So we start with this equation:
[tex]\frac{(3x-2)}{(x^{2}-2x-3)} - \frac{1}{(x-3)} [/tex]
For part A, you need to factor the denominator of the first fraction, which is this:
[tex] x^{2} -2x-3[/tex]
To factor it, we need to find two numbers that add up to '-2' but multiply to '-3'. The only numbers that satisfy this are '-3' and '1', so we can now factor:
[tex]x^{2} -2x-3=(x-3)(x+1)[/tex]
And there's the answer for part A.
I believe for part B, all you need to do is clarify that [tex](x-3)(x+1)[/tex] is the LCD (let me know if you need to do anything else, and I'll be happy to help.)
For part C, we need to make it so that both fractions have the LCD as the denominator. Since the second fraction's denominator is already [tex](x-3)[/tex], we just need to multiply the second fraction by [tex] \frac{(x+1)}{(x+1)} [/tex]:
[tex] \frac{1}{(x-3)}* \frac{(x+1)}{(x+1)} = \frac{(x+1)}{(x-3)(x+1)} [/tex]
And that's your answer for part C.
For part D, we just need to subtract the second fraction from the first:
[tex] \frac{(3x-2)}{(x-3)(x+1)}- \frac{(x+1)}{(x-3)(x+1)} = \frac{3x-2-(x+1)}{(x-3)(x+1)} [/tex]
However, we can still simplify the numerator that results, by distributing the '-' in front of (x+1):
[tex]3x-2-(x+1)=3x-2-x-1=2x-3[/tex]
We end up with:
[tex] \frac{2x-3}{(x-3)(x+1)} [/tex]
I'm not sure if your teacher wants you to simplify any further, so I'm going to leave it there :)
Hope this helped, let me know if you have any questions! :D
