# How to find the value of $a+b$? [closed]

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 $x = 4$.

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shuchang, Daryl, Newb, Claude Leibovici, Gigili
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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 $x=4$, $a+b=8+\frac{2}{3}=\frac{26}{3}$.