How do you solve #(2x+3)/(x^2-9) + x/(x-3)#?

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Answer 1 · 2018-04-03T16:03:16

#(x^2+5x+3)/(x^2-9)#

Add them like fractions (make sure their denominators are the same)

The two denominators are #x^2-9# are #x-3#. Find the least common denominator.

#x^2-9=(x-3)(x+3) rarr# One of the denominators is a factor of the other

Multiply the second fraction's numerator and denominator by #x+3# to get the same denominator as the first fraction

#x/(x-3)*(x+3)/(x+3)#

#(x^2+3x)/(x^2-9)#

Now add the fractions together

#(2x+3+x^2+3x)/(x^2-9)#

Combine like terms:

#(x^2+5x+3)/(x^2-9)#