document.write( "Question 564240: How do I find three consecutive odd integers such that three times the second minus the third is 11 more than the first?
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Algebra.Com's Answer #365287 by nerdybill(7384) $\"\"$ $\"About$

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How do I find three consecutive odd integers such that three times the second minus the third is 11 more than the first?
\n" ); document.write( "Let x = first (smallest) of three consecutive odd integers
\n" ); document.write( "then
\n" ); document.write( "x+2 = second
\n" ); document.write( "x+4 = third
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\n" ); document.write( "3(x+2)-(x+4) = x+11
\n" ); document.write( "(3x+6)-(x+4) = x+11
\n" ); document.write( "3x+6-x-4 = x+11
\n" ); document.write( "2x+6-4 = x+11
\n" ); document.write( "2x+2 = x+11
\n" ); document.write( "x+2 = 11
\n" ); document.write( "x = 9 (first)
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\n" ); document.write( "second:
\n" ); document.write( "x+2 = 9+2 = 11
\n" ); document.write( ".
\n" ); document.write( "third:
\n" ); document.write( "x+4 = 9+4 = 13
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\n" ); document.write( "answer: 9,11,13 \n" ); document.write( "

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