# Equations - Can't solve it!

• Jan 21st 2009, 08:22 AM
Pingu
Equations - 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 AM
Moo
Hello,
Quote:

Originally Posted by Pingu
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?

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 AM
hkerbest
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 AM
princess_21
Quote:

Originally Posted by Pingu
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?

you should factor $\displaystyle 2x-6$ first to get the common factor

$\displaystyle \frac{2(x-3)}{x-3}$

so the common factor is $\displaystyle x-3$

$\displaystyle \frac{3}{x-3}-\frac{1}{2x-6}=\frac{5}{6}$

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

$\displaystyle \frac{6}{2x-6}-\frac{1}{2x-6}=\frac{5}{6}$

$\displaystyle \frac{5}{2x-6}=\frac{5}{6}$

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

$\displaystyle 30=10x-30$

$\displaystyle 60=10x$

$\displaystyle x=6$

:)