document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1059000>Question 1059000</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (1). Bertha is three times as old as Martha but four times as old as Gifty. Fine their respective ages if the sum of their ages is 84 years. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #674078 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=674078\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Bertha is three times as old as Martha but four times as old as Gifty. Find their respective ages if the sum of their ages is 84 years.<BR>\n" );
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document.write( "Equations:<BR>\n" );
document.write( "B = 3*M<BR>\n" );
document.write( "B = 4*G<BR>\n" );
document.write( "B + M + G = 84<BR>\n" );
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document.write( "B + (1/3)B + (1/4)B = 84<BR>\n" );
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document.write( "12B + 4B + 3B = 12*84<BR>\n" );
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document.write( "19B = 12*84<BR>\n" );
document.write( "B = 53.05 years<BR>\n" );
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document.write( "M = (1/3)53.05 = 17.68 years<BR>\n" );
document.write( "B = (1/4)53.05 = 13.26 years<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "------------  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );