1. ## equation

$\displaystyle a^2(x-1)-9=a(3x-6)$
$\displaystyle x*a(a-3)=a^2-6a+9$
$\displaystyle x*a(a-3)=(a-3)^2$
$\displaystyle x=\frac{(a-3)^2}{a(a-3)}$

Now... I am sure that solutioin is not defined for a=0, but can I reduce the fraction to:
$\displaystyle x=\frac{(a-3)}{a}$,
without any notes, or do I have to state that reduction of fraction is only posible for $\displaystyle a$ not equal 3? Or in other words, is a=3 also a solution?

2. Originally Posted by riki_haj
$\displaystyle a^2(x-1)-9=a(3x-6)$
$\displaystyle x*a(a-3)=a^2-6a+9$
$\displaystyle x*a(a-3)=(a-3)^2$
$\displaystyle x=\frac{(a-3)^2}{a(a-3)}$

Now... I am sure that solutioin is not defined for a=0, but can I reduce the fraction to:
$\displaystyle x=\frac{(a-3)}{a}$,
without any notes, or do I have to state that reduction of fraction is only posible for $\displaystyle a$ not equal 3? Or in other words, is a=3 also a solution?
Look at the original equation, you will see that it is inconsistent when a=3.
So you know that if there is a solution a!=3, and you can do the cancellation.
But you should say so.

RonL