# Writing Algebraic Expression as a Single Fraction

• Sep 28th 2010, 08:50 AM
BIOS
Writing Algebraic Expression as a Single Fraction
Need Help writing the following expression as a single fraction:

$\displaystyle \frac {10}{2x^2-3x-2}-\frac {2}{x-2}$

I've tried a few different ways but not getting near the answer in the book :?

Any help would be great!

Cheers

BIOS
• Sep 28th 2010, 08:57 AM
Plato
Factor: $2x^2-3x-2=(2x+1)(x-2)$.
• Sep 28th 2010, 09:05 AM
BIOS
Thanks for the reply Plato. I've tried that bit i think i'm messing up the step after that.

So say i have:

$\frac {10}{(2x-1)(x-2)}-\frac {2}{x-2}$

After that do i multiply like this:

$\frac {10}{(2x-1)(x-2)}-\frac {2(2x+1)}{(x-2)(2x+1)}$
• Sep 28th 2010, 09:06 AM
BIOS
Thanks for the reply Plato. I've tried that bit i think i'm messing up the step after that.

So say i have:

$\frac {10}{(2x-1)(x-2)}-\frac {2}{x-2}$

After that do i multiply like this:

$\displaystyle \frac {10}{(2x-1)(x-2)}-\frac {2(2x-1)}{(x-2)(2x-1)}$
• Sep 28th 2010, 09:08 AM
Plato
Quote:

Originally Posted by BIOS
Thanks for the reply Plato. I've tried that bit i think i'm messing up the step after that.

So say i have:

$\frac {10}{(2x-1)(x-2)}-\frac {2}{x-2}$

After that do i multiply like this:

$\frac {10}{(2x-1)(x-2)}-\frac {2(2x+1)}{(x-2)(2x+1)}$

It should be $\dfrac {10}{(2x+1)(x-2)}-\dfrac {2(2x+1)}{(2x+1)(x-2)}$
• Sep 28th 2010, 09:15 AM
BIOS
Yeah sorry edited that post right after i posted it! Okay so from here where do i go? Haven't done one like this before so unsure what exactly is the best method.
• Sep 28th 2010, 09:21 AM
Plato
It is correct now.
• Sep 28th 2010, 09:26 AM
BIOS
Okay yeah i can see i mistyped the factoring in the first denominator in the original post (the one you quoted). Still don't know where to go from there however. If i could see it done once that would be very helpful :)
• Sep 28th 2010, 09:49 AM
BIOS
Do i do this and then cancel:

$\displaystyle \frac {10-2(2x+1)}{(2x+1)(x-2)}$

?
• Sep 28th 2010, 09:54 AM
Plato
Quote:

Originally Posted by BIOS
Do i do this and then cancel:
$\displaystyle \frac {10-2(2x+1)}{(2x+1)(x-2)}$

Write it as $\displaystyle \frac {8-4x}{(2x+1)(x-2)}$
• Sep 28th 2010, 10:07 AM
BIOS
So once i have:

$\displaystyle \frac {8-4x}{(2x+1)(x-2)}$

Do i look for a common factor ala:

$\displaystyle \frac {8-4x}{1} * \frac {1}{(2x+1)(x-2)}
$

The answer in the book is:

$\displaystyle \frac {-4}{2x+1}$

Which i can't see how to get.

If i divide x -2 into the numerator i don't get the numerator in the books answer?

Obviously i'm missing something! Thanks for your help and patience so far.
• Sep 28th 2010, 03:27 PM
Quacky
Factorise the top:

$\displaystyle \frac {8-4x}{(2x+1)(x-2)}$

$=\displaystyle \frac {4(2-x)}{(2x+1)(x-2)}$

Now, take out a factor of -1 from the numerator:

$=\displaystyle \frac {-4(-2+x)}{(2x+1)(x-2)}$

$=\displaystyle \frac {-4(x-2)}{(2x+1)(x-2)}$

And I'm sure you can finish.
• Sep 29th 2010, 04:59 AM
BIOS
Thanks Quacky! I really needed that explained. Should be able to get through to rest of these now :D