What is his and how do I solve it?

1. Jul 26, 2012

Ashir

x^2+x-6
___________

x^2-7x+10

I need to know the name of an equation like this (someone said they were quadratic equations but those are only say the denominator or numerator of this fraction, so are they quadratic fractions or something?) so I can search it and revise it more. If I'm not mistaken, is the answer:

(x-2)(x+3)
_________

(x-5)(x-2)

Thanks a lot for the help, I need this for my GSCE's in November.

2. Jul 26, 2012

Charmar

I don't know what the equation is called, but I can see that your ansnwer can be simplified a little more, think about what on the top can cancel out something on the bottom, hope this helps.

3. Jul 26, 2012

Ashir

4. Jul 26, 2012

rollcast

I think the gcse spec calls those algebraic fractions.

General way to simplify them is to factor the numerator and denominator and then cancel terms that are the same in both.

Its a bit more complicated if you are adding, subtracting, multiplying or dividing 2 algebraic fractions but the same rules apply.

Factor and multiply to get common denominators and then work with like a normal fraction.

These 2 links should explain more:

http://www.mathsrevision.net/gcse/pages.php?page=1

PS. If you are going to type anything more complicated, try and learn a bit of Latex as maths notation written in plaintext gets confusing quite easily.

EG.

$\large\frac{x^{2}+x-6}{x^{2}-7x+10}$

5. Jul 26, 2012

Mentallic

They're called rational functions.

6. Jul 26, 2012

eumyang

It cannot be an equation, because there is no equal sign.

I think it's more accurate to call this a rational expression. To be a rational function, you would need a "f(x) = " in front, like this:
$f(x) = \frac{x^2 + x - 6}{x^2 - 7x + 10}$

7. Jul 26, 2012

rollcast

If you look at a GCSE maths spec you'll see those referenced to as "Algebraic fractions"

Although I would agree that their proper term would be a rational expression.

8. Jul 29, 2012

Ashir

Thank you.
So, how can I simplify it further from:

(x-2)(x+3)
_______________
(x-5)(x-2)

9. Jul 29, 2012

Ashir

I realize I can cancel out the -2's but let's say x=3. The expression would be equivalent to:

1+6 (7)
__________ = -7.
-2+1 (-1)

If we cancel the 1's, it is:

6
___ = -3
-2

So how could that work?

10. Jul 29, 2012

SammyS

Staff Emeritus
No, you can not cancel out the -2's.

You can cancel out the (x-2) from the numerator & denominator.

11. Jul 29, 2012

Ashir

Sorry that's what I meant.
How can you if the results would be different?

12. Jul 29, 2012

rollcast

The results aren't different for example:

if your fraction is $\frac{(x-2)(x+7)}{(3x+8)(x-2)}$

Then consider that if you follow the usual rules of arithmetic and do anything in brackets first, (x-2) will be the same no matter whether it is on the top or bottom of the fraction.

So you can cancel them out.

for example if x = 6 then (x-2) = (6-2) = 4.

so then your fraction could be rewritten as:

$\frac{(4)(x+7)}{(3x+8)(4)}$

Therefore since you have multiplied the denominator and numerator by 4 this will cancel out.

$\frac{(x+7)}{(3x+8)}$

Hopefully now you should see why they cancel.

Last edited: Jul 29, 2012
13. Jul 29, 2012

Ashir

Confusing. When did I multiply the numerator/denominator?

14. Jul 29, 2012

Ashir

Is the reason basically ratio-involving? So that, whether or not the x-2 was cancelled out or not the ratio would be, say, 3:2 anyways?

15. Jul 29, 2012

Villyer

In the very first step, when you rewrote $x^2+x-6$ as $(x-2)(x+3)$ and $x^2-7x+10$ as $(x-5)(x-2)$.

16. Jul 29, 2012

Ashir

Isn't that more dividing?

17. Jul 29, 2012

Villyer

It can be considered decomposing. Diving refers to $\frac{f(x)}{g(x)}$. Decomposing is more of taking $f(x)$ and rewriting it as $g(x)h(x)$, or taking a function and making it into the product of two functions.

18. Jul 29, 2012

gabbagabbahey

Why are you adding the two terms together in both your numerator and denominator? When you have one bracketed expression directly next to another, you multiply them together, you don't add them:

$$\frac{(3-2)(3+3)}{(3-5)(3-2)} =\frac{(1)(6)}{(-2)(1)}=\frac{6}{-2} = -3$$

19. Jul 30, 2012

Ashir

Oh just realized the mistake, thanks.