document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=757168>Question 757168</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Patty is 3 times as old as her brother, Terry. In 10 years, the sum of their ages will be 36. How old are they now? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #460675 by </B><A HREF=/tutors/aboutme.mpl?userid=LearnWithMajor><B>LearnWithMajor(33)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=LearnWithMajor><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=460675\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P = patty's age now  T = terry's age now\r<BR>\n" );
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document.write( "Patty is 3 times as old as terry.  3T = P<BR>\n" );
document.write( "In 10 years the sum of their ages will be 36. (T+10) + (P+10) = 36\r<BR>\n" );
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document.write( "Now substituting/replacing the P in the second equation with 3T from the first equation we get<BR>\n" );
document.write( "(T+10) + (3T+10) = 36\r<BR>\n" );
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document.write( "Now we combine like terms and then solve for T\r<BR>\n" );
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document.write( "(T+3T) + (10+10) = 36<BR>\n" );
document.write( "4T + 20 = 36  <BR>\n" );
document.write( "4T = 16<BR>\n" );
document.write( "T = 4<BR>\n" );
document.write( "P = 3T = 3(4) = 12 </SPAN>\n" );
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