document.write( "Question 558789: x to -2 power - y to -2 power divided by x to -1 power + y to -1 power
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\n" ); document.write( "answer is y - x divided by xy
\n" ); document.write( "how do you get the answer \n" ); document.write( "
Algebra.Com's Answer #363231 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
As I understand your problem, you are to simplify the problem:
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\n" ); document.write( "\"%28x%5E%28-2%29-y%5E%28-2%29%29%2F%28x%5E%28-1%29%2B+y%5E%28-1%29%29\"
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\n" ); document.write( "Let's work on the numerator first by simplifying:
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\n" ); document.write( "\"%28x%5E%28-2%29-y%5E%28-2%29%29\"
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\n" ); document.write( "From the rules of exponents, if you have a negative exponent you can convert that by dividing 1 by the same term with a positive exponent. Following that rule we can convert:
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\n" ); document.write( "\"x%5E%28-2%29\" to \"1%2Fx%5E2\"
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\n" ); document.write( "and similarly we can convert:
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\n" ); document.write( "\"y%5E%28-2%29\" to \"1%2Fy%5E2\"
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\n" ); document.write( "Substituting these into the expression you were given for the numerator converts the numerator to:
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\n" ); document.write( "\"1%2Fx%5E2+-+1%2Fy%5E2\"
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\n" ); document.write( "The common denominator for this expression is \"x%5E2%2Ay%5E2\". We can multiply the x-term by \"y%5E2%2Fy%5E2\" to change it to \"y%5E2%2F%28x%5E2%2Ay%5E2%29\". Similarly we can multiply the y-term by \"x%5E2%2Fx%5E2\" to change it to \"x%5E2%2F%28x%5E2y%5E2%29\".
\n" ); document.write( "Substituting these changed terms into the numerator makes it become:
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\n" ); document.write( "\"y%5E2%2F%28x%5E2%2Ay%5E2%29-x%5E2%2F%28x%5E2y%5E2%29\" and finally combining the two terms over the common denominator results in the numerator becoming:
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\n" ); document.write( "\"%28y%5E2+-x%5E2%29%2F%28x%5E2%2Ay%5E2%29\" <--- remember this numerator for later
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\n" ); document.write( "Now let's do a similar procedure for the denominator of the problem. We had a denominator of:
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\n" ); document.write( "\"%28x%5E%28-1%29%2B+y%5E%28-1%29%29\"
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\n" ); document.write( "We can convert these to positive exponential terms by placing the \"x%5E%28-1%29\" under 1 to get \"1%2Fx\" and the \"y%5E%28-1%29\" under 1 to get \"1%2Fy\". This makes the denominator become:
\n" ); document.write( ".
\n" ); document.write( "\"1%2Fx+%2B+1%2Fy\"
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\n" ); document.write( "The common denominator for this is \"x%2Ay\" and when we place the two terms over this common denominator (by following the general process we used for the numerator) the result is:
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\n" ); document.write( "\"%28y%2Bx%29%2F%28x%2Ay%29\" <--- this is now the denominator for the original problem
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\n" ); document.write( "We are now going to divide this denominator into the numerator that we remembered for later.
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\n" ); document.write( "The denominator is a fraction. Remember from long ago that you can divide by a fraction by inverting it and multiplying. So we can divide by inverting the denominator and multiplying it by the numerator we remembered. When we invert the denominator we get:
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\n" ); document.write( "\"%28x%2Ay%29%2F%28y%2Bx%29\"
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\n" ); document.write( "Then multiplying this by the numerator the problem becomes:
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\n" ); document.write( "\"%28%28y%5E2+-x%5E2%29%2F%28x%5E2%2Ay%5E2%29%29%2A%28%28x%2Ay%29%2F%28y%2Bx%29%29\"
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\n" ); document.write( "Notice that the \"x%2Ay\" numerator of the second term can be canceled with the \"x%5E2%2Ay%5E2+\" denominator of the first term to just an \"x%2Ay\" in the denominator and make the expression become:
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\n" ); document.write( "\"%28%28y%5E2+-x%5E2%29%2F%28x%2Ay%29%29%2A%28%281%29%2F%28y%2Bx%29%29\"
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\n" ); document.write( "Then notice that the \"y%5E2+-+x%5E2\" is the difference of two squares. It can, therefore, be factored to \"%28y+-+x%29%2A%28y%2Bx%29\" to convert the expression to:
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\n" ); document.write( "\"%28%28%28y+-x%29%2A%28y%2Bx%29%29%2F%28x%2Ay%29%29%2A%28%281%29%2F%28y%2Bx%29%29\"
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\n" ); document.write( "Next notice and cross out the common factor in the numerator with a common factor in the denominator:
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\n" ); document.write( "and as a final step, when you multiply everything out you end up with:
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\n" ); document.write( "\"%28y+-+x%29%2F%28x%2Ay%29\"
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\n" ); document.write( "which is just as you said it should be.
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\n" ); document.write( "I hope that with a little careful tracking you can work your way through this and that it gives you some insight to the problem. Good luck ...
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