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Let $a$ and $b$ be natural numbers. If $a^{2}+a b+a=14$ and $b^{2}+a b+b=28$, then $(2 a+b)$ equals

  1. $8$
  2. $9$
  3. $7$
  4. $10$
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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.
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