college algebra help?

Question

im horrible at this

1 + 1/1-b
over
1 - 1/1+b

i typed it wrong, i guess. sorry, first time ive done this lol

[1+ (1/1-b)] / [1 - (1/1+b)]

im trying to make it look the same lol

1 + ( 1 )
(1-b)
/
1 - ( 1 )
( 1 + b)

and according to the key in the back of the book, yupchagee is right. im just trying to figure out how...

Answer 1

If I'm understanding the problem right it will be simplified to:
(1+1-b) / (1-1+b) which in turn will be further simplified to -b / b which will ultimately give you an answer of -1. I hope this helps!

Answer 2

Change both 1's so you have common denominators:
(1-b)/(1-b) + 1/(1-b) OVER (1+b)/(1+b) - 1/(1+b)

The top part becomes 2 - b / 1 - b
The bottom part becomes b / 1 + b

When you divide a fraction, you take the reciprocal of the bottom one and multiply, so you'd have (2 - b) / (1 - b) * (1 + b)/b

Multiply and you get [(2 - b)(1 + b)]/[b(1 - b)]

When I foil the top and distribute the bottom I get (2 + b - b^2)/(b - b^2)

Turn those around so you have the highest exponent first:
(-b^2 + b + 2) / (-b^2 + b)

To get rid of the negative signs on the first terms, multiply both by -1 and you get -1(b^2 - b - 2) / -1(b^2 - b)

The negative 1 on top and bottom cancel out.

You can write your final answer as (b^2 - b - 2) / (b^2 - b) OR factor them and write it as [(b - 2)(b + 1)]/[b(b - 1)]

Answer 3

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

Answer 4

[1 + 1/(1 - b)]/[1 - 1/(1 + b)]

Multiply (1 + b)(1 - b) to numerator and denominator
= [(1 + b)(1 - b) + (1 + b)]/[(1 + b)(1 - b) - (1 - b)]

Distribute
= [1 - b² + 1 + b]/[1 - b² - 1 + b]

combine like terms, and multiply -1/-1
= (b² - b - 2)/(b² - b)

Factor both numerator and denominator
= (b - 2)(b + 1)/b(b - 1)

Since there is no common factor, then the final answer may be the last expression, or
= (b² - b - 2)/b(b - 1)

^_^

Answer 5

Numerator (N)
1 + 1 / (1 - b)
= [ (1 - b) + 1] / (1 - b)
= (2 - b) / (1 - b)
Denominator (N)
1 - 1 / (1 + b)
= [(1 + b) - 1] / (1 + b)
= b / (1 + b)
N/D = [(2 - b) / (1 - b)] x [(1+b) / b]
N/D = [(2 - b).(1 + b)] / [ b.(1 - b) ]
Nothing cancels unfortunately!

Answer 6

I think you are missing some ( ).

(1+1/(1-b))/(1-1/(1+b)) multiply numerator & denominator by 1-b²

(1-b²+1+b)/(1-b²-(1-b))
(-b²+b+2)/(-b²+b)
((-b+2)(b+1))/(b(-b+1))
((2-b)(b+1))/((b(1-b)) I think this is as far as you can go with this.

Answer 7

CAN BE CLEARLY EXPRESSED AS:

((1+1)/(1-b))/((1-1)/(1+b))

SIMPLIFIED TO:

(2/(-b+1))/((0/(b+1)))

(2/(-b+1))/0

DIVISION BY ZERO ERROR!

:)

ARE YOU SURE YOU DON'T HAVE ANY PARENTHESIS?

WITHOUT ANY PARENTHESIS:

(1 + 1/1 - b) / (1 - 1/1 + b)

(1 + 1 - b) / (1 - 1 + b)

(2 - b) / b

(-b + 2) / b

USING THIS EQUATION:

(1 + (1/1 - b)) / (1 - (1/1 + b))

(1 + (1 - b)) / (1 - (1+b))

(1 + (-b + 1)) / (1 - (b + 1))

(1 - b + 1) / (1 - b - 1)

(-b +1 +1) / (-b +1 -1)

(-b + 2) / (-b)

-(-b + 2) / b

(b - 2) / b

Answer 8

positive times a negative is negative
1 /1 is 1 and a negative one times one is negative one (-1)

Answer 9

what are you trying to do enplane it seems that every thing cancels out. to b/b and this =1

Answer 10

What are you trying to do? Simplify?