help me please

$$\small{\text{$
\begin{array}{rcl}
\dfrac{a+x}{ax} &=& \dfrac{1}{a}+\dfrac{a}{a+x} \\\\
\dfrac{a}{ax}+\dfrac{x}{ax} &=& \dfrac{1}{a}+\dfrac{a}{a+x} \\\\
\dfrac{1}{x}+\dfrac{1}{a} &=& \dfrac{1}{a}+\dfrac{a}{a+x} \\\\
\dfrac{1}{x} &=& \dfrac{a}{a+x} \\\\
a+x &=& xa \\
xa &=& a+x \\
xa-x &=& a\\
x(a-1) &=& a \\\\
\mathbf{x} & \mathbf{=} & \mathbf{\dfrac{a}{a-1}}
\end{array}
$}}$$

note that      $$a\ne0\;\;\;x\ne0\;\;\;a+x\ne0$$

The lowest common denominator is ax(a+x) so I am going to multiply both sides by this

$$\\(a+x)^2=x(a+x)+a^2x\\\\
a^2+x^2+2ax=x^2+ax+a^2x\\\\
a^2+ax=a^2x\\\\
a^2+ax-a^2x=0\\\\
a(a+x-ax)=0\\\\
a\ne0\;\;so\\\\
a+x-ax=0\\\\
x(1-a)=-a\\\\
x=\frac{-a}{1-a}\\\\
x=\frac{a}{a-1}\\\\$$

I saw the pop up!  You beat me Heureka  :))

heuruka if i am writing it this way:

a+x=xa

x-xa=-a

x(1-a)=-a

x=-a/(1-a)

the answer is incomplete then how do i get your answer?i need to divide it or  something?

Do you want to wait longer for Heureka to answer you?

I am reasonabley sure he will answer if you wait long enough.  :)

Your answer is the same as heureka's, sabi92; it just looks slightly different!  Multiply the top and bottom of your expression by -1 to get it to look like heureka's.

.

Yes I know my answer is the same. I did not discuss it because Sabi specifically requested Heureka to answer him :)

Hallo sabi92,

$$\small{\text{Your answer:~~
$
\mathbf{x=-\dfrac{a}{1-a} }
$}}\\ \\
\small{\text{now you can write:~~
$
\mathbf{x=\dfrac{a}{-(1-a)} }
$}}\\\\
\small{\text{and:~~
$
\mathbf{x=\dfrac{a}{-1-(-a)} }
$}}\\\\
\small{\text{finally:~~
$
\mathbf{x=\dfrac{a}{-1+a} }
$}}\\\\
\small{\text{or:~~
$
\mathbf{x=\dfrac{a}{a-1} }
$}}\\\\$$

thanks to everyone)

You are welcome Sabi :)