+0

# anser this question

0
188
6

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

Jun 21, 2021

#1
0

Jun 21, 2021
#2
+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
+287
0

Bginner  Jun 22, 2021
#4
0

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
+114940
0

It is a nice example of LaTex formating too.

Melody  Jun 22, 2021
#5
+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