document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182418>Question 1182418</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A grocery store offers a cheese sampler that includes a pepper cheddar cheese that costs $18 per kilogram and Pennsylvania Jack that costs $8 per kilogram. How many kilograms of each were used to make a 5-kilogram mixture that costs $11.00 per kilogram?<BR>\n" );
document.write( "pepper cheddar    	<BR>\n" );
document.write( " kg<BR>\n" );
document.write( "Pennsylvania Jack    	<BR>\n" );
document.write( " kg<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #812438 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=812438\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The conditions of the problem make for an easy setup of the problem using two variables for the numbers of kilograms of the two kinds of cheese:<br><BR>\n" );
document.write( "x = kg of pepper cheddar<BR>\n" );
document.write( "y = kg of Pennsylvania Jack<br><BR>\n" );
document.write( "x+y=5  the total number of kilograms is 5<BR>\n" );
document.write( "18x+8y=55  the total cost of the mixture is $55 (5 kg at $11 per pound)<br><BR>\n" );
document.write( "Given those two equations, the solution from the other tutor uses substitution to solve the problem.<br><BR>\n" );
document.write( "That is a valid method -- but when the two equations are both in Ax+By=C form, I think elimination is much easier.<br><BR>\n" );
document.write( "8x+8y=40  (1st equation, multiplied by 8)<BR>\n" );
document.write( "18x+8y=55<BR>\n" );
document.write( "10x=15  (difference between the two equations, eliminating y)<BR>\n" );
document.write( "x=1.5<br><BR>\n" );
document.write( "ANSWER: x=1.5 pounds of pepper cheddar; the other 3.5 pounds of Pennsylvania Jack<br><BR>\n" );
document.write( "And here is a very quick and easy non-algebraic method for solving this or any other 2-part mixture problem.<br><BR>\n" );
document.write( "Look at the three prices on a number line -- 8, 11, and 18 -- and observe/determine/calculate that 11 is 3/10 of the way from 8 to 18.<br><BR>\n" );
document.write( "That means 3/10 of the mixture is the more expensive cheese.<br><BR>\n" );
document.write( "ANSWER: 3/10 of 5 kg, or 1.5 kg, of the pepper cheddar and the other 3.5 kg of Pennsylvania Jack.<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );