Respuesta :

[tex] \frac{x^2+2x+9}{3x+9} [/tex]

with the work being

[tex] \frac{x+7}{x+3} + \frac{x-4}{3} [/tex]

multiply the left term by 3/3 and the right by (x+3)/(x+3), so it looks like [tex] \frac{3(x+7)}{3(x+3)} + \frac{(x+3)(x-4)}{(x+3)3} [/tex]

then you can multiply out the tops of the fractions and add together, and you have a common denominator

[tex] \frac{3x+21}{3x+9} + \frac{x^2+3x-4x-12}{3x+9} = \frac{x^2+2x+9}{3x+9} [/tex]