document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=213838>Question 213838</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this equation: Three times a first number decreased by a second number is 1. The first number increased by twice the second number is 12. Find the numbers\r<BR>\n" );
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document.write( "I tried: Let x equal 1st number<BR>\n" );
document.write( "         Let y equal 2nd number<BR>\n" );
document.write( "         3x-y=1 <BR>\n" );
document.write( "         3x+2y=12<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #161533 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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=161533\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 3x-y=1 multiply by 2 & add.<BR>\n" );
document.write( "x+2y=12<BR>\n" );
document.write( "6x-2y=2<BR>\n" );
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document.write( "7x=14<BR>\n" );
document.write( "x=14/7<BR>\n" );
document.write( "x=2 ans.<BR>\n" );
document.write( "2+2y=12<BR>\n" );
document.write( "2y=12-2<BR>\n" );
document.write( "2y=10<BR>\n" );
document.write( "y=10/2<BR>\n" );
document.write( "y=5 ans.<BR>\n" );
document.write( "Proof:<BR>\n" );
document.write( "3*2-5=1<BR>\n" );
document.write( "6-5=1<BR>\n" );
document.write( "1=1 </SPAN>\n" );
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