document.write( "Question 1032589: A chemist has two solutions of HNO3. One has a 40% concentraiton and the other has a 25% concentration to obtain 58 liters of a 35% solution. How many liters of each solution must be mixed to obtain 58 liters of a 35% solution? \n" ); document.write( "

Algebra.Com's Answer #647259 by Edwin McCravy(20059) $\"\"$ $\"About$

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document.write( "Let the number of liters of the stronger solution be x\r\n" );
document.write( "Let the number of liters of the weaker solution be y\r\n" );
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document.write( "                      % pure     Liters\r\n" );
document.write( " Type      Liters       of         of\r\n" );
document.write( "  of         of        each       PURE\r\n" );
document.write( "solution   liquid    solution     HNO3\r\n" );
document.write( "-------------------------------------------\r\n" );
document.write( "stronger      x       0.40       0.40x\r\n" );
document.write( "weaker        y       0.25       0.25y\r\n" );
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document.write( "mixture      58       0.35       0.35(58) = 20.3 liters\r\n" );
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document.write( " The first equation comes from the liters of liquid column.\r\n" );
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document.write( "  \r\n" );
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document.write( "                 x + y = 58\r\n" );
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document.write( " The second equation comes from the last column.\r\n" );
document.write( "  \r\n" );
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document.write( "         0.40x + 0.25y = 20.3\r\n" );
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document.write( "Get rid of decimals by multiplying every term by 100:\r\n" );
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document.write( "             40x + 25y = 2030\r\n" );
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document.write( " So we have the system of equations:\r\n" );
document.write( "           .\r\n" );
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document.write( "We solve by substitution.  Solve the first equation for y:\r\n" );
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document.write( "           x + y = 58\r\n" );
document.write( "               y = 58 - x\r\n" );
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document.write( "Substitute (58 - x) for y in 40x + 25y = 2030\r\n" );
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document.write( "    40x + 25(58 - x) = 2030\r\n" );
document.write( "    40x + 1450 - 25x = 2030\r\n" );
document.write( "          15x + 1450 = 2030\r\n" );
document.write( "                 15x = 580\r\n" );
document.write( "                   x = 38.6666667 = the number of liters of the stronger.\r\n" );
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document.write( "Substitute in y = 58 - x\r\n" );
document.write( "              y = 58 - (38.6666667)\r\n" );
document.write( "              y = 19.3333333 = the number of liters of the weaker.\r\n" );
document.write( "Edwin

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