Answer:
[tex]\dfrac{3x^2+34x+51}{2x^2+12x+18}[/tex]
Step-by-step explanation:
[tex]\dfrac{3x+9}{2x+6}+\dfrac{8x+12}{x^2+6x+9}= \\\\\\\dfrac{3(x+3)}{2(x+3)}+\dfrac{8x+12}{(x+3)(x+3)}[/tex]
To add these two fractions, you need to make the denominators equal:
[tex]\dfrac{3(x+3)(x+3)}{2(x+3)(x+3)}+\dfrac{16x+24}{2(x+3)(x+3)}= \\\\\\\dfrac{(3x^2+18x+27)+(16x+24)}{2(x+3)(x+3)}= \\\\\\\dfrac{3x^2+34x+51}{2(x+3)(x+3)}= \\\\\\\dfrac{3x^2+34x+51}{2x^2+12x+18}[/tex]
Hope this helps!
