document.write( "Question 483727: at the local grocery store,lemons are 52 cents each and limes are 21 cents each.how many lemons and limes can you buy for exactly $3.75? \n" ); document.write( "
Algebra.Com's Answer #331084 by Edwin McCravy(20055)\"\" \"About 
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document.write( "let x = the number of lemons\r\n" );
document.write( "let y = the number of limes\r\n" );
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document.write( "$0.52x + $0.21y = $3.75\r\n" );
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document.write( "      52x + 21y = 375\r\n" );
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document.write( "Write 52 and 375 in terms of their nearest multiple of 21\r\n" );
document.write( "So we write 52 as (42+10) and 375 as 378 - 3\r\n" );
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document.write( "(42 + 10)x + 21y = 378 - 3\r\n" );
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document.write( " 42x + 10x + 21y = 378 - 3\r\n" );
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document.write( "Divide through by 21\r\n" );
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document.write( "  2x + \"10x%2F21\" + y = 18 - \"3%2F21\"\r\n" );
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document.write( "Isolate fractions on the left:\r\n" );
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document.write( "  \"10x%2F21\" + \"3%2F21\" = 18 - 2x - y\r\n" );
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document.write( "The right side is an integer and the left side is positive, so\r\n" );
document.write( "both sides equals some positive integer A\r\n" );
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document.write( "  \"10x%2F21\" + \"3%2F21\" = A and  18 - 2x - y = A\r\n" );
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document.write( "Clear of fractions:\r\n" );
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document.write( "        10x + 3 = 21A\r\n" );
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document.write( "Write 21 in terms of its nearest multiple of 10\r\n" );
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document.write( "        10x + 3 = (20 + 1)A\r\n" );
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document.write( "        10x + 3 = 20A + A\r\n" );
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document.write( "Divide through by 10\r\n" );
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document.write( "          x + \"3%2F10\" = 2A + \"A%2F10\"\r\n" );
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document.write( "Isolate fractions on the left\r\n" );
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document.write( "          \"3%2F10\" - \"A%2F10\" = 2A - x\r\n" );
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document.write( "The right side is an integer, so the left side is too, say B\r\n" );
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document.write( "\"3%2F10\" - \"A%2F10\" = B and  2A - x = B\r\n" );
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document.write( "Clear of fractions:\r\n" );
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document.write( "3 - A = 10B\r\n" );
document.write( "   -A = -3 - 10B\r\n" );
document.write( "    A = 3 + 10B\r\n" );
document.write( "    A = 10B + 3\r\n" );
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document.write( "2A - x = B\r\n" );
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document.write( "2(10B + 3) - x = B\r\n" );
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document.write( "20B + 6 - x = B\r\n" );
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document.write( "19B + 6 = x\r\n" );
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document.write( "and since\r\n" );
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document.write( "18 - 2x - y = A,\r\n" );
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document.write( "18 - 2(19B + 6) - y = 10B + 3\r\n" );
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document.write( "18 - 38B - 12 - y = 10B + 3\r\n" );
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document.write( "               -y = 48B - 3\r\n" );
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document.write( "                y = -48B + 3\r\n" );
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document.write( "                y = 3 - 48B\r\n" );
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document.write( "we must buy more than -1 but less than 18 limes\r\n" );
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document.write( "            -1 < y < 18\r\n" );
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document.write( "         -1 < 3 - 48B < 18\r\n" );
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document.write( "          -4 < -48B < 18\r\n" );
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document.write( "Divide through by -48 which reverses inequalities:\r\n" );
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document.write( "          \"%28-4%29%2F%28-48%29\" > B > \"18%2F%28-48%29\"\r\n" );
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document.write( "          \"1%2F12\" > B > \"-3%2F8\"\r\n" );
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document.write( "Since B is an integer then B = 0 because 0 is\r\n" );
document.write( "the only integer between those values.\r\n" );
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document.write( "Therefore since \r\n" );
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document.write( "         y = 3 - 48B\r\n" );
document.write( "         y = 3 - 48(0)\r\n" );
document.write( "         y = 3 - 0\r\n" );
document.write( "         y = 3     \r\n" );
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document.write( "and since \r\n" );
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document.write( "       19B + 6 = x\r\n" );
document.write( "     19(0) + 6 = x\r\n" );
document.write( "         0 + 6 = x\r\n" );
document.write( "             6 = x\r\n" );
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document.write( "Therefore 6 lemons and 3 limes is the only possibility.\r\n" );
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document.write( "Edwin
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