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

3 Answers
Jul 4, 2018

factorise top and bottom

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

Cancel the #x+n#

#c/(x-n)#

Jul 4, 2018

#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!

Jul 4, 2018

#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#