Express \[ \frac{3}{2x + 3} - \frac{1}{2x - 3} + \frac{6}{4x^2 - 9} \] as a single fraction in its simplest form. - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 6

The common denominator for these fractions is ((2x - 3)(2x + 3)). Therefore, we can express each fraction as follows:

[ \frac{3(2x - 3)}{(2x + 3)(2x - 3)} - \frac{1(2x + 3)}{(2x - 3)(2x + 3)} + \frac{6}{(2x - 3)(2x + 3)} ]

Now that we have a common denominator, we can combine the fractions:

[ \frac{3(2x - 3) - (2x + 3) + 6}{(2x - 3)(2x + 3)} ]

This simplifies to:

[ \frac{6x - 9 - 2x - 3 + 6}{(2x - 3)(2x + 3)} = \frac{4x - 6}{(2x - 3)(2x + 3)} ]

The numerator can be factored further:

[ 4x - 6 = 2(2x - 3) ]

So our expression now looks like:

[ \frac{2(2x - 3)}{(2x - 3)(2x + 3)} ]

We can cancel ((2x - 3)) from the numerator and denominator (assuming (x \neq \frac{3}{2})), which gives us:

[ \frac{2}{2x + 3} ]