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November 22, 2014

November 22, 2014

Posted by **Cassidy** on Sunday, September 30, 2007 at 10:33pm.

(the exponents are after a carrot and are in parentheses so there is no confusion)

14(2x)^(3)y^(5)

----------------

28x^(7)y^(3)

-3x^(-5)y^(-3)z

----------------

6x^(-3)y^(-5)z^(-2)

Sorry it's looks so confusing...

- Math -
**DrBob222**, Sunday, September 30, 2007 at 10:50pm14(2x)^(3)y^(5)

---------------- =

28x^(7)y^(3)

rewriting:

14(2x)^{3}*y^{5}

--------------------------------- =

28x^{7}*y^{3}

14*8x^{3}*y^{5}

-------------------------------- =

28*x^{7}y^{3}

14 into 28 = 2 and 8/2 = 4.

4x^{3}*y^{5}

----------------------------- =

x^{7}*y^{3}

4*y^{5-3}

------------------ =

x^{7-3}

4y^{2}

--------------- =

x^{4}

I hope this looks ok and I didn't goof with the superscripts.

- Math -
**Nick**, Monday, October 1, 2007 at 1:21amOne 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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