document.write( "Question 1163850: Find two consecutive odd integers such that their product is 143 more than 5 times their sum. \n" ); document.write( "

Algebra.Com's Answer #788068 by Edwin McCravy(20060) $\"\"$ $\"About$

You can put this solution on YOUR website!

Instead of doing it for you, I'll do one just like it:

\n" ); document.write( "Find two consecutive odd integers such that their product is 1 less than 6 times their sum.
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document.write( "Let the smaller one be x and larger be x+2.\r\n" );
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document.write( "x(x+2) = 6[x+(x+2)]-1\r\n" );
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document.write( "x²+2x = 6[x+x+2]-1\r\n" );
document.write( "x²+2x = 6[2x+2]-1\r\n" );
document.write( "x²+2x = 12x+12-1\r\n" );
document.write( "x²+2x = 12x+11\r\n" );
document.write( "x²-10x-11 = 0\r\n" );
document.write( "(x-11)(x+1) = 0\r\n" );
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document.write( "x-11=0;  x+1=0\r\n" );
document.write( "   x=11;   x=-1\r\n" );
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document.write( "One answer is 11 and 13, the other answers are -1 and +1\r\n" );
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document.write( "Checking 11 and 13.  Their product is 143, which is 1 less than\r\n" );
document.write( "6 times their sum, which is 6 times 11+13=24 and 6×24 = 144, and\r\n" );
document.write( "143 is 1 less than 144. So 11 and 13 checks.\r\n" );
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document.write( "Checking -1 and 1.  Their product is -1, which is 1 less than\r\n" );
document.write( "6 times their sum, which is 6 times -1+1 = 0 and 6×0 = 0, and\r\n" );
document.write( "-1 is 1 less than 0. So -1 and 1 checks. \r\n" );
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document.write( "Now use this as a guide and solve yours.  This problem has a\r\n" );
document.write( "\"less than\" and yours has a \"more than\".  So you'll have to add\r\n" );
document.write( "instead of subtract.\r\n" );
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document.write( "Edwin

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