document.write( "Question 161279This question is from textbook algebra i
\n" ); document.write( ": one type of antifreeze is 40% glycol, and another type of antifreeze is 60% glycol. How much of each kind should be used to make 100 gallons of antifreeze that is 48% glycol? \n" ); document.write( "
Algebra.Com's Answer #118773 by vleith(2983)\"\" \"About 
You can put this solution on YOUR website!
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Solved by pluggable solver: Mixing Acids

\n" ); document.write( "This problem can be solved by a system of equations. The first step though is to define your variables correctly. Because our unknowns are volumes, we can call the volume of solution A a and the volume of B b.
\n" ); document.write( "We know from the problem that the total volume is 100. Since that total is made up of combining A & B, we can write it as their sum:
\n" ); document.write( "\"a%2Bb=100\"\r
\n" ); document.write( "\n" ); document.write( "The other equation is devised by realizing that the amount of acid contributed by each solution is equal to its volume times its percentage. Thus acid a would contribute a times 40 and b would contribute b * 60. The final volume of acid will also be equal to the (100) times the final percentage of this solution (48). This gives:
\n" ); document.write( "\"40%2Aa%2B60%2Ab=100%2A48\"\r
\n" ); document.write( "\n" ); document.write( "We have now successfully converted from English to math and can solve our two equations. In this example, it's easier done with substitution. Rewriting our first equation solving for a, we get:
\n" ); document.write( "\"a=100-b\"\r
\n" ); document.write( "\n" ); document.write( "Substituting this into to equation 2 for a gives:
\n" ); document.write( "\"40+%2A+%28100-b%29+%2B+60%2Ab+=+100%2A48\" Distributing on left & solving right side gives:
\n" ); document.write( "\"4000+-+40%28b%29+%2B+60%2Ab+=+4800\" Rearrange combining b's and constants to get:
\n" ); document.write( "\"20%2Ab=800\"Dividing both sides by 20 gives:
\n" ); document.write( "\"b=40\" don't forget the units mL.
\n" ); document.write( "This can then be put back into the equation:
\n" ); document.write( "\"a=100-b\"substituting for b gives:
\n" ); document.write( "\"a=100-40\" giving the volume a as:
\n" ); document.write( "\"a=60\" mL.\r
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