1. ## Simplifying Algebraic Fractions

I've given this a try but I don't think I'm doing it right. If someone could explain what the best way to go about this that would be great. Thanks in advance.

Question (simplify this fraction):
(x/x-2)-(10/(x+3)(x-2))

I tried to get a common denominator and expand out of the brackets but I have a feeling I'm going about this wrong.

I did:

(x^3+x^2-6x+10x-20)/(x^3+x^2-6x-2x^2-2x+12)
(x^3+x^2+4x-20)/(x^3-x^2-8x+12)
(4x-20)/(-8x+12)

However I'm pretty sure I'm doing it wrong.

Thanks!

2. Originally Posted by hiyarearl
I've given this a try but I don't think I'm doing it right. If someone could explain what the best way to go about this that would be great. Thanks in advance.

Question (simplify this fraction):
(x/x-2)-(10/(x+3)(x-2))

I tried to get a common denominator and expand out of the brackets but I have a feeling I'm going about this wrong.

I did:

(x^3+x^2-6x+10x-20)/(x^3+x^2-6x-2x^2-2x+12)
(x^3+x^2+4x-20)/(x^3-x^2-8x+12)
(4x-20)/(-8x+12)

However I'm pretty sure I'm doing it wrong.

Thanks!

$\frac{x}{x-2}-\frac{10}{(x+3)(x-2)}$

$=\frac{x^2+3x-10}{(x+3)(x-2)}$

$=\frac{(x+5)(x-2)}{(x+3)(x-2)}$

Do some canceling ..

3. Originally Posted by hiyarearl
I've given this a try but I don't think I'm doing it right. If someone could explain what the best way to go about this that would be great. Thanks in advance.

Question (simplify this fraction):
(x/x-2)-(10/(x+3)(x-2))

I tried to get a common denominator and expand out of the brackets but I have a feeling I'm going about this wrong.

I did:

(x^3+x^2-6x+10x-20)/(x^3+x^2-6x-2x^2-2x+12)
(x^3+x^2+4x-20)/(x^3-x^2-8x+12)
(4x-20)/(-8x+12)

However I'm pretty sure I'm doing it wrong.

Thanks!
Mathaddict, you beat me, but I've added a step at the beginning that might be useful, so I'll post it anyway.

$\frac{x}{x-2}-\frac{10}{(x+3)(x-2)}$

Your common denominator will be $(x+3)(x-2)$

$\frac{x(x+3)-10}{(x+3)(x-2)}$

$\frac{x^2+3x-10}{(x+3)(x-2)}$

$\frac{(x+5)(x-2)}{(x+3)(x-2)}$

$\frac{x+5}{x+3}$

4. Ah yes of course so:

as when subtracting normal fractions we find the common denominator here effectively we have:

$\frac{x(x+3)}{(x-2)(x+3)}-\frac{10}{(x+3)(x-2)}$

When multiplied by $(x+3)$ to achieve the common denominator.

I understand now - thank you two very much for your help.

Originally Posted by masters
Mathaddict, you beat me, but I've added a step at the beginning that might be useful, so I'll post it anyway.

$\frac{x}{x-2}-\frac{10}{(x+3)(x-2)}$

Your common denominator will be $(x+3)(x-2)$

$\frac{x(x+3)-10}{(x+3)(x-2)}$

$\frac{x^2+3x-10}{(x+3)(x-2)}$

$\frac{(x+5)(x-2)}{(x+3)(x-2)}$

$\frac{x+5}{x+3}$