document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=207890>Question 207890</A>: <I><SPAN STYLE=\"word-break: break-all;\"> this is the given problem: Solution X is 2% alcohol and solution Y is 6% alcohol. A drugstore owner wants to mix them in order to get a 60-L of solution that is 3.2% alcohol. How many liters of each should the owner use?<BR>\n" );
document.write( "Solution:<BR>\n" );
document.write( "         Let x = the number of liters of X<BR>\n" );
document.write( "             y = the number of liters of Y<BR>\n" );
document.write( "-----------------------------------------------------------------<BR>\n" );
document.write( "                     +  solution   +  solution  +  mixture      +<BR>\n" );
document.write( "                     +    X        +    Y       +               +<BR>\n" );
document.write( "---------------------+-------------+------------+---------------+<BR>\n" );
document.write( "amt. of solution(L)  +    x        +    y       +   60          +   <BR>\n" );
document.write( "---------------------+-------------+------------+---------------+<BR>\n" );
document.write( "percent of alcohol   +    2%       +    6%      +   3.2%        + <BR>\n" );
document.write( "---------------------+-------------+------------+---------------+<BR>\n" );
document.write( "amt. of alcohol in   +   0.02%     +    0.06%   +   0.032(60)or + 0.02x + 0.06y<BR>\n" );
document.write( "the mixture(L)       +             +            +       1.92    +   = 1.9<BR>\n" );
document.write( "----------------------------------------------------------------+\r<BR>\n" );
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document.write( "note: to get the amount of a alcohol, we multiply the amount of solution by the percentages.<BR>\n" );
document.write( "equations:<BR>\n" );
document.write( "                 (1) x + y = 60<BR>\n" );
document.write( "                 (2) 0.02x + 0.06y = 1.92<BR>\n" );
document.write( "multiplying eq.(2) by 100, we now have this system:<BR>\n" );
document.write( "                 (1) x + y = 60<BR>\n" );
document.write( "                 (2)2x +6y = 192<BR>\n" );
document.write( "solving the system using elimination, we first multiply eq. by -2 and add the result to eq. (2)<BR>\n" );
document.write( "                    -2x-2y =-120<BR>\n" );
document.write( "                     2x+6y = 192<BR>\n" );
document.write( "                    ------------<BR>\n" );
document.write( "                        4y = 72<BR>\n" );
document.write( "                         y = 18<BR>\n" );
document.write( "solving for x, we substitute 18 in place of y in eq. (1)<BR>\n" );
document.write( "                     x + 18 = 60<BR>\n" );
document.write( "                          x = 42\r<BR>\n" );
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document.write( "Please help me why it was multiplied by negative 2, and where this -2 came from?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #157244 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=157244\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (1) x + y = 60<BR>\n" );
document.write( "(2)2x +6y = 192\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The second equation has a 2x, while the first equation has only x.  You need to make the x terms subtract out, so in order to subtract out the 2x, you need a -2x.  To get that, you have to multiply the first equation by -2.  You could also have eliminated the y terms by multiplying the first equation by -6.  We usually try to eliminate using the smaller number.\r<BR>\n" );
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document.write( "R^2 </SPAN>\n" );
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