1/a-b - 1/a+b?

Question

How do i combine and simplify this algebra question?

Answer 1

multiply the two denomenator to get its denomenator. then, cross multiply the numerator of the first term to the denomenator of the second term, and also, cross multiply the numerator of the second term to the denomenator of the first term.... try to find out the commen variables of the numerator, after that try to cancel those term that are similar in both numerator and denomenator!

i hope that i'mble to help you!

Answer 2

1/a-b -1/a+b =0/(a-b)(a+b)=0

Answer 3

multiply the first term by (a+b)/(a+b) and multiply the second term by (a-b)/(a-b). You should be left with (a+b)/(a-b)(a+b) - (a-b)/(a+b)(a-b). Subtract the numerators (don't forget to distribute the negative), and FOIL the denominators. You should be left with:

2b/(a^2-b^2)

Answer 4

Answer: 2b/(a^2 - b^2)

Reason:
Multiply the equation by (a+b)*(a-b)/(a+b)*(a-b). Notice that for any a and b, this equation is equal to one.

So, ((a+b)*(a-b)/(a+b)*(a-b))*
(((1/(a+b))-(1/(a-b))) =
((a+b)*(a-b)/(a+b)*(a-b)*(a-b))-
((a+b)*(a-b)/(a+b)*(a-b)*(a-b)) =
((a+b)/(a+b)*(a-b))-
((a-b)/(a+b)*(a-b)) =
(a+b)-(a-b)/(a+b)*(a-b) =
a+b-a+b/(a+b)*(a-b) =
2b/(a+b)*(a-b) =
2b/(a^2 -ab + ab -b^2) =
2b/(a^2 - b^2)

where ^2 = to the second power

Answer 5

First, you find the LCD. it is (a-b)(a+b)

Then, you have to multiply numerators by the factor u multiplied the denominator by to find the LCD:

(a+b) / (a-b)(a+b) - (a-b) / (a-b)(a+b)

subtract the numerator and
ur answer is:

2b / (a-b)(a+b)

OR

2b / (a^2 - b^2)

Answer 6

1/(a-b) -1/(a+b)
=(a+b)-(a-b)/(a-b)(a+b)
=(a+b-a+b)/a^2-b^2
2b/(a^2-b^2) ans

Answer 7

= 0
no thats wrong it would be = -b