Help! t-1 over 1-t squared.....?

Question

They want it simplified, and their answer is -1 over 1+t and I have no idea how they got it... I don't know if it is 1-(t squared) or (1-t)squared. I ended up with -1 over 1-t, not 1+t . What am I doing wrong?

Answer 1

t - 1
------ = ........... NOTE: 1 - t² = 1² - t² = (1 - t)(1 + t) ≠(1 - t)²
1 - t²

-1(1 - t)
--------------- = ........... NOTE: Now divide top and bottom by (1 - t)
(1 - t)(1 + t)

-1
------ ..................... QED
1 + t

Answer 2

It's definitely -1/1+t if its assuming 1-(t squared) on the bottom as you simply and you get :

1
---------
-1(t+1)

as the (-t+1) cancels out with the (1-t) on the bottom...
hope that made sense!

Answer 3

(t-1) / 1-t^2 = [ I am assuming it is 1-(t squared)]
t -1 / (1-t)(1+t) [ This is a conversion from a difference of squares]
-(1-t)/-(1-t)(1+t) [ Multiply numerator and denominator by -1]
-1 /-(1+t) [ cancelled common factor (1-t)]
1/(1+t) [Multiply numerator and denominator by -1]

Now assuming it is (1-t) squared:
(t-1)/(1-t)^2
-(t-1)/-(1-t)^2 [Multiply numerator and denominator by -1]
(1-t)/-(1-t)^2 [Simplifying the numerator]
1/-(1-t)^1 [Cancelling (1-t) from numerator and denominator
1/(t-1) or -1/(1-t)

As you can see, from both assumptions, none of the expected answer shows up.

Answer 4

t-1 over 1-t² =
t -1 over (t -1)(t+1) =
Cancelling t-1 by t-1:
1 over t+1
>><

Answer 5

(t - 1)/(1 - t)

*Rewrite the denominator in alpha descending order > the variable "t" is first:

(t - 1)/(-t + 1)

First: factor the denominator > factor out a (-1):

(t - 1)/[-1(t - 1)]

Second: cancel "like" terms > cancel both sets of (t - 1):

= 1/-1
= -1

Answer 6

(t - 1)/(1 - t^2)
(t - 1)/(-t^2 + 1)
(t - 1)/(-(t^2 - 1))
-(t - 1)/(t^2 - 1)
-(t - 1)/((t - 1)(t + 1))
-1/(t + 1)

ANS : -1/(t + 1)