document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=859007>Question 859007</A>: <I><SPAN STYLE=\"word-break: break-all;\"> new brand of plant fertilizer is to be made from three different types of chemicals (A, B, C).<BR>\n" );
document.write( "The mixture includes 80% of chemicals A and B.<BR>\n" );
document.write( "Chemical B and C must be in ratio of 3 to 4 by weight.<BR>\n" );
document.write( "How much of each type of chemical is needed to make 600kg of plant fertilizer?<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 #517539 by </B><A HREF=/tutors/aboutme.mpl?userid=Awesom3guy><B>Awesom3guy(31)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Awesom3guy><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=517539\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Set up a system of three equations with three unknowns.<BR>\n" );
document.write( "new brand of plant fertilizer is to be made from three different types of chemicals (A, B, C).<BR>\n" );
document.write( "A + B + C = 600\r<BR>\n" );
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document.write( "The mixture includes 80% of chemicals A and B.<BR>\n" );
document.write( "A + B = 600*0.8<BR>\n" );
document.write( "A + B = 480\r<BR>\n" );
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document.write( "Chemical B and C must be in ratio of 3 to 4 by weight.<BR>\n" );
document.write( "B = (3/4)C\r<BR>\n" );
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document.write( "Now plug in the value of any variable from one equation to the other two, in the place of that variable. Say, from the last equation.<BR>\n" );
document.write( "We will plug in (3/4)C in both the top two equations.\r<BR>\n" );
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document.write( "A + B + C = 600<BR>\n" );
document.write( "A + (3/4)C + C = 600<BR>\n" );
document.write( "A + (7/4/)C = 600\r<BR>\n" );
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document.write( "A + B = 480<BR>\n" );
document.write( "A + (3/4)C = 480\r<BR>\n" );
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document.write( "Now we have the two equations with two variables (B was eliminated):<BR>\n" );
document.write( "A + (7/4)C = 600<BR>\n" );
document.write( "A + (3/4)C = 480\r<BR>\n" );
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document.write( "From the first equation:<BR>\n" );
document.write( "A = 600 - (7/4)C<BR>\n" );
document.write( "Plug this into the second equation:<BR>\n" );
document.write( "600 - (7/4)C + (3/4)C = 480<BR>\n" );
document.write( "Now we collect like terms and:<BR>\n" );
document.write( "-(4/4)C = -120<BR>\n" );
document.write( "C = 120\r<BR>\n" );
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document.write( "Now we have that it takes 120 kg of chemical C for the mixture.<BR>\n" );
document.write( "Next, we solve for A and B.\r<BR>\n" );
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document.write( "Now you know C, so from the equation:<BR>\n" );
document.write( "B = (3/4)C<BR>\n" );
document.write( "B = 3/4 * 120<BR>\n" );
document.write( "B = 90\r<BR>\n" );
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document.write( "And finally, solve for A.<BR>\n" );
document.write( "A + B = 480<BR>\n" );
document.write( "A = 480 - B<BR>\n" );
document.write( "A = 480 - 90<BR>\n" );
document.write( "A = 390\r<BR>\n" );
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document.write( "It takes 390 kg of chemical A, 90 kg of B and 120 kg of C.\r<BR>\n" );
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document.write( "We also want to check.<BR>\n" );
document.write( "They must add to 600.<BR>\n" );
document.write( "390 + 90 + 120 = 600<BR>\n" );
document.write( "A and B together must make 80% of 600.<BR>\n" );
document.write( "(390 + 90)/600 = 0.8<BR>\n" );
document.write( "And B must equal 3/4 of C<BR>\n" );
document.write( "90 * (3/4) = 120.\r<BR>\n" );
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document.write( "Great. Hope that helps:) </SPAN>\n" );
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