^{2 }- 6j - 27

2j+6 over j^{2 }- 6j - 27

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2 (j + 3)

(J - 9)(J + 3)

this simplifies to

2 / (j - 9)

the value 9 for j is not allowed because dividing by zero is not permitted

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.

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