document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=896263>Question 896263</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the sum of three consecutive odd numbers is 63.Find them.<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 #543517 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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/cgi-bin/show-face.mpl?user=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=543517\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> B=A+2<BR>\n" );
document.write( "C=A+4<BR>\n" );
document.write( "where A is the smallest number (they're consecutive odd so they'll be 2 apart)<BR>\n" );
document.write( "A + B + C = 63<BR>\n" );
document.write( "Substitute the known values of B & C from the top equations into the bottom<BR>\n" );
document.write( "A + B + C = 63<BR>\n" );
document.write( "A + (A+2) + (A+4) = 63<BR>\n" );
document.write( "3A + 6 = 63<BR>\n" );
document.write( "Subtract 6 from each side<BR>\n" );
document.write( "3A = 57<BR>\n" );
document.write( "Divide each side by 3<BR>\n" );
document.write( "A = 19<BR>\n" );
document.write( "If A is 19, then B is 21, and C is 23<BR>\n" );
document.write( "19+21+23=63<BR>\n" );
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document.write( "There's a faster way.<BR>\n" );
document.write( "Take 63 and divide it by the number of values<BR>\n" );
document.write( "63/3=21<BR>\n" );
document.write( "That will be your middle number, with a number on each side of it. </SPAN>\n" );
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