Math help desperetly needed?

Question

Simplify:
1/(x^2-4 ) + 2/(3x-6)

Need to know answer and why?

This is to help study for an upcoming exam, not homework

Answer 1

First, lets factor x^2-4. It turns out that this is equal to
(x+2)(x-2)

Let's also factor (3x-6). By pulling out the 3, you get 3(x-2)

So, the lowest common denomninator is 3(x+2)(x-2)

So, for the first term, multiply by 3/3 and you get:

3/3(x+2)(x-2)

For the second term, multiply by (x+2)/(x+2)
This will yield:
2(x+2)/3(x+2)(x-2) or 2x+4/3(x+2)(x-2)

Now, add the two terms and you get:

2x+4+3/3(x+2)(x-2) = 2x+7/3(x+2)(x-2)

Answer 2

Your problem 1/(x^2-4 ) + 2/(3x-6)
which simplifies to ANSWER: (2x+7) / [ 3 (x-2)(x+2)]

is an addition. You must have a common denominator before you can add.

The first fraction's denominator factors into (x-2)(x+2).

Your second fraction's denominator factors into 3(x-2).

So your common denominator is the denominator 3(x-2)(x+2). To change the first fraction to this common denominator you'd multiply top and bottom by 3. To change the second fraction's denominator to this common denominator you'd multiply top and bottom by (x+2); so you end up with

3/[3(x-2)(x+2)] + 2(x+2)/[3(x-2)(x+2)] = the ANSWER provided in the first sentence at top.

Answer 3

Must factor both sides to get

1/((x - 2)(x + 2)) + 2/(3 (x - 2))

Notice that there is a common factor that can be removed using the distributive property to become

1/(x-2)(1/(x+2) + 2/3)

To be honest I am not sure that this is any simpler than the original, but it depends on what the teacher is looking for.

Answer 4

1/(x^2 - 4) + 2/(3x - 6) can be written as

1/(x + 2) (x - 2) + 2/3(x - 2)

= 1/(x - 2) [1/(x + 2) + 2/3]

= 1/(x - 2) [(3 + 2x + 4) / 3(x + 2)]

= (2x + 7) / (x - 2) ( x + 2)

= 2x + 7 / x^2 - 4

Answer 5

1/(x²-4) factors into 1/((x+2)*(x-2) and
2/(3x-6) can be written as 2/(3*(x-2)). Now you can factor a 1/(x-2) out of each term to get
(1/(x-2))*((1/(x+2)) + 2/3)

Doug

Answer 6

1/(x^2 - 4) + 2/(3x - 6)

1/[(x - 2)*(x + 2)] + 2/[3*(x - 2)]

1/(x - 2)*1/(x + 2) + 2/3*1/(x - 2)

1/(x - 2)*[1/(x + 2) + 2/3]

1/(x - 2)*[1/(x + 2) + 2/3*(x + 2)/(x + 2)

1/(x - 2)*[1 + 2/3x + 4/3]/(x + 2)

1/(x - 2)*[7/3 + 2/3x]/(x + 2)

(2/3x + 7/3)*[(x - 2)*(x + 2)

(2x + 7)/[3*(x^2 - 4)

Answer 7

1/(x^2-4) + 2/(3x-6)

=1/(x^2-2^2) + 2/[3(x-2)]

= 1/[(x-2)x(x+2)] + 2/[3(x-2)]

=1/(x-2) X { 1/(x+2) + 2/3 }


= 1/(x-2) X { [3+2(x+2)] / 3 (x+2) }

= 1/ (x-2) X { [2x+7] / 3[x+2]}

= [2x+7] / {3(x-2)(x+2)}

= [2x+7] / {3(x^2-4)}

Answer 8

2x+7/ 3(x+2)(x-2)

Need to factor both denominators and then get common denominator that you see in answer

Answer 9

(2x^2 + 3x - 14) / (3x^3 - 6x^2 - 12x + 24)

.....That's about how far i can go, u on your own from there, if its correct.

Answer 10

take out common factors..

first equations can be turned into a perfect square:
1
--------------------------------
(x+2)(x-2)

the second equation can be simplified:
2
-------------------------------
3(x-2)

now cross multiply
and simplify...

voila!