Suppose y varies as the sum of two quantities of which one varies directly as x and the other inversely as x. If y = 6 when x = 4 and y = $${\text{3}}\frac{{\text{1}}}{{\text{3}}}$$ when x = 3, then the relation between x and y is -

Solution(By Examveda Team)

$$\eqalign{ & {\text{y }}\alpha {\text{ }}\left( {{\text{x}} + \frac{1}{{\text{x}}}} \right) \cr & \Rightarrow {\text{y}} = {\text{kx}} + \frac{{\text{m}}}{{\text{x}}}{\text{,}} \cr} $$
Where k and m are constants,
Then,
$$\eqalign{ & = 4{\text{k}} + \frac{{\text{m}}}{4} = 6......({\text{i}}) \cr & and \cr & = 3{\text{k}} + \frac{{\text{m}}}{3} = 10......(ii) \cr} $$
Multiplying (i) by 3 and (ii) by 4, we get :
$$\eqalign{ & = 12{\text{k}} + \frac{{3{\text{m}}}}{4} = 18....({\text{iii}}) \cr & and \cr & = 12{\text{k}} + \frac{{4{\text{m}}}}{3} = \frac{{40}}{3}....({\text{iv}}) \cr} $$
Subtracting (iv) from (iii), we get :
$$\eqalign{ & = \frac{{3{\text{m}}}}{4} - \frac{{4{\text{m}}}}{3} = 18 - \frac{{40}}{3} \cr & \Rightarrow - \frac{{7{\text{m}}}}{{12}} = \frac{{14}}{3} \cr & \Rightarrow {\text{m}} = - 8. \cr & {\text{Putting m}} = - {\text{8 in (i),}} \cr & {\text{we get}}: \cr & 4{\text{k}} + \frac{{\left( { - 8} \right)}}{4} = 6 \cr & \Rightarrow 4{\text{k}} = 8 \cr & \Rightarrow {\text{k}} = 2 \cr & \therefore {\text{y}} = 2{\text{x}} - \frac{8}{{\text{x}}} \cr} $$