document.write( "Question 1106740: Simplify the following expression: 48y^13/12xy^7 \n" ); document.write( "
Algebra.Com's Answer #721723 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
\"48y%5E13%2F12xy%5E7\"\r\n" );
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document.write( "\"%2848%2Ay%5E13%29%2F%2812%2Ax%2Ay%5E7%29\"\r\n" );
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document.write( "48 = 2∙24 = 2∙2∙12 = 2∙2∙2∙6 = 2∙2∙2∙2∙3 \r\n" );
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document.write( "12 = 2∙6 = 2∙2∙3\r\n" );
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document.write( "y13 = y∙y∙y∙y∙y∙y∙y∙y∙y∙y∙y∙y∙y\r\n" );
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document.write( "y7 = y∙y∙y∙y∙y∙y∙y\r\n" );
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document.write( "\"%2848%2Ay%5E13%29%2F%2812%2Ax%2Ay%5E7%29\"\"%22%22=%22%22\"\r\n" );
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document.write( "Cancel the 2 factors of 2 in the bottom into 2 of the factors\r\n" );
document.write( "of 2 in the numerator:\r\n" );
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document.write( "Cancel the 7 factors of y in the bottom into 7 of the factors\r\n" );
document.write( "of y in the numerator:\r\n" );
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document.write( "And all that's left is\r\n" );
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document.write( "\"%282%2A2%2Ay%2Ay%2Ay%2Ay%2Ay%2Ay%2Ay%29%2Fx\"\r\n" );
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document.write( "and since 2∙2 = 4, and y∙y∙y∙y∙y∙y = y6\r\n" );
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document.write( "\"%284%2Ay%5E6%29%2Fx\"  <-- final answer.\r\n" );
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document.write( "There is a much shorter way that involves subtracting the\r\n" );
document.write( "exponents to see how many factors would cancel if you did\r\n" );
document.write( "it this long way.  See if you can figure out the short way \r\n" );
document.write( "by yourself, so you can do problems like this is just one \r\n" );
document.write( "or two steps.\r\n" );
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document.write( "Edwin
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