document.write( "Question 146833: a barrel contains 30 gallons of 40% antifreeze. how many gallons of an 80% antifreeze solution should be added to make the barrel contain 50% antifreeze? \n" ); document.write( "

Algebra.Com's Answer #107228 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of 80% antifreeze needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure antifreeze in the 40% barrel(0.40*30) plus the amount of pure antifreeze in the amount that was added (0.80*x) has to equal the amount of pure antifreeze in the final solution (0.50(30+x)). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.40*30+0.80x=0.50(30+x) get rid of parens and simplify\r
\n" ); document.write( "\n" ); document.write( "12+0.80x=15+0.50x subtract 0.50x and also 12 from each side\r
\n" ); document.write( "\n" ); document.write( "12-12+0.80x-0.50x=15-12+0.50x-0.50x collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.30x=3 divide both sides by 0.30\r
\n" ); document.write( "\n" ); document.write( "x=10 gal-----------------------------amount of 80% antifreeze needed\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "0.40*30+0.80*10=0.50*40\r
\n" ); document.write( "\n" ); document.write( "12+8=20
\n" ); document.write( "20=20\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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