2 Math Problems! HELP!!?

Question

1. Simplify, State any restrictions on the variable(s).

2xy
-----
x^2y-xy^2

2.Multiply. Assume the domains of the variables do not include values for which any denominator is zero.

2a-6
------
3

divided by (3-a/6)^2


THanks!!! :)

Answer 1

You will restrict the variables by making sure that nothing in the denominator can produce a result of 0 (0 in the denominator is what you want to avoid - it makes the result undefined).
Look for that at the beginning. So for the first problem which I read as '2xy divided by ((x squared) times y) minus (x times (y squared))' specify that x and y can't be 0. You don't want to multiply anything by 0 in the denominator - if you multiply y times 0 (if 0 were x) the result is 0 which is important because the next part of the expression is ' minus x times y squared which means you would still be multiplying by 0 if x were 0 in the expression - pretty much guaranteeing 0 in the denominator. Y is the same situation. If y is 0 but x is not, you will still get 0 - 0 after substituting values.
That done, 2/x-y is the answer, as others have shown. (x squared times y) minus (x times (y squared)) can be factored to xy(x-y). when you put that under 2xy as the denominator, the xy's cancel out on the top and bottom and what you have left in the numerator is 2 and what you have left in the denominator is (x-y).


2. For your second problem, it was harder to understand the way the actual problem was laid out. What I think I see is:
((2 a minus 6) over (3)) divided by (((3 minus a) over (6))squared)

You multiply out the denominator. This means squaring (multiplying by itself) ((3 minus a) over (6)). When you multiply this out, you get (a squared minus 6 a plus 9). That is the numerator but you also had to multiply the denominator by itself (6 times 6) to have the denominator of 36.
So you have: ((a^2 - 6a +9)/36). It so happens that this can be factored also to be not only (3-a) squared but also (a-3) squared. Try it yourself. You will see that it factors both ways.
This works out well - you use ((a-3) (a-3) over 36). Since you are dividing the two rational expressions, flip one (multiply the reciprocal is what you do when you divide). Might as well flip this so it is 36/((a-3) (a-3)). Multiply times (2a-6)/3. Factor the 2a-6 to be 2(a-3). Now you have 2(a-3)/3 multiplied times 36/((a-3)(a-3).
Cancel out the (a-3) next to the 2 and one of the (a-3)s in the other expression. Cancel the 3 and the 36 to be 1 and 12 respectively. What you are left with after all this are the numerators 2 and 12 which you multiply to make 24, your answer's numerator. You have a 1 and an (a-3) in the denominators to multiply. They give you (a-3) for your answer's denominator. So the answer is 24/(a-3).

Answer 2

1.2xy/x^y-xy^2
=2xy/xy(x-y)
=2/(x-y)

2.2a-6/3
=2(a-3)/3
=(2/3)(a-3

Answer 3

2xy
---------------
x^2y-xy^2




2x y
=---------------
xy(x-y)


2
=---------
(x-y)