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June 28th, 2014, 02:02 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, 05:45 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Aman Verma! Quote:
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, 07:13 AM  #3  
Member Joined: Dec 2013 Posts: 86 Thanks: 1  Quote:
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, 07:39 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,636 Thanks: 2080 
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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