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

We have two equations with natural numbers $a$ and $b$:

$a^2 + ab + a = 14$

$b^2 + ab + b = 28$

We need to find the value of $(2a + b)$.

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We can factor each equation:

For the first equation: $a^2 + ab + a = 14$

Factor out $a$: $a(a + b + 1) = 14$

For the second equation: $b^2 + ab + b = 28$

Factor out $b$: $b(a + b + 1) = 28$

When we have terms like $a^2 + ab + a$, we can factor out the common factor $a$ from all terms: $a \cdot a + a \cdot b + a \cdot 1 = a(a + b + 1)$

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Now we have:

$a(a + b + 1) = 14$ ... (1)

$b(a + b + 1) = 28$ ... (2)

Both equations have the same factor $(a + b + 1)$!

Dividing equation (2) by equation (1):

$\frac{b(a + b + 1)}{a(a + b + 1)} = \frac{28}{14}$

The $(a + b + 1)$ terms cancel out:

$\frac{b}{a} = \frac{28}{14} = 2$

Therefore: $b = 2a$

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Now that we know $b = 2a$, substituting this into the first equation:

$a(a + b + 1) = 14$

$a(a + 2a + 1) = 14$

$a(3a + 1) = 14$

$3a^2 + a = 14$

$3a^2 + a - 14 = 0$

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We need to solve: $3a^2 + a - 14 = 0$

Since we're looking for natural numbers, let's check small positive integer values:

Try $a = 2$:

$3(2)^2 + 2 - 14 = 3(4) + 2 - 14 = 12 + 2 - 14 = 0$

So $a = 2$ works!

Since $b = 2a$, we have $b = 2(2) = 4$.

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We need to find $(2a + b)$:

$2a + b = 2(2) + 4 = 4 + 4 = 8$

Therefore, $(2a + b) = 8$

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When we have two equations with similar structures, we look for common factors! This problem became much easier once we factored and noticed that both equations shared the factor $(a + b + 1)$, allowing us to find the relationship between $a$ and $b$ quickly.