document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1141003>Question 1141003</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find three consecutive odd integers such that the sum of two times the first, three times the second, and one time the third is 412. List the numbers in order from smallest to largest.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #761544 by </B><A HREF=/tutors/aboutme.mpl?userid=4419875><B>4419875(21)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=4419875><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=761544\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the orientation will be like this:<BR>\n" );
document.write( "2(x+1)+3(x+3)+(x+5) = 412\r<BR>\n" );
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document.write( "where:<BR>\n" );
document.write( "2 times the 1st -> 2(x+1)<BR>\n" );
document.write( "3 times the 2nd -> 3(x+3)<BR>\n" );
document.write( "1 times the 3rd -> (x+5)\r<BR>\n" );
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document.write( "Thus,\r<BR>\n" );
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document.write( "6x + 16 = 412<BR>\n" );
document.write( "6x = 396<BR>\n" );
document.write( "x = 66\r<BR>\n" );
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document.write( "so the numbers would be\r<BR>\n" );
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document.write( "[66+1, 66+3, 66+5]<BR>\n" );
document.write( "[67, 69, 71]\r<BR>\n" );
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document.write( "to check if it's true<BR>\n" );
document.write( "2(66+1)+3(66+3)+(66+5)= 412<BR>\n" );
document.write( "134 + 207 + 71 = 412\r<BR>\n" );
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document.write( "412 = 412\r<BR>\n" );
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document.write( "Hope to do your homework well. Goodluck </SPAN>\n" );
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