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

Question

Julia Adams · Accepted Answer

#c/(x-n)#

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!

Eli Assouline · Answer

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

Avery Alexander · Answer

factorise top and bottom #(c(x+n))/((x+n)(x-n))# Cancel the #x+n# #c/(x-n)#

Nathan Axel · Answer

undefined To simplify the complex fraction \frac { c x + c n } { x ^ { 2} - n ^ { 2} }, you can factor the denominator as the difference of squares: x^2 - n^2 = (x + n)(x - n). Then, cancel out the common factor of c in the numerator and denominator. The simplified form of the complex fraction is \frac {c(x + n)} {(x + n)(x - n)}. Finally, you can further simplify by canceling out the common factor of (x + n) in the numerator and denominator, resulting in the simplified form \frac {c} {x - n}.