document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=267552>Question 267552</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Mr. Atkins is 33 years old and Mr.s Speyer is 27 years old.  How many years ago was Mr. Atkins 1.5 as old as Mr Spyer was then? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #196311 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=196311\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let A = Mr. Atkin's present age (33) and S = Mr. Speyer's present age (27).<BR>\n" );
document.write( "x years ago, Mr. Atkin's age was 1.5 times Mr. Speyer's age then.<BR>\n" );
document.write( "This can be expressed algebraically as:<BR>\n" );
document.write( "(A-x) = 1.5(S-x) Substitute A = 33 and S = 27.<BR>\n" );
document.write( "(33-x) = 1.5(27-x) Simplify and solve for x.<BR>\n" );
document.write( "33-x = 40.5-1.5x Add 1.5x to both sides.<BR>\n" );
document.write( "33+0.5x = 40.5 Subtract 33 from both sides.<BR>\n" );
document.write( "0.5x = 7.5 Divide both sides by 0.5<BR>\n" );
document.write( "x = 15<BR>\n" );
document.write( "15 years ago, Mr. Atkin's age was 1.5 times Mr. Speyer's age then.<BR>\n" );
document.write( "  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );