If x+y varies inversely as 1/x + 1/y, show that xy is prop. to (x+y)^2 and

[VincentTong]

29. If (x + y) varies inversely as \(\displaystyle \left(\dfrac{1}{x}\, +\, \dfrac{1}{y}\right)\) show that:

(a) \(\displaystyle (x\, +\, y)^2\, \propto\, xy\)

(b) \(\displaystyle xy\, \propto\, (x\, +\, y)^2\)

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29. It is given that z varies directly as y when x is constant, and varies inversely as x when y is constant. Show that

. . .\(\displaystyle z\, \propto\, y^2\, \mbox{ if }\, y\, \propto\, \dfrac{1}{x}\)

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I couldn't find the answer. The topic i was in is variations. And the Question 29 containing a and b i solved a but not b. Could i be answered?

 

[MarkFL]

If \(\displaystyle (x+y)\) varies inversely as \(\displaystyle \displaystyle \left(\frac{1}{x}+\frac{1}{y}\right)\), then we may state:

\(\displaystyle \displaystyle x+y=\frac{k}{\dfrac{1}{x}+\dfrac{1}{y}}\)

\(\displaystyle \displaystyle x+y=\frac{kxy}{x+y}\)

Can you proceed now to reach the given conclusions?

 

[VincentTong]

If \(\displaystyle (x+y)\) varies inversely as \(\displaystyle \displaystyle \left(\frac{1}{x}+\frac{1}{y}\right)\), then we may state:

\(\displaystyle \displaystyle x+y=\frac{k}{\dfrac{1}{x}+\dfrac{1}{y}}\)

\(\displaystyle \displaystyle x+y=\frac{kxy}{x+y}\)

Can you proceed now to reach the given conclusions?

[FONT=MathJax_Math-italic](x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]y)²=kxy
x²+y²[/FONT][FONT=MathJax_Math-italic]=(k-2)xy
then xy=(x²+y²)/(k-2)
So it cannot be proved?

[/FONT]

 

[stapel]

[FONT=MathJax_Math-italic](x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]y)²=kxy

x²+y²[/FONT][FONT=MathJax_Math-italic]=(k-2)xy[/FONT]

[FONT=MathJax_Math-italic]then xy=(x²+y²)/(k-2)

So it cannot be proved?[/FONT]

What does it mean for one thing to "be proportional to" another thing?

Please be specific, including a clear statement of why you feel your last line above violates this definition. Thank you! ;-)

 

[HallsofIvy]

"A is proportional to B" if and only if there exist a number k such that A= kB. But if k is a number so is k- 2! A= (k-2)B also says that A is proportional to B.