How do you evaluate #(3a -9i +2ai +6)/(a^2+9) + (3-9i+3i+9)/(9+a^2)#?

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Answer 1 · 2017-08-14T17:15:39

#(3a+2ai-15i+18)/(9+a^2)#

The first thing we notice with the two expression here is that the denominators are the same since #a^2+9=9+a^2#.

This means that we can "combine" the two fractions to add the numerators:

#(3a-9i+2ai+6+3-9i+3i+9)/(9+a^2)#

Now, we can combine like terms on the numerator. The like terms have been put in the same color, so you just add/subtract them:

#(3a-color(green)(9i)+2ai+color(red)6+color(red)3-color(green)(9i)+color(green)(3i)+color(red)9)/(9+a^2)#

Rearranging to bring like terms closer together and to make it easier to simplify:

#(3a+2ai-color(green)(9i)-color(green)(9i)+color(green)(3i)+color(red)6+color(red)9+color(red)3)/(9+a^2)#

#(3a+2ai-15i+18)/(9+a^2)#