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A quantity $a$ varies directly as $x$ and another quantity $b$ varies inversely as $x$. When $x = 2$, then sum of $a$ and $b$ is $7$, and when $x = 3$, the sum is $8$. Find the value of $a+b$ when $x = 4$.

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closed as off-topic by Shuchang, Daryl, Newb, Claude Leibovici, Gigili Jan 26 '14 at 5:19

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What do you mean by $b$ varies inversely as $x$? – akinn Jan 19 '14 at 16:36
Isn't this $a(x) = k_1x$ and $b(x) = \frac{k_2}{x}$? Then you just solve for $k_1$ and $k_2$ given the values of $a(x)$ and $b(x)$ at $x=2$ and $x=4$. – Alec Jan 19 '14 at 16:42

1 Answer 1

As it's pointed out in the comments: $a=k_1x$ and $b=\frac{k_2}{x}$. We have two equations: $2k_1+\frac{k_2}{2}=7$ and $3k_1+\frac{k_2}{3}=8$ For which the solution is $k_2=6$ and $k_1=2$. So when $x=4$, $a+b=8+\frac{2}{3}=\frac{26}{3}$.

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