# can someone help me start this?

• Sep 4th 2008, 11:22 PM
NotEinstein
can someone help me start this?
f(x)= 3+(12/(x-3)) over 2-16(x/x^2-9)
• Sep 5th 2008, 12:21 AM
Prove It
Quote:

Originally Posted by NotEinstein
f(x)= 3+(12/(x-3)) over 2-16(x/x^2-9)

$\displaystyle f(x) = \frac{3+\frac{12}{x-3}}{2 - 16\frac{x}{x^2 - 9}}$

Let's get the top and bottom to have the same denominator to start with...

$\displaystyle =\frac{\frac{3(x-3)+12}{x-3}}{\frac{2(x^2-9)-16x}{x^2-9}}$

And I don't like having fractions in a numerator and denominator, so let's write this as a product instead...

$\displaystyle =\frac{3(x-3)+12}{x-3}\times \frac{x^2-9}{2(x^2-9)-16x}$

Expand...

$\displaystyle =\frac{3x+3}{x-3}\times \frac{x^2-9}{2x^2-18-16x}$

Factorise everything possible...

$\displaystyle \frac{3(x+1)}{x-3}\times \frac{(x+3)(x-3)}{2(x+1)(x-9)}$

Cancelling everything possible leaves...

$\displaystyle \frac{3(x+3)}{2(x-9)}$

There you go...
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