document.write( "Question 212066: how many quarts of pure antifreeze must be added to 5 quarts of 10% antifreeze solution to obtain a 30% antifreeze solution \n" ); document.write( "

Algebra.Com's Answer #161205 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Let x=amount of pure antifreeze that needs to be added\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure antifreeze added(x) plus the amount of pure antifreeze in the 5 qts of 10% antifreeze solution ((0.10)(5)) has to equal the amount of pure antifreeze in the final mixture ((0.30)(5+x)). So our equation to solve is:
\n" ); document.write( "x+0.10*5=0.30(5+x) get rid of parens
\n" ); document.write( "x+0.5=1.5+0.30x subtract 0.30x and also0.5 from each side
\n" ); document.write( "x-0.30x+0.5-0.5=1.5-0.5+0.30x-0.30x collect like terms
\n" ); document.write( "0.70x=1 divide each side by 0.70
\n" ); document.write( "x=1.429 qts--------------------------amount of pure antifreeze that needs to be added\r
\n" ); document.write( "\n" ); document.write( "ACTUALLY, IT MIGHT BE A LITTLE EASIER IN THIS CASE TO DEAL WITH THE WATER RATHER THAN THE ANTIFREEZE:\r
\n" ); document.write( "\n" ); document.write( "Amount of water before the mixture((0.90*5) equals amount of water after the mixture ((0.70*5+x), or:
\n" ); document.write( "0.90*5=0.70(5+x)
\n" ); document.write( "4.5=3.5+0.70x subtract 3,5 from each side
\n" ); document.write( "4.5-3.5=3.5-3.5+0.70x
\n" ); document.write( "0.70x=1------------------------same as before\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "1.429 +0.5=0.30(5+1.429)
\n" ); document.write( "1.929+=1.5 + 0.429
\n" ); document.write( "1.929=1.929\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

\n" );