document.write( "Question 900947: how much pure acid should be mixed with 2 gallons of a 40% acid solution in order to get a 70% acid solution? \n" ); document.write( "
Algebra.Com's Answer #546357 by richwmiller(17219)\"\" \"About 
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x+2*.40=.7*(2+x)
\n" ); document.write( "x=2 gallons
\n" ); document.write( "check
\n" ); document.write( "2+.8=.7*4
\n" ); document.write( "2.8=2.8
\n" ); document.write( "from view of water
\n" ); document.write( "2*0.6=0.3*(2+x)
\n" ); document.write( "1.2=0.6+0.3*x
\n" ); document.write( "1.2-0.6=0.3*x
\n" ); document.write( "0.6=0.3*x
\n" ); document.write( "x=2
\n" ); document.write( "check
\n" ); document.write( "2*0.6=0.3*(2+x)
\n" ); document.write( "2*0.6=0.3*(4)
\n" ); document.write( "1.2=1.2
\n" ); document.write( "ok\r
\n" ); document.write( "\n" ); document.write( "or from view of acid
\n" ); document.write( "1*y+0.4*2=0.7(2+y)
\n" ); document.write( "1*y+0.8=1.4+0.7y
\n" ); document.write( "1*y-0.7y=1.4-0.8
\n" ); document.write( "0.3y=0.6
\n" ); document.write( "y=0.6/0.3
\n" ); document.write( "y=2\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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