document.write( "Question 640968: three times the first of three consecutive odd integers is 3 more than twice the third. find the third integer? \n" ); document.write( "
Algebra.Com's Answer #403515 by KMST(5328)\"\" \"About 
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Let the last/greatest of the three consecutive odd integers be \"n\" .
\n" ); document.write( "(I like \"n\". You could use \"x\" if you like that better).
\n" ); document.write( "We state that \"n\" has to be odd.
\n" ); document.write( "(If the \"n\" we find turns to be something that is not an odd integer, then the problem has no solution).
\n" ); document.write( "The odd integer before \"n\" is \"n-2\", and the odd integer before that is \"n-4\".
\n" ); document.write( "The three consecutive odd integers, in order, are:
\n" ); document.write( "\"n-4\" , \"n-2\" , and \"n\" .
\n" ); document.write( "Three times the first is \"3%28n-4%29\".
\n" ); document.write( "Twice the third one is \"2n\", and 3 more than that is \"2n%2B3\" .
\n" ); document.write( "Our equation is
\n" ); document.write( "\"3%28n-4%29=2n%2B3\" .
\n" ); document.write( "To solve, we first apply the distributive property to do the multiplication on the left side of the equal sign.
\n" ); document.write( "\"3%28n-4%29=2n%2B3\" --> \"3n-3%2A4=2n%2B3\" --> \"3n-12=2n%2B3\"
\n" ); document.write( "Next, we subtract \"2n\" from both sides of the equal sign:
\n" ); document.write( "\"3n-12=2n%2B3\" --> \"3n-12-2n=2n%2B3-2n\" --> \"n-12=3\"
\n" ); document.write( "Then we add 12 to both sides of the equal sign:
\n" ); document.write( "\"n-12=3\" --> \"n-12%2B12=3%2B12\" --> \"highlight%28n=15%29\"
\n" ); document.write( "The third of the three consecutive odd integers is \"highlight%2815%29\" .
\n" ); document.write( "The three consecutive odd integers are 11, 13, and 15, and 3 times the first, \"3%2A11=33\" , is 3 more than twice the third (\"2%2A15=30\" and \"30%2B3=33\"). \n" ); document.write( "
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