Hi,

I need to get from

[(1-p)f+p(1-f)](1+v)-[(1-p)(1-f)+pf] = x

to (2+v)(f+p-2pf)-1 = x

but I'm stuck. I'd appreciate any tips on what I should I do after the following.

(f+p-2pf)(1+v) + (f + p - 2pf) - 1 = x

Thanks in advance.

Hi,

I need to get from

[(1-p)f+p(1-f)](1+v)-[(1-p)(1-f)+pf] = x

to (2+v)(f+p-2pf)-1 = x

but I'm stuck. I'd appreciate any tips on what I should I do after the following.

(f+p-2pf)(1+v) + (f + p - 2pf) - 1 = x

Thanks in advance.

I need to get from

[(1-p)f+p(1-f)](1+v)-[(1-p)(1-f)+pf] = x

to (2+v)(f+p-2pf)-1 = x

but I'm stuck. I'd appreciate any tips on what I should I do after the following.

(f+p-2pf)(1+v) + (f + p - 2pf) - 1 = x

Thanks in advance.

- stapel_eliz
**Posts:**1687**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

tobriain wrote:I need to get from

[(1-p)f+p(1-f)](1+v)-[(1-p)(1-f)+pf] = x

to (2+v)(f+p-2pf)-1 = x

but I'm stuck. I'd appreciate any tips on what I should I do after the following.

(f+p-2pf)(1+v) + (f + p - 2pf) - 1 = x

You've done the hard part. Now do the easy last step: Factor out the common expression from each of the first two "terms":

. . . . .(1 + v)(f + p - 2pf) + (1)(f + p - 2pf)

Simplify what's left, and you're done!