# Thread: Looks simple but I keep messing things up

1. ## Looks simple but I keep messing things up

Any idea where I go wrong? The two equations in the system are:
(1) $x/y-y/x = 1$

(2) $1/x-1/y = 1/2$

Now trying to simplify:

(1;2) $(xy)(x/y-y/x) = x^2 - y^2 = xy$

(2;2) $(xy)(1/x-1/y) = y-x = xy/2 -> /*2 -> 2y-2x = xy$

Substituting 2;2 into 1;2 $-> x^2 -y^2 = 2y-2x$ ->

$x^2 +2x = y^2 +2y ->$ then $x=y$??? what??? IF x = y, then the original equation doesn´t hold: $x=y -> x/y-y/x =1 -> x/x -x/x = 1?? 1-1 = 1$???
Solving this seemingly rediculously easy equationssystems for last 40min,

Oh the correct answer should be $x = 1+sqrt5 -> y = -3-sqrt5$
$x =1-sqrt5 -> 7 = -3+sqrt5$

2. Originally Posted by Henryt999
Any idea where I go wrong? The two equations in the system are:
(1) $x/y-y/x = 1$

(2) $1/x-1/y = 1/2$

Now trying to simplify:

(1;2) $(xy)(x/y-y/x) = x^2 - y^2 = xy$

(2;2) $(xy)(1/x-1/y) = y-x = xy/2 -> /*2 -> 2y-2x = xy$

Substituting 2;2 into 1;2 $-> x^2 -y^2 = 2y-2x$ ->

$x^2 +2x = y^2 +2y ->$ then $x=y$??? what??? IF x = y, then the original equation doesn´t hold: $x=y -> x/y-y/x =1 -> x/x -x/x = 1?? 1-1 = 1$???
Solving this seemingly rediculously easy equationssystems for last 40min,
Hint:

$\frac{x}
{y} - \frac{y}
{x} = 1 \Leftrightarrow \frac{x}
{y} - {\left( {\frac{x}
{y}} \right)^{ - 1}} = 1 \Leftrightarrow \left\{ {\frac{x}
{y} = a} \right\} \Leftrightarrow a - {a^{ - 1}} = 1 \Leftrightarrow$

$\Leftrightarrow {a^2} - a - 1 = 0 \Leftrightarrow {a_{1,2}} = \frac{{1 \pm \sqrt 5 }}
{2} \Leftrightarrow \frac{x}
{y} = \frac{{1 \pm \sqrt 5 }}
{2} \Leftrightarrow \frac{1}
{y} = \frac{{1 \pm \sqrt 5 }}
{{2x}}.$

3. ## Thanks

Do you know where I went wrong?

4. From the first equation you have:

$\frac{x}{y} - \frac{y}{x} = 1 ~ \Leftrightarrow ~ \frac{x}
{y} - \left(\frac{x}{y}\right)^{-1} = 1 ~ \Leftrightarrow ~ \left\{ {\frac{x}{y} = a} \right\} ~ \Leftrightarrow ~ a - a^{-1} = 1 ~ \Leftrightarrow$

$\Leftrightarrow ~ a^2 - a - 1 = 0 ~ \Leftrightarrow ~ a_{1,2} = \frac{1 \pm \sqrt 5}{2} ~ \Leftrightarrow ~ \frac{x}{y} = \frac{1 \pm \sqrt 5}
{2} ~ \Leftrightarrow ~ \frac{1}{y} = \frac{1 \pm \sqrt 5}{{2x}}.$

