document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=46764>Question 46764</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the first and third of three consecutive odd integers are added together, the result is 63 less than five times the second integer.  Find the third integer. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #30945 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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=30945\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the first odd integer, x+2 the next consecutive odd integer, and x+4 the third consecutive odd integer.  From the problem description, you can write:\r<BR>\n" );
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document.write( "x + (x+4) = 5(x+2)-63 Simplify and solve for x.<BR>\n" );
document.write( "2x+4 = 5x+10-63<BR>\n" );
document.write( "2x+4 = 5x-53 Subtract 2x from both sides of the equation.<BR>\n" );
document.write( "4 = 3x-53 Add 53 to both sides.<BR>\n" );
document.write( "3x = 57 Divide both sides by 3.<BR>\n" );
document.write( "x = 19<BR>\n" );
document.write( "x+2 = 21<BR>\n" );
document.write( "x+4 = 23 </SPAN>\n" );
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