Simplify this

Simplify $\frac{{1/3}(x-3)}{x^2-9}$

Here's my working out:

$\frac{{1/3}(x-3)}{x^2-9}$

$\frac {3x-9}{3x^2-27}$

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

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

$\frac {1}{x+3}$

The books answer is: $\frac{1}{3(x+3)}$

Where have I gone wrong? I can't seem to see my silly error?

Simplify $\frac{{1/3}(x-3)}{x^2-9}$

Here's my working out:

$\frac{{1/3}(x-3)}{x^2-9}$

Here is where you went wrong:
$\frac {3x-9}{3x^2-27}$

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

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

$\frac {1}{x+3}$

The books answer is: $\frac{1}{3(x+3)}$

Where have I gone wrong? I can't seem to see my silly error?

Where have I gone wrong? I can't seem to see my silly error?

Is that (1/3)(x-3), or (x-3)/3?
Assuming it's the second one, factorise x^2-9 and cancel.
In your working you seem to have multiplied the fracrion by 9/3 rather than 3/3.

Error on second line of working. You've multiplied the numerator by three as well as the denominator.

The bottom part of the fraction is the difference of two squares, it will factorise to give (x+3)(x-3), and the 3 from the 1/3 on the top part of the fraction can be written on the bottom of it. Then you can cancel the two (x-3)s from the top and bottom and you get the answer.

You are not multiplying the $\frac{1}{3}$ correctly.

You'll kick yourself

On the first step. You multiplied the bottom, and the top by 3.

Edit: Way too slow!

Is that (1/3)(x-3), or (x-3)/3?
Assuming it's the second one, factorise x^2-9 and cancel.
In your working you seem to have multiplied the fracrion by 9/3 rather than 3/3.

(1/3)(x-3) = (x-3)/3

The solution has already been posted but an easier way to do this question:

$\frac{\frac{1}{3}(x-3)}{x^2 - 9}$

Multiply top and bottom by 3 to get rid of our multi-layered fractions:

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

Difference of two squares on the bottom - factorise:

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

Divide both sides of the fraction by (x-3):

$\frac{1}{3(x+3)}$