Simplify the expression, state excluded values if possible

Question

2j+6 over j² - 6j - 27

Answer 1

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

Answer 2

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)/(x²-6x-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:

 

ax²+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

 

x²-6x-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

 

x₁=-3

x₂=9

 

This means that we can write

 

x²-6x-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.