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Question 11
Let a and b be natural numbers. If $$a^2 + ab + a = 14$$ and $$b^2 + ab + b = 28$$, then $$(2a + b)$$ equals
Solution
a(a + b + 1) = 14 ...... (1)
b(a + b + 1) = 28 ...... (2)
$$\frac{a}{b}=\frac{1}{2}$$
b = 2a
Substituting in (1), we get
a(3a + 1) = 14
$$3a^2+a-14=0$$
$$3a^2-6a+7a-14=0$$
$$3a\left(a-2\right)+7\left(a-2\right)=0$$
Given, a and b are natural numbers.
Therefore, a = 2 and b = 4
2a + b = 2(2) + 4 = 8
The answer is option A.
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