document.write( "Question 886344: find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two. \n" ); document.write( "
Algebra.Com's Answer #535870 by MathTherapy(10556)\"\" \"About 
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\n" ); document.write( "find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.
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\n" ); document.write( "Let first integer be F
\n" ); document.write( "Then 2nd is: F + 2, and 3rd is: F + 4
\n" ); document.write( "Therefore, 2(F)(F + 2) = (F + 2)(F + 4) + 7
\n" ); document.write( "\"2F%5E2+%2B+4F+=+F%5E2+%2B+6F+%2B+8+%2B+7\"
\n" ); document.write( "\"2F%5E2+-+F%5E2+%2B+4F+-+6F+-+15+=+0\"
\n" ); document.write( "Solve for F, the 1st integer, and then determine the other 2 integers. Note that 2 values will ensue for
\n" ); document.write( "the 1st integer, F, thus 2 values for the 2nd and 3rd integers. \n" ); document.write( "
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