9514 1404 393
Answer:
(4x^2 +21x +18)/(3x^2 +34x -24)
Step-by-step explanation:
I find it helpful to keep in mind the form ...
a/b + c/d = (ad +bc)/(bd)
When we apply that to this case, we get ...
[tex]\dfrac{x+3}{x+12}+\dfrac{x+2}{3x-2}=\dfrac{(x+3)(3x-2)+(x+12)(x+2)}{(x+12)(3x-2)}\\\\=\dfrac{(3x^2+7x-6)+(x^2+14x+24)}{(x+12)(3x-2)}=\boxed{\dfrac{4x^2+21x+18}{3x^2+34x-24}}[/tex]
