document.write( "Question 1054852: a woman is 4 times older than her daughter. 5 years ago, the product of their ages was 175. find their present ages
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Algebra.Com's Answer #670057 by Edwin McCravy(20056)\"\" \"About 
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a woman is 4 times older than her daughter.
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document.write( "W = 4D\r\n" );
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\n" ); document.write( "5 years ago,
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document.write( "That was when the woman was W-5\r\n" );
document.write( "That was when the daughter was D-5\r\n" );
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\n" ); document.write( "the product of their ages was 175.
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document.write( "(W-5)(D-5) = 175\r\n" );
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document.write( "\"system%28W=4D%2C%28W-5%29%28D-5%29=175%29\"\r\n" );
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document.write( "Substitute 4D for W in the second equation:\r\n" );
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document.write( "(4D-5)(D-5) = 175\r\n" );
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document.write( "4Dē-20D-5D+25 = 175\r\n" );
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document.write( "4Dē-25D+25 = 175\r\n" );
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document.write( "4Dē-25D-150 = 0\r\n" );
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document.write( "(4D+15)(D-10) = 0\r\n" );
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document.write( "4D+15=0;      D-10=0\r\n" );
document.write( "   4D=-15;       D = 10\r\n" );
document.write( "    D=-15/4;\r\n" );
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document.write( "Ignore the negative fraction answer.\r\n" );
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document.write( "The Daughter is 10.\r\n" );
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document.write( "The woman is 4D = 4(10) = 40.\r\n" );
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document.write( "Checking: 5 years ago the woman was 35 and her daughter was 5.\r\n" );
document.write( "And sure enough 5 times 35 is 175.\r\n" );
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document.write( "Edwin
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