Question

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

Answer 1

factorise top and bottom

`(c(x+n))/((x+n)(x-n))`

Cancel the `x+n`

`c/(x-n)`

Answer 2

`c/(x-n)`

Explanation:

We have the following:

`color(blue)(cx+cn)/color(purple)(x^2-n^2)`

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

`color(blue)(c(x+n))`

The denominator is a difference of squares, which factors as

`color(purple)((x+n)(x-n))`. If you multiply this out, you will indeed get `x^2-n^2`. Putting it together, we now have

`(color(blue)(c(x+n)))/(color(purple)((x+n)(x-n)))`

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

`(c cancel((x+n)))/(cancel(x+n)(x-n))`

`=>c/(x-n)`

Hope this helps!

Answer 3

`c/(x-n)`

Explanation:

`"factor the numerator/denominator and cancel common"`
`"factor"`

`"numerator "c(x+n)larrcolor(blue)"common factor"`

`"the denominator is a "color(blue)"difference of squares"`

`x^2-n^2=(x-n)(x+n)`

`(cx+cn)/(x^2-n^2)`

`=(c cancel((x+n)))/((x-n)cancel((x+n)))`

`=c/(x-n)to"with restriction "x!=n`