document.write( "Question 1144999: System of equation problem
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\n" ); document.write( "\n" ); document.write( "A 12-L cooling system is filled with 25% antifreeze. How many liters must be replaced with 100%
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Algebra.Com's Answer #766180 by ikleyn(52931)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "            In my view,  there is some misunderstanding with this problem,  how it is presented/worded in this post.\r
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\n" ); document.write( "\n" ); document.write( "            This problem is for one unknown variable,  which is the volume under the question.\r
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\n" ); document.write( "\n" ); document.write( "            Accordingly,  the standard and traditional way to solve it is using one single equation.\r
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\n" ); document.write( "\n" ); document.write( "            NO system of equations is needed,  and NO Substitution or Elimination methods are used.\r
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\n" ); document.write( "\n" ); document.write( "            Below is this traditional solution method.\r
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document.write( "Let V be the volume to drain off from 12 liters of the 25%-antifreeze.\r\n" );
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document.write( "Step 1:  Draining.  After draining,  you have 12-V liters of the 25% antifreeze.\r\n" );
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document.write( "                    It contains 0.25*(12-V) of pure antifreeze.\r\n" );
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document.write( "Step 2:  Replacing.  Then you add V liters of the pure antifreeze (the replacing step).\r\n" );
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document.write( "                     After the replacing,  you have the same total liquid volume of 12 liters.\r\n" );
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document.write( "                     It contains (0.25(12-V) + V) liters of pure antifreeze.\r\n" );
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document.write( "So, the antifreeze concentration after replacement is  \"%280.25%2A%2812-V%29%2BV%29%2F12\". \r\n" );
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document.write( "It is the ratio of the pure antifreeze volume to the total volume.\r\n" );
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document.write( "Therefore, your \"concentration equation\" is\r\n" );
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document.write( "    \"%280.25%2A%2812-V%29%2BV%29%2F12\" = 0.45.    (1)    \r\n" );
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document.write( "The setup is done and completed.\r\n" );
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document.write( "To solve the equation (1), multiply both sides by 12. You will get\r\n" );
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document.write( "    0.25*(12-V) + V= 0.45*12,\r\n" );
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document.write( "    3 - 0.25V + V= 5.4,\r\n" );
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document.write( "    0.75V = 5.4 - 3 = 2.4  ====>  V = \"2.4%2F0.75\" = 3.2 liters.\r\n" );
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document.write( "Answer.  3.2 liters of the 25% antifreeze must be drained and replaced by 3.2 liters of pure antifreeze.\r\n" );
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document.write( "Check.   \"%280.25%2A%2812-3.2%29%2B3.2%29%2F12\" = 0.45.    ! PRECISELY Correct !\r\n" );
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\n" ); document.write( "\n" ); document.write( "In this site, there is entire bunch of introductory lessons covering various types of mixture problems\r
\n" ); document.write( "\n" ); document.write( "    - Mixture problems\r
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\n" ); document.write( "\n" ); document.write( "    - Word problems on mixtures for antifreeze solutions (*)\r
\n" ); document.write( "\n" ); document.write( "    - Word problems on mixtures for alloys \r
\n" ); document.write( "\n" ); document.write( "    - Typical word problems on mixtures from the archive\r
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\n" ); document.write( "\n" ); document.write( "Read them and become an expert in solution mixture word problems.
\n" ); document.write( "Notice that among these lessons there is one on antifreeze solutions marked by (*).\r
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\n" ); document.write( "\n" ); document.write( "Also, you have this free of charge online textbook in ALGEBRA-I in this site\r
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\n" ); document.write( "\n" ); document.write( "The referred lessons are the part of this textbook in the section \"Word problems\" under the topic \"Mixture problems\".\r
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\n" ); document.write( "\n" ); document.write( "Save the link to this online textbook together with its description\r
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\n" ); document.write( "\n" ); document.write( "Free of charge online textbook in ALGEBRA-I
\n" ); document.write( "https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson\r
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\n" ); document.write( "\n" ); document.write( "to your archive and use it when it is needed.\r
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