(a + 1\right) \left(a - 1\right)$?

$\textcolor{g r e e n}{\text{If we multiply it by 1 but in the form of "(a-1)/(a-1)}}$ we change the way it looks but not its actual value. Think of $2 \times \frac{3}{3} = \frac{6}{3}$.

~~~~~~~~~~~~~~~~~~~~~~ end of method introduction ~~~~~~~

$\left[\frac{a}{a + 1} \times \frac{a - 1}{a - 1}\right] - \left[\frac{1}{a - 1} \times \frac{a + 1}{a + 1}\right]$

$\left[\frac{a \left(a - 1\right)}{\left(a + 1\right) \left(a - 1\right)}\right] - \left[\frac{a + 1}{\left(a + 1\right) \left(a - 1\right)}\right]$

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

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

but ${a}^{2} - {1}^{2} = {a}^{2} - 1 = \left(a + 1\right) \left(a - 1\right)$

So by substitution we have:

$\frac{{a}^{2} - 2 a - 1}{{a}^{2} - 1}$