Posted by Cassidy on Sunday, September 30, 2007 at 10:33pm.
14(2x)^(3)y^(5)
---------------- =
28x^(7)y^(3)
rewriting:
14(2x)3*y5
--------------------------------- =
28x7*y3
14*8x3*y5
-------------------------------- =
28*x7y3
14 into 28 = 2 and 8/2 = 4.
4x3*y5
----------------------------- =
x7*y3
4*y5-3
------------------ =
x7-3
4y2
--------------- =
x4
I hope this looks ok and I didn't goof with the superscripts.
One trick that might help you out is to remember that powers subtract when dividing (and add when multiplying) So starting with the original equation, you can pull out all the constants (by putting them to the exponent power) then simply subtract the powers. For instance, the top half of the fraction 14 (2x)^3 * y^5 can be rewritten as 14*2^3 * x^3 * y^5 you can then just subtract the exponents on the bottom of the fraction which gives you (ignoring the constants) x^(-4) * y^(2). All thats left is dealing with the constants (which should be pretty easy) and putting the negative powers on the bottom of the fraction for the final solution.
Best of luck!
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