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If (x+1/x):(x-1/x)=5:4, then find the value of x.

 Jun 21, 2021
 #1
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Nvm I found the answer 

 Jun 21, 2021
 #2
avatar+287 
+1

You may want to see a slick way of solving this problem.  It depends on knowing this fact.
If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.  For your problem it goes: We first note that $\frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}$.  so
\begin{eqnarray*}
\dfrac{x+1}{x-1}
&=& \frac{5}{4} \\
\dfrac{(x+1)+(x-1)}{(x+1)-(x-1)}
&=& \dfrac{5+4}{5-4} \\
\dfrac{2x}{2}
&=& \dfrac{9}{1} \\
x &=& 9
\end{eqnarray*}

 Jun 21, 2021
 #3
avatar+287 
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Please delete that.   I read the question wrong.  Sorry.

Bginner  Jun 22, 2021
 #4
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There is no need to delete this.

Though it may not be a direct solution to the posted problem it’s still a good presentation of mathematical problem solving.  

 

--. .-

Guest Jun 22, 2021
 #6
avatar+114940 
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It is a nice example of LaTex formating too.

Melody  Jun 22, 2021
 #5
avatar+287 
+2

OK then.   You can still use my technique for this problem.
Note that \[\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}} =
\dfrac{x^2+1}{x^2-1}.\]
So
\begin{eqnarray*}
    \dfrac{x^2+1}{x^2-1} &=& \dfrac{5}{4} \\
    \dfrac{(x^2+1)+(x^2-1)}{(x^2+1)-(x^2-1)} &=& \dfrac{5+4}{5-4} \\
    \dfrac{2x^2}{2} &=& \dfrac{9}{1} \\
    x^2 &=& 9 \\
    x &=& \pm 3.
\end{eqnarray*}

 Jun 22, 2021

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