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 June 28th, 2014, 03:02 AM #1 Member   Joined: Dec 2013 Posts: 86 Thanks: 1 Variations Q: If y varies as the sum of two quantities of which one varies directly as x and the other inversely as x and if y=6 when x=4 and $y=3\frac{1}{3}$ when x=3, then the relation between x and y is ?
June 28th, 2014, 06:45 AM   #2
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Hello, Aman Verma!

Quote:
 If $y$ varies as the sum of two quantities of which one varies directly as $x$ and the other inversely as $x$ and if $y=6$ when $x=4$ and $y=3\tfrac{1}{3}$ when $x=3$, then the relation between $x$ and $y$ is?

We have: $\:y \:=\:ax + \dfrac{b}{x}$

We are given: $\:y(4) = 6,\;y(3) = 3\frac{1}{3}$

Hence: $\:\begin{Bmatrix} 4a + \dfrac{b}{4} \:=\:6 & \Rightarrow & 16a + b \:=\:24 \\ 3a + \dfrac{b}{3} \:=\:\dfrac{10}{3} & \Rightarrow & 9a + b \:=\:10 \end{Bmatrix}$

Subtract: $\:7a \:=\:14 \quad\Rightarrow\quad a \,=\,2 \quad\Rightarrow\quad b \,=\, \text{-}8$

Therefore: $\:y \:=\:2x - \dfrac{8}{x}$

June 28th, 2014, 08:13 AM   #3
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Quote:
 Originally Posted by soroban Hello, Aman Verma! We have: $\:y \:=\:ax + \dfrac{b}{x}$ We are given: $\:y(4) = 6,\;y(3) = 3\frac{1}{3}$ Hence: $\:\begin{Bmatrix} 4a + \dfrac{b}{4} \:=\:6 & \Rightarrow & 16a + b \:=\:24 \\ 3a + \dfrac{b}{3} \:=\:\dfrac{10}{3} & \Rightarrow & 9a + b \:=\:10 \end{Bmatrix}$ Subtract: $\:7a \:=\:14 \quad\Rightarrow\quad a \,=\,2 \quad\Rightarrow\quad b \,=\, \text{-}8$ Therefore: $\:y \:=\:2x - \dfrac{8}{x}$
Hi Sir,

Many thanks! I understood the entire solution except one small thing. Why is:y(4)=6 and y(3)=$3\frac{1}{3}$ ? Is x the proportionality constant ? Why are we multiplying x with y? If y is directly and indirectly variable to x then why should we only multiply x with y? Why not divide ?

 June 28th, 2014, 08:39 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,977 Thanks: 1851 There's no need to quote the entire preceding post when responding to it. You asked a good question. The notation $\color{#00AA00}{\textbf{soroban}}$ used does not denote multiplication (though the same symbols could denote multiplication in a different context). The notation "y(4) = 6" means "the value of y (when x = 4) equals 6". This meaning applies only when it's known that the value of y depends on the value of x. This is a major (potential) ambiguity in algebraic notation, but you can easily get used to it.

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