Question 77424

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6X TO THE POWER OF 2 Y TO THE POWER OF 3 + 9X TO THE POWER OF 2 Y TO THE POWER OF 3 OVER 3X TO THE POWER OF 2 Y TO THE POWER OF 2 =

{{{(6x^2Y^3+9X^2Y^3)/(3X^2Y^2)}}}
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The two term in the top are like terms because they differ
only by their coefficients 6 and 9, so we can just add the 
two terms in the numerator by adding their coefficients and
get 15x²y³ on top, so we have

{{{(15x^2Y^3)/(3X^2Y^2)}}}

1. The 3 in the bottom cancels into the 15 on top,
leaving 5 on top. 
2. The x² in the bottom cancels out with the x² on
the top.
3. The y² in the bottom cancels into the y³ on top,
leaving just y¹ (subtract exponents) which is just y.

So we end up with

{{{5y}}}

Edwin</pre>