document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161320>Question 1161320</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the sum of three numbers such that the 3rd number is twice the 2nd, and the 2nd exceeds the 1st number which is 80 by 70. Thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #784834 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=784834\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let the 3 numbers be represented by a,b,c.<BR>\n" );
document.write( "the third number is equal to twice the second number, therefore:<BR>\n" );
document.write( "c = 2b<BR>\n" );
document.write( "the second number exceeds the first number, which is 80, by 70.<BR>\n" );
document.write( "a = 80<BR>\n" );
document.write( "b = 80 + 70 = 150<BR>\n" );
document.write( "since c = 2b, then c = 300<BR>\n" );
document.write( "a + b + c = 80 + 150 + 300 = 530 is what i get.<BR>\n" );
document.write( "530 should be your solution if i did it correctly.<BR>\n" );
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