So you get

$\left\{ \begin{gathered}
\frac{x}
{y} - \frac{y}
{x} = 1, \hfill \\
\frac{1}
{x} - \frac{1}
{y} = \frac{1}
{2}; \hfill \\
\end{gathered} \right.~\Leftrightarrow{\color{white}...}$
$\left[\begin{gathered}\left\{\begin{gathered}\frac{1}{y} = \frac{1 - \sqrt 5}{2x}, \hfill \\ \frac{1}{x} - \frac{1 - \sqrt 5}{2x} = \frac{1}{2}; \hfill \\ \end{gathered} \right. \hfill \\ \left\{\begin{gathered}\frac{1}{y} = \frac{1 + \sqrt 5 }{2x}, \hfill \\ \frac{1}{x} - \frac{1 + \sqrt 5}{2x} = \frac{1}{2}; \hfill \\ \end{gathered}\right. \hfill \\ \end{gathered}\right. ~\Leftrightarrow{\color{white}...}$ $\left[ \begin{gathered}\left\{ \begin{gathered}\frac{1}
{y} = \frac{1}
{2}\frac{1 - \sqrt 5}
{{1 + \sqrt 5 }}, \hfill \\
x = 1 + \sqrt 5 ; \hfill \\
\end{gathered} \right. \hfill \\
\left\{ \begin{gathered}
\frac{1}
{y} = \frac{1}
{2}\frac{{1 + \sqrt 5 }}
{{1 - \sqrt 5 }}, \hfill \\
x = 1 - \sqrt 5 ; \hfill \\
\end{gathered}\right. \hfill \\
\end{gathered}\right.~\Leftrightarrow{\color{white }...}$
$\left[\begin{gathered}\left\{ \begin{gathered}x_1 = 1 + \sqrt 5 , \hfill \\y_1 = - 3 - \sqrt 5 . \hfill \\ \end{gathered}\right. \hfill \\ \left\{\begin{gathered}x_2 = 1 - \sqrt 5 , \hfill \\ y_2 = - 3 + \sqrt 5 . \hfill \\ \end{gathered}\right. \hfill \\ \end{gathered}\right.$

_________________
$\frac{1}{y} = \frac{1}{2}\frac{1 - \sqrt 5}{1 + \sqrt 5} ~\Leftrightarrow{\color{white}...}$ $y = \frac{2\left(1 + \sqrt 5\right)}{1 - \sqrt 5} = \frac{2\left(1 + \sqrt 5\right)^2}{\left(1 - \sqrt 5\right)\left(1 + \sqrt 5\right)} = \frac{2\left(1 + 2\sqrt 5 + 5\right)}{1 - 5} = - \frac{6 + 2\sqrt 5}{2} = -3 - \sqrt 5 .$

$\frac{1}{y} = \frac{1}{2}\frac{1 + \sqrt 5}{1 - \sqrt 5}~\Leftrightarrow{\color{white}...}$ $y = \frac{2\left(1 - \sqrt 5\right)}{1 + \sqrt 5} = \frac{2\left(1 - \sqrt 5\right)^2}{\left(1 + \sqrt 5\right)\left(1 - \sqrt 5\right)} = \frac{2\left(1 - 2\sqrt 5 + 5\right)}{1 - 5} = - \frac{6 - 2\sqrt 5}{2} = -3 + \sqrt 5 .$

Do you understand this?

5. Hello, Henryt999!

This is how I solved it . . .

Solve the system: . $\begin{array}{cccc}\dfrac{x}{y}-\dfrac{y}{x} &=& 1 & [1] \\ \\[-3mm]
\dfrac{1}{x}-\dfrac{1}{y} &=& \dfrac{1}{2} & [2]\end{array}$

$\begin{array}{cccccc}
\text{Multiply [1] by }xy: & x^2 - y^2 &=& xy & [3] \\ \\
\text{Multiply [2] by }2xy: & 2y - 2x &=& xy & [4] \end{array}$

Equate [3] and [4]: . $x^2-y^2 \:=\:2y-2x \quad\Rightarrow\quad x^2 - y^2 + 2x - 2y \:=\:0$

Factor: . $(x-y)(x+y) + 2(x-y) \:=\:0$

Factor: . $(x-y)(x+y+2) \:=\:0$

We have two equations to solve:

. . $x - y \:=\:0 \quad\Rightarrow\quad x \:=\:y$ . . . which does not check (extraneous root)

. . $x+y+2\:=\:0 \quad\Rightarrow\quad y \:=\:-x-2\;\;[5]$

Substitute into [4]: . $2(-x-2) - 2x(-x-2) \:=\:x(-x-2) \quad\Rightarrow\quad x^2-2x-4 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{2\pm\sqrt{20}}{2} \:=\:\frac{2\pm2\sqrt{5}}{2} \quad\Rightarrow\quad\boxed{ x\:=\:1\pm\sqrt{5}}$

Substitute into [5]: . $y \:=\:-(1\pm\sqrt{5}) - 2 \:=\:-1 \mp\sqrt{5} -2 \quad\Rightarrow\quad\boxed{ y \:=\:-3\mp\sqrt{5}}$