document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=161501>Question 161501</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Pop's Cycle Shop sells bicycles and tricycles. The number of bicycles is 1 less than 5 times the number of tricycles. All the bicycles and tricycles together have a total of 154 wheels. How many bicycles are there? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #118971 by </B><A HREF=/tutors/aboutme.mpl?userid=gonzo><B>gonzo(654)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=gonzo><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=118971\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let b = number of bicycles<BR>\n" );
document.write( "let t = number of tricycles<BR>\n" );
document.write( "number of bicycles is 5 times the number of tricycles minus 1 becomes<BR>\n" );
document.write( "b = 5*t - 1<BR>\n" );
document.write( "subtract b from both sides of the equation and add 1 to both sides of the equation and it becomes<BR>\n" );
document.write( "1 = 5*t - b<BR>\n" );
document.write( "which is the same as<BR>\n" );
document.write( "5*t - b = 1<BR>\n" );
document.write( "since a bicycle has 2 wheels and a tricycle has 3 wheels, 2 times the number of bicycles plus 3 times the number of tricycles totals the number of wheels which is 154, so the equation becomes<BR>\n" );
document.write( "2*b + 3*t = 154 which is the same as 3*t + 2*b = 154<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "2 equations to solve simultaneously are:<BR>\n" );
document.write( "5*t - b = 1<BR>\n" );
document.write( "3*t + 2*b = 154<BR>\n" );
document.write( "multiply both sides of the first equation by 2 to change the first equation to<BR>\n" );
document.write( "10*t - 2*b = 2 which will allow us to cancel out the b's.<BR>\n" );
document.write( "simultaneous equations are now<BR>\n" );
document.write( "10*t - 2*b = 2<BR>\n" );
document.write( "3*t + 2*b = 154<BR>\n" );
document.write( "add first equation to second equation to get<BR>\n" );
document.write( "13*t + 0*b = 156<BR>\n" );
document.write( "which becomes<BR>\n" );
document.write( "13*t = 156<BR>\n" );
document.write( "which becomes<BR>\n" );
document.write( "t = 156 / 13 = 12<BR>\n" );
document.write( "so,<BR>\n" );
document.write( "t = 12<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "original equations copied here for ease of reference:<BR>\n" );
document.write( "2 equations to solve simultaneously are:<BR>\n" );
document.write( "5*t - b = 1<BR>\n" );
document.write( "3*t + 2*b = 154<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "substitute 12 for t in the first equation to get<BR>\n" );
document.write( "5*12 - b = 1<BR>\n" );
document.write( "which becomes<BR>\n" );
document.write( "60 - b = 1<BR>\n" );
document.write( "adding b to both sides of the equation and subtracting 1 from both sides of the equation gets<BR>\n" );
document.write( "60 - 1 = b<BR>\n" );
document.write( "which becomes<BR>\n" );
document.write( "b = 59<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "substituting 59 for b and 12 for t in the second equation and it becomes<BR>\n" );
document.write( "3*12 + 2*59 = 154<BR>\n" );
document.write( "which becomes<BR>\n" );
document.write( "36 + 118 = 154<BR>\n" );
document.write( "which becomes<BR>\n" );
document.write( "154 = 154 proving the answer is correct.<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "answer is:<BR>\n" );
document.write( "12 tricycles and 59 bicycles\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );