document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=412950>Question 412950</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There are three consecutive odd integers, such that three times the first, minus two times the second, plus 13, equals negative three times the sum of the first and the third. What is the first integer? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #289980 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=289980\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> With a little thought about consecutive odd integers, it should be easy to figure out that they are two apart from each other. So if<BR>\n" );
document.write( "x = the first odd integer<BR>\n" );
document.write( "then the next odd integer will be two more:<BR>\n" );
document.write( "x+2 = the second odd integer<BR>\n" );
document.write( "and the next odd integer will be two more than the second one:<BR>\n" );
document.write( "(x+2) + 2 = x + 4 = the third odd integer<br><BR>\n" );
document.write( "Now that we have expressions for the three odd integers, we can go about writing expressions for the other parts of this problem:<BR>\n" );
document.write( "\"three times the first\" translates to: 3x<BR>\n" );
document.write( "\"two times the second\" translates to: 2(x+2) or 2x+4<BR>\n" );
document.write( "\"The sum of the first and third\" translates to: x + (x+4) or 2x+4<BR>\n" );
document.write( "\"negative three times the sum of the first and third\" translates to: -3(2x+4) or -6x-12<br><BR>\n" );
document.write( "We can now use these expressions to translate \"three times the first, minus two times the second, plus 13, equals negative three times the sum of the first and the third\"<BR>\n" );
document.write( "(3x) - (2x+4) + 13 = (-6x-12)<BR>\n" );
document.write( "Note the use of parentheses. It is a good habit to use them. Sometimes they don't make any difference. Sometimes they make a hugh difference. In this equation the parentheses are the 3x and the -6x-12 are unnecessary. But the parentheses around the 2x+4 is critical! Without the parentheses we sould just subtract the 2x but not the 4. But the whole expression, 2x+4, represents \"two times the second odd integer\" and if we're supposed to subtract \"two times the second odd integer\" as the problem says, then <i>you need to subtract the entire expression, not just the 2x part of it!</i><br><BR>\n" );
document.write( "Now that we have an equation that models the problem we can solve it. We start by simplifying:<BR>\n" );
document.write( "3x - 2x- 4 + 13 = -6x-12<BR>\n" );
document.write( "x + 9 = -6x-12<BR>\n" );
document.write( "Next we get the variable on just one side of the equation. Adding 6x to each side we get:<BR>\n" );
document.write( "7x + 9 = -12<BR>\n" );
document.write( "Next we isolate x, starting with subtracting 9 from each side:<BR>\n" );
document.write( "7x = -21<BR>\n" );
document.write( "And lastly we divide both sides by 7:<BR>\n" );
document.write( "x = -3<br><BR>\n" );
document.write( "Since the problem asks just for the first odd integer and since x was defined to be the first odd integer (see back near the start where we defined what x stood for,), then the answer to your problem is -3. </SPAN>\n" );
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