How do you add #q/(q^2+5q+6)# and #1/(q^2+3q+2)#?

1 Answer
Apr 22, 2016

You must put on an equal denominator.

Explanation:

#q/((q + 3)(q + 2)) + 1/((q + 2)(q + 1))#

#(q(q + 1))/((q + 3)(q + 2)(q + 1)) + (1(q + 3))/((q + 2)(q + 1)(q + 3))#

#(q^2 + q)/((q + 3)(q + 2)(q + 1)) + (q + 3)/((q + 3)(q + 2)(q + 1))#

#(q^2 + 2q + 3)/((q + 3)(q + 2)(q + 1))#

Hopefully this helps!