document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=209881>Question 209881</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There are three consecutive odd integers such that twice their sum is 30 less than 8 times the third integer. Find these odd integers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #159484 by </B><A HREF=/tutors/aboutme.mpl?userid=math+problem+solving><B>math problem solving(30)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math+problem+solving><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=159484\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the middle number is n. <BR>\n" );
document.write( "Then the three numbers can be expressed as n-2,n,n+2.\r<BR>\n" );
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document.write( "Their sum is n-2 + n + n + 2 = 3n.\r<BR>\n" );
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document.write( "Now use the given relationship to set up an equation:<BR>\n" );
document.write( "twice their sum is 30 less than 8 times the third integer\r<BR>\n" );
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document.write( "The relationship can be written as:<BR>\n" );
document.write( "2* their sum = 8*the third number - 30\r<BR>\n" );
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document.write( "2*3n =8(n+2)-30<BR>\n" );
document.write( "Solving for n, we have<BR>\n" );
document.write( "6n = 8n + 16 -30<BR>\n" );
document.write( "0 = 2n - 14<BR>\n" );
document.write( "2n = 14<BR>\n" );
document.write( "n = 7<BR>\n" );
document.write( "So the three numbers are 5, 7, 9. </SPAN>\n" );
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