Question 178160
It is not an equation.
It is an expression and you want to simplify it.
{{{x/(x^2-2x-3)-(3x)/(x^2-9)}}} 
The expressions are both fractions.
When subtracting (or adding) fractions, you need to have a common denominator. 
Let's look to see if we can factor the denominators first to simplify this task.
{{{x^2-2x-3=(x-3)(x+1)}}}
and 
{{{x^2-9)=(x-3)(x+3)}}}
OK, let's make this substitution,
{{{x/(x^2-2x-3)-(3x)/(x^2-9)=x/((x-3)(x+1))-(3x)/((x-3)(x+3))}}}
The common denominator is 
{{{(x-3)(x+3)(x+1)}}}
You need to multiply the first term by {{{ (x+3)/(x+3) }}} and the second term by {{{ (x+1)/(x+1) }}} to get the common denominators. 
{{{x/(x^2-2x-3)-(3x)/(x^2-9)=(x/((x-3)(x+1)))((x+3)/(x+3))-((3x)/((x-3)(x+3)))((x+1)/(x+1))}}}
{{{x/(x^2-2x-3)-(3x)/(x^2-9)=(x(x+3))/((x-3)(x+1)(x+3))-((3x)(x+1))/((x-3)(x+3)(x+1))}}}
{{{x/(x^2-2x-3)-(3x)/(x^2-9)=((x(x+3))-((3x)(x+1)))/((x-3)(x+3)(x+1))}}}
{{{x/(x^2-2x-3)-(3x)/(x^2-9)=((x^2+3x)-(3x^2+3x))/((x-3)(x+3)(x+1))}}}
{{{x/(x^2-2x-3)-(3x)/(x^2-9)=(-2x^2)/((x-3)(x+3)(x+1))}}}
That's the final answer, it's as simplified as it can be.