document.write( "Question 180957This question is from textbook College Algebra
\n" ); document.write( ": How many gallons of acid must be mixed with 40 gallons of 60% acid solution to get a 75% solution? \n" ); document.write( "
Algebra.Com's Answer #135658 by ptaylor(2198)\"\" \"About 
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Let x= amount of pure acid that is needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure acid that is needed(x) plus the amount of pure acid in the 60% solution(0.60*40) has to equal the amount of pure acid in the final mixture(0.75(40+x)). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "x+0.60*40=0.75(40+x) get rid of parens and simplify
\n" ); document.write( "x+24=30+0.75x subtract 0.75x and also 24 from each side
\n" ); document.write( "x-0.75x+24-24=30-24+0.75x-0.75x collect like terms
\n" ); document.write( "0.25x=6 divide each side by 0.25
\n" ); document.write( "x=24 gal-------------------------------amount of pure acid needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "24+0.60*40=0.75*64
\n" ); document.write( "24+24=48
\n" ); document.write( "48=48\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r
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