document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4692>Question 4692</A>: <I><SPAN STYLE=\"word-break: break-all;\"> USING A COMBINATION OF 24% NITROGEN AND 12% NITROGEN TO ACHIEVE AN END RESULT OF 100 GRAMS OF 21% NITROGEN. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #2209 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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=2209\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Set up your problem this way:\r<BR>\n" );
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document.write( "Assign the variable(s):<BR>\n" );
document.write( "Let x = # of grams of 24% nitrogen and 100 - x = # of grams of 12% nitrogen.<BR>\n" );
document.write( "Convert the percents to decimals.\r<BR>\n" );
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document.write( "x(.24)+ (100 - x)(.12) = 100((.21)  Simplify.<BR>\n" );
document.write( ".24x + 12 - .12x = 21<BR>\n" );
document.write( ".12x = 9<BR>\n" );
document.write( "x = 9/(.12)<BR>\n" );
document.write( "x = 75 grams\r<BR>\n" );
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document.write( "100 - x = 100 - 75 = 25 grams\r<BR>\n" );
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document.write( "You would need 75 grams of 24% nitrogen plus 25 grams of 12% nitrogen to achieve a mixture of 100 grams of 21% nitrogen. </SPAN>\n" );
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