Basic fractional equation

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January 4th 2010, 02:27 AM
Mukilab

Basic fractional equation

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

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

Yes or no?
January 4th 2010, 02:36 AM
earboth

Quote:

Originally Posted by Mukilab

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

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

Yes or no?

Definitely No!

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

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.
January 4th 2010, 02:37 AM
mr fantastic

Quote:

Originally Posted by Mukilab

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

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

Yes or no?

Sorry, but this is completely wrong. By your reasoning, you would simplify something like $\frac{x^2 - 4}{x^2 - 1}$ as $\frac{-4}{-1} = 4$ .

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

Can you factorise $x^2 + x - 12$ ? Can you factorise $x^2 - 9x + 18$ ?
January 4th 2010, 02:40 AM
Mukilab

Thanks for the answer earboth ^^ worked it out now
January 4th 2010, 04:22 AM
dkaksl

Quote:

Originally Posted by earboth

Determine the domain of the term at the LHS.

[...]

Multiply both sides of the equation by the denominator after you've determined the domain of the complete term.

Hey, I have no clue what the domain is. Tried wikipedia but it didn't help too much. Mind clarifying?
January 4th 2010, 04:26 AM
e^(i*pi)

Quote:

Originally Posted by dkaksl

Hey, I have no clue what the domain is. Tried wikipedia but it didn't help too much. Mind clarifying?

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 $\frac{(x+4)}{(x-6)}$ which means any value that makes $x-6 = 0$ is not part of the domain
January 4th 2010, 04:50 AM
dkaksl

So... it's safe to assume the following; $x+4=0$ , $x+6\neq{0}$ ; $x=-4$ ?

then how does this work out?

Quote:

Originally Posted by earboth

Multiply both sides of the equation by the denominator after you've determined the domain of the complete term.

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

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

does this not grant two answers for $x$ ?

Edit: O wait, I think I got it now. The domain also includes $x-3\neq{0}$ which in the end means you do have to factorise both nominator and denominator of the original equation.
January 4th 2010, 04:58 AM
e^(i*pi)

I actually missed out something in my answer and that is that 3 is not in the domain either.