Given that:
$a^2+ab+a=14\rightarrow a(a+b+1)=14……….(i)$
$b^2+ab+b=28\rightarrow b(a+b+1)=28…..….(ii)$
form eq (i),(ii) we get:
$(a+b+1)=14/a,(a+b+1)=28/b$
from above $\frac{14}{a}=\frac{28}{b}\rightarrow b=2a$
put $b=2a$ into eq(i) we get:
$a(a+2a+1)=14$
$\implies a(3a+1)=14$
$\implies 3a^2+a-14=0$
$\implies 3a^2-6a+7a-14=0$
$\implies a=2,b=4$ (a,b is natural number)
$\therefore$ the value of $(2a+b)=2*2+4=8$
Option (A) is correct.