document.write( "Question 159080This question is from textbook Intermediate Algebra
\n" ); document.write( ": The cooling system of a car has a capacity of 15 liters. If the system currently has a mixture of 40% antifreeze how much of this mixture should be drained and replaced with pure antifreeze so that the system is filled with 50% antifreeze?\r
\n" ); document.write( "\n" ); document.write( "I think that the answer is 1.5 liters. I just don't know how to get it in an Algebraic expression. I used the equation 15(.5)=15(.4)+ a
\n" ); document.write( "Let a=the amount of antifreeze to be added. In case you couldn't tell I am severely confused. \r
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Algebra.Com's Answer #117195 by jojo14344(1513)\"\" \"About 
You can put this solution on YOUR website!
*See this one, thank you.
\n" ); document.write( "Let \"a\", amount in Liters to be drained
\n" ); document.write( "\"b\", amount of 100% A/F to be added
\n" ); document.write( "Therefore,
\n" ); document.write( "\"highlight%2815%280.40%29-a%280.40%29%2Bb%281.00%29=15%280.50%29%29\" ------> working eqn
\n" ); document.write( "The above eqn means the 15L capacity with 40% A/F is drained with \"a\" liters amount with the same mixture of %40 A/F of course.Then you're adding 100% A/F in \"b\" liters amount that will result to the same capacity of 15L with 50% A/F
\n" ); document.write( "Also, take note:
\n" ); document.write( "\"15L-a%28L%29%2Bb%28L%29=15L\"
\n" ); document.write( "The cooling system 15L has taken out \"a%28liters%29\" amount plus \"b%28liters%29\" amount equals the same capacity of 15L
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\n" ); document.write( "Continuing,
\n" ); document.write( "\"cross%2815L%29-a%28L%29%2Bb%28L%29=cross%2815L%29\"
\n" ); document.write( "\"a%28L%29=b%28L%29\" ---> substitute in working eqn:
\n" ); document.write( "\"15%280.40%29-a%280.40%29%2Bhighlight%28a%29%281.00%29=15%280.50%29\"
\n" ); document.write( "\"6%2B0.60a=7.5\"
\n" ); document.write( "\"0.60a=7.5-6=1.5\"
\n" ); document.write( "\"cross%280.60%29a%2Fcross%280.60%29=cross%281.5%292.5%2Fcross%280.60%29\"
\n" ); document.write( "\"highlight%28a=2.5L%29\" --------> amount to be drained = amount to be added
\n" ); document.write( "In doubt? go back working eqn:
\n" ); document.write( "\"15%280.40%29-2.5%280.40%29%2B2.5%280.40%29=15%280.50%29\"
\n" ); document.write( "\"6-1%2B2.5=7.5\"
\n" ); document.write( "\"7.5=7.5\"
\n" ); document.write( "thank you,
\n" ); document.write( "Jojo \n" ); document.write( "
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