Algebra help ? (Fractions)?

Question

(-2x^-1y^-2)^-2 / (x^-4y^2) times (x^4y^2 / x^2y^-1)^-1 ?

Answer 1

I'll rewrite this with a little more spacing and bracketing to clarify, and hopefully I understand this the way you meant it:

[(-2 x^-1 y^-2)^-2 / (x^-4 y^2)] . [(x^4 y^2) / (x^2 y^-1)]^-1

OK, first thing to do is use the rules (xy)^a = (x^a)(y^a) and (x^a)^b = x^(ab) wherever a bracketed expression is raised to a power. As an example, (x^2 y^-3)^-2 = (x^-4)(y^6). This gives us

[((-2)^-2 x^2 y^4) / (x^-4 y^2)] . [(x^-4 y^-2) / (x^-2 y^1)].

Now, use the rule that 1/x^n = x^-n to bring everything to the top:

[(-2)^-2 x^2 y^4 . x^4 y^-2] . [x^-4 y^-2 . x^2 y^-1]

Combine like bases on each side using the rule x^(a+b) = x^a . x^b, in reverse:

[(-2)^-2 x^6 y^2] . [x^-2 y^-3]

And again:

(-2)^-2 x^4 y^-1

Or in a more natural format:

x^4 / (4y)

Answer 2

First, multiply indices where we have a power raised to a power. This the first numerator becomes
(-2)^(-2)*(x^2)*(y^4), and the second part, the parentheses raised to the power -1, becomes
(x^(-4)*y^(-2)/(x^(-2)*y^1)

Now get rid of those negative indices. Easy, as the numerators and denominators are all products, so all we have to do is make each negative index positive and switch it and its base from bottom to top or vice versa. e.g. for the first fraction, take (-2)^(-2) to the bottom as (-2)^2, and at the same time, move x^(-4) from the bottom to the top as x^4:
So it's
(x^4)*(x^2)*(y^4)/((-2)^(2)*y^(2)), which reduces to

(x^6)(y^4)/(4y^2), or (x^6)(y^2)/4

The second part becomes
(x^2)/((x^4)(y^2)y) which is
1/((x^2)(y^3))

Multiplying these gives
(x^4)/4y

Answer 3

thats a lot of numbers... it can be pretty hard to read if you are just staring blankley at the screen... man, this is tough...

Answer 4

Try http://www.algebrahelp.com/

Hope it helps.