This one is a bit tricky. First lets write it using all math symbols. I'll call the whole quantity y, and let j=x.
y=(2x+6)/(x^{2}-6x-27)
Now, ordinarily this might be quite the troublesome problem. Our hope for a simple solution depends on factoring the denominator. It is a quadratic, so unless we can "see" how it factors, we will need to use the quadratic formula.
For a general quadratic equation:
ax^{2}+bx+c=0
the solution is given by
x=-(b/2a)±sqrt((b/2a)^2-c/a)
where 'sqrt( )' is the square root of a quantity.
So we must substitute in our values of a, b, and c into this equation. This will give us the factors of the denominator.
The denominator reads
x^{2}-6x-27
so the values of a, b, and c are
a=1
b=-6
c=-27
Thus the value of x is given by
x=-(-6/2)±sqrt((-6/2)^2+27)
x=3±sqrt(9+27)
x=3±sqrt(36)
x=3±6
So the two values of x are
x_{1}=-3
x_{2}=9
This means that we can write
x^{2}-6x-27=(x+3)(x-9)
(you can check that if you multiply these out, they are indeed equal).
Now we can rewrite our quantity:
y=(2x+6)/((x+3)(x-9))
Now we can notice the the numerator can be transformed as follow:
(2x+6)=2(x+3)
Thus we can write
y=2(x+3)/((x+3)(x-9))
Now we can cancel out the x+3 !
y=2/(x-9)
This is our final answer, except we must make the note that x≠9. That is, we cannot have x=9, otherwise we are dividing by 0.
I guess rewriting using j we have
2/(j-9), with j≠9
Hope this helps.