i need to add (a - 1/a + 1) + (a + 1/a - 1)?

Answer 1

Since this is under "Programming and Design" and not "Mathematics" I'll give you a suggestion on how to program it:

Declare answer as integer = 0;
Declare a as integer = 0;

a = Integer.convert(**input value from a form,string)
answer = (a-1/a+1)+(a+1/a-1)

WriteLine(answer);

But if you're looking for a mathematical solution the answers we're all getting are more or less the same.

(a-1) + (a+1)
------- --------
(a+1) (a-1)

LCD = (a+1)(a-1)

(a-1) * (a-1) + (a+1) * (a+1)
------ ------ ------ ------
(a+1) (a-1) (a-1) (a+1)


(a^2 - a - a +1) + (a^2 +a +a + 1)
---------------------------------------------
(a+1)(a-1)


2a^2 +2
------------
(a+1)(a-1)


2(a^2+1)
------------
(a+1)(a-1)


2(a^2+1)
------------
a^2-1

Answer 2

So plug in the number for A, then write out the equation. Mutliply the denominators by each other to find the common denominator. Then multiply the numerators by the opposite denominator. Then add them up.

For example: If a = 5, you would do this:

4/5 +5/4 = 16/20 + 25/20 = 41/20 or 2 and 1/20.

Update: Argh, my answer was totally wrong. Let me try again.

If a = 5, you would do this:

4/6 +6/4 = 2/3 + 3/2, now solve for the common denominator
= 4/6 + 9/6 = 13/6 or 2 1/6.

How some of the others got 2a, I'm not sure, but it doesn't work when you plug in an actual number for a and then solve.

Answer 3

I assume that you mean add (a - 1)/(a + 1) + (a + 1)/(a - 1)

In this case, the result is: 2*(a^2+1)/(a^2-1)

Answer 4

The answer is 2a.

Remember MDAS? Since your parens are around the main parts first, they really aren't needed and the equasion becomes equally:

a - 1/a +1 + a + 1/a - 1

And reducing, we get

a + a or 2a

(-1/a and +1/a cancel each other, +1 and -1 cancel each other and your left with a + a.)

I think - it's been YEARS since I've done algebra. :)

Answer 5

here are the steps:

(a-1)/(a+1) + (a+1)/(a-1)

= [(a-1)^2+(a+1)^2] / [(a+1)(a-1)]

(expand numerator and keep the denominator as it is)

= (2a^2 + 2)/[(a+1)(a-1)]

--- is the answer

Answer 6

treat it like a fraction. so, you need to set it up with a common denominator:
Note - this answer assumes that the original problem was two fractions as opposed to the version you have written i.e.,
(a-1)/(a+1) + (a+1)/(a-1)

(a - 1)......(a + 1)
--------- + --------
(a + 1).....(a - 1)


...(a-1) * (a-1).....(a+1) * (a+1)
= ----------------- + -------------------
...(a+1) * (a-1).....(a+1) * (a-1)

...(a^2 - 2a +1) + (a^2 + 2a +1)
= -----------------------------------
..........(a^2 + a - a - 1)

...(a^2 +2)
= ------------
...(a^2 - 1)

Answer 7

(a -1/a+1) The answer is 2a
(a+1/a -1)
2a

Answer 8

its a very simple.....may be u know how to add it....but asked this just for fun :)
well.........take the LCM of denominators which is (a*a - 1 ) and then divide this by each denominator and then multiply it to its numerator and then add both values.....

Answer 9

ok, so do it...