2j+6 over j

^{2 }- 6j - 272j+6 over j^{2 }- 6j - 27

Tutors, please sign in to answer this question.

North Providence, RI

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

Seattle, WA

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.

Mark D.

Quantitative research SPECIALIST with academic/professional experience

New York, NY

5.0
(148 ratings)

John P.

Tutor of math and physics, recent college graduate

Short Hills, NJ

5.0
(21 ratings)

- Algebra 2 3444
- Algebra 5022
- Math 9726
- Prealgebra 170
- Algebra Help 965
- Geometry 1876
- Algebra Word Problem 2463
- Math Help 5393
- Word Problem 5059
- Precalculus 1531