Fully simplify

$\displaystyle \frac{x^2+x-12}{x^2-9x+18}$

$\displaystyle \frac{x-12}{-9x+18}=0$?

$\displaystyle x-12=0*-9x+18=0$

$\displaystyle x=12$

Yes or no?

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- Jan 4th 2010, 02:27 AMMukilabBasic fractional equation
Fully simplify

$\displaystyle \frac{x^2+x-12}{x^2-9x+18}$

$\displaystyle \frac{x-12}{-9x+18}=0$?

$\displaystyle x-12=0*-9x+18=0$

$\displaystyle x=12$

Yes or no? - Jan 4th 2010, 02:36 AMearboth
Definitely No!

$\displaystyle \frac{x^2+x-12}{x^2-9x+18} = \frac{(x+4)(x-3)}{(x-6)(x-3)}$

1. Cancel the common factor.

2. Determine the domain of the term at the LHS.

3. Set the numerator equal to zero, solve for x.

By the way: The simplification isn't necessary here: Multiply both sides of the equation by the denominator after you've determined the domain of the complete term. - Jan 4th 2010, 02:37 AMmr fantastic
Sorry, but this is completely wrong. By your reasoning, you would simplify something like $\displaystyle \frac{x^2 - 4}{x^2 - 1}$ as $\displaystyle \frac{-4}{-1} = 4$.

You are expected to factorise the numerator and the denominator and then cancel the common factor.

Can you factorise $\displaystyle x^2 + x - 12$? Can you factorise $\displaystyle x^2 - 9x + 18$? - Jan 4th 2010, 02:40 AMMukilab
Thanks for the answer earboth ^^ worked it out now

- Jan 4th 2010, 04:22 AMdkaksl
- Jan 4th 2010, 04:26 AMe^(i*pi)
The domain is which input values give real output values. Essentially this boils down to a few simple rules of which you should use the first

The denominator can never equal 0

Values inside logarithms must be greater than 0

Values inside even roots (such as square root) must be greater than 0

In your case $\displaystyle \frac{(x+4)}{(x-6)}$ which means any value that makes $\displaystyle x-6 = 0$ is not part of the domain - Jan 4th 2010, 04:50 AMdkaksl
So... it's safe to assume the following; $\displaystyle x+4=0$ , $\displaystyle x+6\neq{0}$ ; $\displaystyle x=-4$?

then how does this work out?

Quote:

Originally Posted by**earboth**

$\displaystyle x^2-9x+18\neq{0}$ ; $\displaystyle x^2+x-12=0$

does this not grant two answers for $\displaystyle x$?

Edit: O wait, I think I got it now. The domain also includes $\displaystyle x-3\neq{0}$ which in the end means you do have to factorise both nominator and denominator of the original equation. - Jan 4th 2010, 04:58 AMe^(i*pi)
I actually missed out something in my answer and that is that 3 is not in the domain either.