document.write( "Question 174976: Could someone help me out?
\n" ); document.write( "How much of a 40% antifreeze solution must a mechanical mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are need?\r
\n" ); document.write( "\n" ); document.write( "I tried the following and was marked wrong.
\n" ); document.write( "Let c represent the 40% solution
\n" ); document.write( "___ y _____________ 80% ________\r
\n" ); document.write( "\n" ); document.write( "x + y= 20
\n" ); document.write( "40x+80y= 50
\n" ); document.write( "------------
\n" ); document.write( "40x+40y=8
\n" ); document.write( "40x+40y=.5
\n" ); document.write( "-4y=7.5
\n" ); document.write( "y=18.75
\n" ); document.write( "------------
\n" ); document.write( "x+18.75=20
\n" ); document.write( "------------
\n" ); document.write( "x=1.25 y= 18.75
\n" ); document.write( "

Algebra.Com's Answer #130039 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are needed?
\n" ); document.write( "------------------------
\n" ); document.write( "active + active = active
\n" ); document.write( "0.40x + 0.80(20-x) = 0.50*20
\n" ); document.write( "40x + 1600 - 80x = 1000
\n" ); document.write( "-40x = -600
\n" ); document.write( "x = 15 liters (amount of 40% antifreeze needed in the mixture)
\n" ); document.write( "-------------
\n" ); document.write( "20-x = 5 liters (amount of 80% antifreeze needed in the mixture)
\n" ); document.write( "=========================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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