How would you add this rational expression?

Question

a - b
-----------------------
a2 + 2ab - 3b2

plus

a + b
---------------------
a2 - 2ab - 3b2

a2 means a squared, b2 means b squared

The answer in my textbooks says:

2a
------------------------
(a - 3b)(a + 3b)

I don't how one would get that
Could someone show me?
Please and thanks

Answer 1

Yes, I can.

First factor the denominators of both items you wish to add together...

The denominators become:

(a-b)(a+3b) AND (a+b)(a-3b)

Now, if you wish to add them you have to make the
denominators of both of them exactly the same...
So...

multiply the numerator and denominator of first phrase by (a+b)(a-3b)

multiply the numerator and denominator of the second
phrase by
(a-b)(a+3b)

This gets kind of long to write out but you should get:

(a-b)(a+b)(a-3b) + (a+b)(a-b)(a+3b)
- -- - - - - - - - - - - - - - - - - - - - - - - - -
(a-b) (a+b) (a+3b)(a-3b)


And you can see that you can cancel (a-b) out of the whole thing.

You can also cancel (a+b) out of the whole thing.

Which leaves you with:

(a-3b) + (a+3b)
--------------------
(a+3b) (a-3b)

combine the items in the numerator.......

and you are done...

2a
-----------------
(a+3b) (a-3b)


Cheers,
Zah

Answer 2

"a2 + 2ab - 3b2" factors to (a+3b)(a-b)

"a2 - 2ab - 3b2" factors to (a-3b)(a+b)

so in the first, (a-b) drops out and you get 1/(a+3b)

in the second, (a+b) drops out and you get 1/(a-3b)

so now you go for a common denominator...

and you get a common denominator of (a+3b)(a-3b)

numerator is a - 3b + a + 3b

so fraction is

a - 3b + a + 3b
------------------
(a+3b)(a-3b)

simplify to:

2a
------------------
(a+3b)(a-3b)

Answer 3

so long because the denominators are equivalent, you could combine words in the numerator. that's a similar as once you're coping with fractions. 2/3 + 2/3 = (2+2)/3 = 4/3 2x/3 + 2/3 = (2x+2)/3 7x/3 + x/3 = 8x/3 So, #a million is 9x/13 and #2 is 9x/15, that may be decreased to 3x/5.

Answer 4

(a^2 + 2ab - 3b^2) factors to (a+3b)(a-b)
(a^2 - 2ab - 3b^2) factors to (a-3b)(a+b)

so your first equation simplifies to: 1 / (a+3b)
and your second equation simplifies to: 1 / (a-3b)

so adding them together gives you:

[(a-3b) / (a+3b)(a-3b)] + [(a+3b) / (a+3b/a-3b)]

Which is equal to:
[(a-3b) + (a+3b)] / [ (a+3b)(a-3b) ]

Which is equal to:
2a / [ (a+3b)(a-3b) ]

Answer 5

a-b a+b
----------------- + ----------------------
a2+2ab-3b2 a2-2ab-3b2

just get the factors....

(a-b) (a+b)
--------------------- + ----------------------
(a+3b)(a-b) (a-3b)(a+b)

cancel common....

1 1
--------------------- + -----------------------
(a+3b) (a-3b)


get common factor by mulitplying

(a-3b) + (a+3b)
-----------------------
(a+3b) (a-3b)


then simple addtion, subtraction

you'll get:

2a
-------------------------
(a-3b) (a+3b)

Answer 6

Like this:
[(a-b)/(a-b)(a+3b)]+[(a+b)/(a+b)(a-3b)]=
[1/(a+3b)]+[1/(a-3b)]=
[(a-3b)/(a+3b)(a-3b)]+[(a+3b)/(a-3b)(a+3b)]=
(a-3b+a+3b)/(a²-9b²)=
2a/(a²-9b²)