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?

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- Jan 21st 2009, 08:22 AMPinguEquations - Can't solve it!
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? - Jan 21st 2009, 10:24 AMMoo
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. - Jan 22nd 2009, 02:11 AMhkerbest
Maybe you should do it like this.

$\displaystyle

\frac{6*3}{6(x-3)}-\frac{3*1}{3*2(x-3)}=\frac{5(x-3)}{6(x-3)}

$ - Jan 22nd 2009, 02:22 AMprincess_21
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} $

$\displaystyle 5(6)=5(2x-6)$

$\displaystyle 30=10x-30$

$\displaystyle 60=10x$

$\displaystyle x=6$

:)