document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=34443>Question 34443</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find two numbers, given that the second is 3 times the first. If you multiply the first number by the second number and the second number by 4, the sum of the products is 420. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #20699 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=20699\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find two numbers, given that the second is 3 times the first. If you multiply the first number by the second number and the second number by 4, the sum of the products is 420.<BR>\n" );
document.write( "LET FIRST NUMBER =X<BR>\n" );
document.write( "SEOND NUMBER =3 TIMES FIRST =3*X<BR>\n" );
document.write( "MULTIPLY THE FIRST NUMBER BY SECOND WE GET =X*3X=3X^2<BR>\n" );
document.write( "MULTIPLY SECOND NUMBER BY 4 WE GET =4*3*X=12X<BR>\n" );
document.write( "SUM OF 2 PRODUCTS =3X^2+12X=420...DIVIDING WITH 3 WE GET<BR>\n" );
document.write( "X^2+4X=140<BR>\n" );
document.write( "X^2+4X-140=0<BR>\n" );
document.write( "X^2+14X-10X-140=0<BR>\n" );
document.write( "X(X+14)-10(X+14)=0<BR>\n" );
document.write( "(X-10)(X+14)=0<BR>\n" );
document.write( "X-10=0 ....OR....X=10<BR>\n" );
document.write( "SO THE 2 NUMBER ARE 10 AND 30....\r<BR>\n" );
document.write( "\n" );
document.write( "OTHER POSSIBILITY IS X+14=0...OR X=-14...THE SECOND NUMBER IS -42 </SPAN>\n" );
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