How do you evaluate #3/(16-a^2) - 5/(12+3a)#?

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Answer 1 · 2015-07-23T20:40:53

#(5a-11)/(3(4+a)(4-a))#

Assuming you want to evaluate to a single expression:

Factoring the denominators we have:
#3/(16-a^2)-5/(12+3a)=3/((4+a)(4-a))-5/(3(4+a))#

we can then make the denominators identical:
#3/((4+a)(4-a))-5/(3(4+a))=#

#3/((4+a)(4-a))-(5/3(4-a))/((4+a)(4-a))=#

#(3-5/3(4-a))/((4+a)(4-a))=(5a-11)/(3(4+a)(4-a))#

Answer 2 · 2015-07-23T20:48:08

I tried factorizing and using a common denominator:

You can write it as:
#3/((4+a)(4-a))-5/(3(4+a))=#
#=(9-5(4-a))/(3(4+a)(4-a))=#
#=(9-20+5a)/(3(4+a)(4-a))=(5a-11)/(3(4+a)(4-a))#