document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=827196>Question 827196</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello, <BR>\n" );
document.write( "It costs 55c for domestic mail within Australia and $2.05 to post a letter to Europe. If I spent $10.00 to post 10 letters, how many of each type did I post?<BR>\n" );
document.write( "Could you assist by providing algebra equation?<BR>\n" );
document.write( "Thank you, <BR>\n" );
document.write( "Bridgit </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #498555 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=498555\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let x be the number of .55 cent letters and y be the number of $2.05 letters, then<BR>\n" );
document.write( "x +y = 10   and<BR>\n" );
document.write( ".55x +(2.05)y = 10.00<BR>\n" );
document.write( "solve first equation for x and we get x = 10 -y, then substitute in second equation<BR>\n" );
document.write( ".55(10-y) +(2.05)y = 10.00<BR>\n" );
document.write( "5.50 -.55y +(2.05)y = 10.00<BR>\n" );
document.write( "(1.5)y = 4.50<BR>\n" );
document.write( "y = 3  and x = 7<BR>\n" );
document.write( "we have 7 letters at .55 cents each and 3 letters at $2.05 each<BR>\n" );
document.write( "let's check the answer<BR>\n" );
document.write( "7(.55) + 3(2.05) = 10.00<BR>\n" );
document.write( "3.85 + 6.15 = 10.00<BR>\n" );
document.write( "10.00 = 10.00<BR>\n" );
document.write( "our answer checks   :-)<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );