document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=165268>Question 165268</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There are two numbers whose sum is 50. Three times the first is five more than twice the second. What are the numbers?\r<BR>\n" );
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document.write( "Please DO answer quickly. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #121824 by </B><A HREF=/tutors/aboutme.mpl?userid=Mathtut><B>Mathtut(3670)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Mathtut><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=121824\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> call the numbers a and b.  a+b=50 so a=50-b    (3 times a)(3a)=5+(2 times b)(2b)\r<BR>\n" );
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document.write( "re written as  3a=5+2b.     Now  substituting a's value from first equation into 2nd equation we get    3(50-b)= 5+2b  or 150-3b=5+2b  adding 3b and subtracting 5 from each side we get 5b=145  so b =29    so since a=50-b(29) the a =21\r<BR>\n" );
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document.write( "answer:  a=21 and b=29 </SPAN>\n" );
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