document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=203148>Question 203148</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The sum of two numbers is 88. Three times the smaller equals twice the larger number,plus 29. Find the numbers. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #153270 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=153270\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let a = the smaller number and b = the larger number, then...<BR>\n" );
document.write( "1) a+b = 88  \"The sum of two numbers is 88\"<BR>\n" );
document.write( "2) 3a = 2b+29  \"Three times the smaller equals twice the larger plus 29\"<BR>\n" );
document.write( "Rewrite equation 1) as a = 88-b and substitute into equation 2).<BR>\n" );
document.write( "3(88-b) = 2b+29  Simplify and solve for b.<BR>\n" );
document.write( "264-3b = 2b+29  Add 3b to both sides.<BR>\n" );
document.write( "264 = 5b+29  Subtract 29 from both sides.<BR>\n" );
document.write( "235 =5b  FInally, divide both sides by 5.<BR>\n" );
document.write( "47 = b and a = 88-47 = 41<BR>\n" );
document.write( "The two numbers are:  41 and 47<BR>\n" );
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