Expert: Scott A Wilson - 7/18/2010

Question
5(xneg1power) -2(yneg1power) divided by 25(xneg2power) - 4(yneg2power)... At the end when cross canceling,25ysq-4xsq is a difference of squares, so how can the book have xy over 2x + 5y when I have xy over 5y - 2x? I just don't see it?

Answer
It is known that a(negbpower) = a^(-b) = 1/a^b.  Using this ican be seen that
5(xneg1power) = 5/x, 2(yneg1power) = 2/7, 25(xneg2power) = 25/x², and 4(yneg2power) = 4/y².

So, the problem is [(5/x) - (2/y)]/[(25/x²) - (4/y²)].
Multiplying this by (x²y²)/(x²y²) gives us (x²y²)[(5/x) - (2/y)]/{(x²y²)[(25/x²) - (4/y²)]}.
Reducing it gives (5xy² - 2x²y)/(25y² - 4x²).

Note that the denominator has a difference of squares.
That is, (25y² - 4x²) = [(5y)²-(2x)²] = (5y-2x)(5y+2x).
This is since a²-b² = (a-b)(a+b), and here, a = 5y an b = 2x.

This factors into [xy(5y-2x)]/[(5y-2x)(5y+2x)], and the (5y-2x) term cancels.
This leaves xy/(5y+2x).