# How do you simplify the complex fraction \frac { c x + c n } { x ^ { 2} - n ^ { 2} }?

Jul 4, 2018

factorise top and bottom

$\frac{c \left(x + n\right)}{\left(x + n\right) \left(x - n\right)}$

Cancel the $x + n$

$\frac{c}{x - n}$

Jul 4, 2018

$\frac{c}{x - n}$

#### Explanation:

We have the following:

$\frac{\textcolor{b l u e}{c x + c n}}{\textcolor{p u r p \le}{{x}^{2} - {n}^{2}}}$

In the numerator, both terms have a $c$ in common, so we can factor that out to get

$\textcolor{b l u e}{c \left(x + n\right)}$

The denominator is a difference of squares, which factors as

$\textcolor{p u r p \le}{\left(x + n\right) \left(x - n\right)}$. If you multiply this out, you will indeed get ${x}^{2} - {n}^{2}$. Putting it together, we now have

$\frac{\textcolor{b l u e}{c \left(x + n\right)}}{\textcolor{p u r p \le}{\left(x + n\right) \left(x - n\right)}}$

Same terms in the numerator and denominator cancel. We're left with

$\frac{c \cancel{\left(x + n\right)}}{\cancel{x + n} \left(x - n\right)}$

$\implies \frac{c}{x - n}$

Hope this helps!

Jul 4, 2018

$\frac{c}{x - n}$

#### Explanation:

$\text{factor the numerator/denominator and cancel common}$
$\text{factor}$

$\text{numerator "c(x+n)larrcolor(blue)"common factor}$

$\text{the denominator is a "color(blue)"difference of squares}$

${x}^{2} - {n}^{2} = \left(x - n\right) \left(x + n\right)$

$\frac{c x + c n}{{x}^{2} - {n}^{2}}$

$= \frac{c \cancel{\left(x + n\right)}}{\left(x - n\right) \cancel{\left(x + n\right)}}$

$= \frac{c}{x - n} \to \text{with restriction } x \ne n$