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

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8 — Easy Both equations factor neatly: $a^2+ab+a=a(a+b+1)$ and $b^2+ab+b=b(a+b+1)$. They share the factor $(a+b+1)$, so dividing them kills it and gives $a:b$ instantly — then plug back. 1 Factor both equations , exposing the common factor: $$\begin{aligned} &a(a+b+1)=14\ &b(a+b+1)=28 \end{aligned}$$ 2 Divide to remove $(a+b+1)$: $$\frac{b}{a}=\frac{28}{14}=2\ \Rightarrow\ b=2a$$ 3 Substitute $b=2a$ into $a(a+b+1)=14$: $$\begin{aligned} &a(a+2a+1)=14\ &a(3a+1)=14\ \Rightarrow\ 3a^2+a-14=0\ &(3a+7

CAT 2022 Slot 1 — QA Question 6

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