Let $$ a $$ and $$ b $$ be natural numbers. If
$$ a^2 + ab + a = 14 $$
and
$$ b^2 + ab + b = 28 $$,
then $$ 2a + b $$ equals:

Question

Let $$ a $$ and $$ b $$ be natural numbers. If  
$$ a^2 + ab + a = 14 $$  
and  
$$ b^2 + ab + b = 28 $$,  
then $$ 2a + b $$ equals:

Accepted Answer

To solve the given equations:

1. **Add the Two Equations**:  
   $$
   a^2 + ab + a + b^2 + ab + b = 14 + 28
   $$  
   Simplify:  
   $$
   a^2 + b^2 + ab + (a + b) = 42
   $$  
   Factorize:  
   $$
   (a + b)^2 + (a + b) = 42
   $$

2. **Let $$ k = a + b $$**:  
   Substitute $$ k $$:  
   $$
   k^2 + k = 42
   $$  
   Solve for $$ k $$:  
   $$
   k(k+1) = 42
   $$

3. **Find $$ k $$**:  
   The product of two consecutive natural numbers is 42, so:  
   $$
   k = 6
   $$

4. **Find $$ a $$ and $$ b $$**:  
   From $$ a + b = k = 6 $$, substitute into the first equation:  
   $$
   a(a+6-a+1) = 14 \implies a(6+1) = 14 \implies a = 2.
   $$  
   Therefore, $$ b = 6 - a = 4 $$.

5. **Calculate $$ 2a + b $$**:  
   $$
   2a + b = 2(2) + 4 = 8.
   $$

Thus, the answer is **8**.

Answer

Answer

Answer