How can you evaluate #(4a-2)/(3a+12) - (a-2)/(a+4) #?

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Answer 1 · 2015-08-03T15:04:53

#(4a-2)/(3a+12)-(a-2)/(a+4)=1/3#

In order to calculate the difference between the two terms, you need to write them with the same denominator. Notice that:

#3a+12=3*(a+4)#

Therefore:

#(a-2)/(a+4)=1*(a-2)/(a+4)=3/3*(a-2)/(a+4)=(3(a-2))/(3(a+4))=(3a-6)/(3a+12)#

Therefore:

#(4a-2)/(3a+12)-(a-2)/(a+4)=(4a-2)/(3a+12)-(3a-6)/(3a+12)#

#=((4a-2)-(3a-6))/(3a+12)#

#=(4a-2-3a+6)/(3a+12)=(4a-3a+6-2)/(3a+12)=(a+4)/(3a+12)#

#=(a+4)/(3(a+4))=1/3#