Here's the thing:
$\displaystyle
\frac{3}{x-3}-\frac{1}{2x-6}=\frac{5}{6}
$
For this to add up, I need the bottom(?) to be the same, but how can I do that?
Hello,
Note that $\displaystyle 2x-6=2(x-3)$
So $\displaystyle \frac{3}{x-3}-\frac{1}{2x-6}=\frac{3}{x-3}-\frac{1/2}{x-3}$
And this easily adds up
In general, it's not that simple, you have to multiply both fractions by a factor such that the denominator becomes the highest common factor between the denominators.
you should factor $\displaystyle 2x-6$ first to get the common factor
$\displaystyle \frac{2(x-3)}{x-3}$
so the common factor is $\displaystyle x-3$
$\displaystyle \frac{3}{x-3}-\frac{1}{2x-6}=\frac{5}{6} $
$\displaystyle \frac{3*2}{2(x-3)}-\frac{1}{2x-6}=\frac{5}{6} $
$\displaystyle \frac{6}{2x-6}-\frac{1}{2x-6}=\frac{5}{6} $
$\displaystyle \frac{5}{2x-6}=\frac{5}{6} $
$\displaystyle 5(6)=5(2x-6)$
$\displaystyle 30=10x-30$
$\displaystyle 60=10x$
$\displaystyle x=6$