SOLUTION: "x to the second power, y to the third power; OVER x to the negative first power, y; all to the negative first power via parenthesis... times... x to the fourth power, y; OVE

Algebra ->  Exponents -> SOLUTION: "x to the second power, y to the third power; OVER x to the negative first power, y; all to the negative first power via parenthesis... times... x to the fourth power, y; OVE      Log On


   

Question 312575: "x to the second power, y to the third power; OVER x to the negative first power, y; all to the negative first power via parenthesis...
times...
x to the fourth power, y; OVER x , y to the second power; all to the second power via parenthesis.
Any assistance would be appreciated. Sending from a mobile phone, so limited on my ability to present this problem any more clearly. Thanks!
Found 2 solutions by nyc_function, JBarnum:
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Can you express the words using math symbols?
We can then proceed.
Write back....

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!

we will start in the division of variables with exponets we simply subtract.
subtract exponents
%28%28x%5E3y%5E2%29%5E%28-1%29%29%2A%28%28x%5E3y%5E%28-1%29%29%5E2%29multiply exponents to the raised exponent
%28x%5E%283%2A-1%29y%5E%282%2A-1%29%29%2A%28x%5E%283%2A2%29y%5E%28-1%2A2%29%29multiply exponents
%28x%5E%28-3%29y%5E%28-2%29%29%2A%28x%5E6y%5E%28-2%29%29 when multiplying exponents you add them together
%28x%5E%28-3%2B6%29y%5E%28-2-2%29%29 add
highlight%28x%5E3y%5E%28-4%29%29