Math Problem - I'm drawing a blank....

I'm hoping someone can help here. I'm drawing a complete blank as to why you get this answer:

1/2[1/(s-b)] + 1/2[1/(s+b)] = 1/2[(2s)/(s^2-b^2)] = s/(s^2-b^2)

I'm good with the denominator but for the life of me I can't rationalize why you get "s" in the numerator.

Any help is appreciated.

-Keith

Probably all wrong since the last math class I took was in 1963.

1/2 of 2s = s

Alrighty then. I'll see if I can help you out with this.

Hal is right. What you get is 2s/2(s^2-b^2), then 2 cancels out, leaving you with s/(s^2-b^2)

EDIT: See below post for intelligent, useful answer.

You multiply the whole thing by S/S (which = 1). That's how you get it in the numerator. Right?

He wants to know how you come about with S in the numerator to begin with.

confuZion3 wrote: You multiply the whole thing by S/S (which = 1). That's how you get it in the numerator. Right?

Actually you have to multiply the 2 denominators together:

(s+b)(s-b) = s^2-sb + bs - b^2 = s^2 - b^2

I can't remember what you do to the numerators tho and how you wind up with 2s

The idea is you can multiply any one part of an equation by 1 and it doesn't change the equation.

In this case you multiply 1/2(s-b) by (s+b)/(s+b) and 1/2(s+b) by (s-b)/(s-b)

you now have (s+b)/2(s^2 - b^2) + (s-b)/2(s^2 - b^2)

which is the same as (s+b+s-b)/2(s^2 - b^2)

you can probably get it from there

Okay, I figured it out, I think.......

when you multiply the denom of the one you have to multiply the num of that one also thus you wind up with....

[(s+b)/s^2-b^2] + [(s-b)/s^2-b^2] = (s+b+s-b)/(s^2-b^2)

The b terms in the num cancel and you are left with 2s/s^2-b^2

I think that's how it works.

Like I said, my nerves are just getting frazzled

Bingo, you got it.

mith612 wrote: The idea is you can multiply any one part of an equation by 1 and it doesn't change the equation. In this case you multiply 1/2(s-b) by (s+b)/(s+b) and 1/2(s+b) by (s-b)/(s-b) you now have (s+b)/2(s^2 - b^2) + (s-b)/2(s^2 - b^2) which is the same as (s+b+s-b)/2(s^2 - b^2) you can probably get it from there

Yep. Thanks. For the life of me could not think of this.

yeah....................

42..........oh wait

Miata. Isn't the answer Miata? Or E30? That's kind of a number.

P71!!

Oooo Math fun. Can we have another?

I think this problem was mostly bs.

I can't believe I was first with that line.

Morons.

The answer is Mx5

billy3esq wrote: I think this problem was mostly bs. I can't believe I was first with that line.

Actually, differential equations is bs

CamaroKeith wrote:

billy3esq wrote: I think this problem was mostly bs. I can't believe I was first with that line.

Actually, differential equations is bs

(N-M) (M-N) What is so hard about that?

I loved DifeScrew after I ahd to take Transfermation of Linear Maticies.

Uhhhhh, Miata?

Ketchup

John Brown wrote: Morons. The answer is Mx5