How do you combine #(3n + 8)/(n^2 + 6n + 8) - ( 4n + 2)/(n^2 + n - 12)#?

Question

1830 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T01:36:03

#(3n+8)/(n^2+6n+8)-(4n-2)/(n^2+n-12)=(n^2+7n+20)/((n+4)(n+2)(3-n))#

#(3n+8)/(n^2+6n+8)-(4n-2)/(n^2+n-12)#

First factor the denominators.

#(3n+8)/((n+4)(n+2))-(4n-2)/((n+4)(n-3))#

The common denominator is #(n+4)(n+2)(n-3)#

Multiply each quotient by the appropriate factor to display the common denominator

#((3n+8))/((n+4)(n+2))((n-3))/((n-3))-((4n-2))/((n+4)(n-3))((n+2))/((n+2))#

Expand the numerators using the distributive property (or FOIL if you like).

#(3n^2-n-24)/((n+4)(n+2)(n-3))-(4n^2+6n-4)/((n+4)(n+2)(n-3))#

We can combine the quotients because they have a common denominator.

#(3n^2-n-24-4n^2-6n+4)/((n+4)(n+2)(n-3))#

Combine like terms.

#(-n^2-7n-20)/((n+4)(n+2)(n-3))=(n^2+7n+20)/((n+4)(n+2)(3-n))#