# What are the excluded values and how do you simplify the rational expression (3y-27)/(81-y^2)?

Jul 23, 2017

$\frac{3 y - 27}{81 - {y}^{2}} = - \frac{3}{9 + y}$

$y \ne 9 \mathmr{and} y \ne - 9$

#### Explanation:

$\frac{3 y - 27}{81 - {y}^{2}} = \frac{3 \left(y - 9\right)}{{9}^{2} - {y}^{2}}$

$= \frac{3 \left(y - 9\right)}{\left(9 - y\right) \left(9 + y\right)} = \frac{- 3 \left(9 - y\right)}{\left(9 - y\right) \left(9 + y\right)}$

$- \frac{3}{9 + y}$

Excluded values are $y = 9 \mathmr{and} y = - 9$

Jul 23, 2017

$y = - 9 \mathmr{and} y = + 9$ are the excluded values

Simplified $\to - \frac{3}{9 + y}$

#### Explanation:

$\textcolor{b l u e}{\text{Determining the excluded values}}$

You are not mathematically 'allowed' do divide by 0. If this situation exists the equation / expression is called 'undefined'

When you get very close to a denominator of 0 the graph forms asymptotes.

So the excluded values are such that ${y}^{2} = 81$

Thus $y = - 9 \mathmr{and} y = + 9$ are the excluded values
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Simplifying the expression}}$

$\textcolor{b r o w n}{\text{Consider the denominator:}}$

As above; ${9}^{2} = 81$ so $81 - {y}^{2} \text{ "->" } {9}^{2} - {y}^{2}$ thus we have

$\frac{3 y - 27}{{9}^{2} - {y}^{2}} \text{ "=" } \frac{3 y - 27}{\left(9 - y\right) \left(9 + y\right)}$

$\text{ }$.......................................................................................

$\textcolor{b r o w n}{\text{Consider the numerator:}}$

$3 y - 27$ this is the same as $3 y - \left[3 \times 9\right]$

Factor out the 3 giving: $3 \left(y - 9\right)$

$\text{ }$..........................................................................................

$\textcolor{b r o w n}{\text{Putting it all together:}}$

$\frac{3 \left(y - 9\right)}{\left(9 - y\right) \left(9 + y\right)} \leftarrow \text{ can not cancel out yet}$

Note that $\left(9 - y\right)$ is the same as $\left[- \left(y - 9\right)\right]$

so by substitution we have:

$- \frac{3 \left(y - 9\right)}{\left(y - 9\right) \left(9 + y\right)}$ giving

$- \frac{y - 9}{y - 9} \times \frac{3}{9 + y}$

but $\frac{y - 9}{y - 9} = 1 \leftarrow \text{ This is what cancelling is all about!}$

Giving: $- 1 \times \frac{3}{9 + y} \text{ "=" } - \frac{3}{9 + y}$