document.write( "Question 164388: Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 800mL of a solution that is 60% alcohol. How much of each solution should you use? \n" ); document.write( "

Algebra.Com's Answer #121099 by ptaylor(2198) $\"\"$ $\"About$

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Let x=amount of 30% solution needed
\n" ); document.write( "Then 800-x=amount of 80% solution needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure alcohol in the 30% solution (0.30x) plus the amount of pure alcohol in the 80% solution (0.80(800-x)) has to equal the amount of pure alcohol in the final mixture (0.60*800). So, our equation to solve is:
\n" ); document.write( "0.30x+0.80(800-x)=0.60*800 get rid of parens (distributive law)
\n" ); document.write( "0.30x+640-0.80x=480 subtract 640 from each side
\n" ); document.write( "0.30x+640-640-0.80x=480-640 collect like terms
\n" ); document.write( "-0.50x=-160 divide each side by -0.50
\n" ); document.write( "x=320ml------------------------------------amount of 30% alcohol needed
\n" ); document.write( "800-x=800-320=480ml----------------------------amount of 80% alcohol needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "0.30*320+0.80*480=0.60*800
\n" ); document.write( "96+384=480
\n" ); document.write( "480=480
\n" ); document.write( "This problem could just as well have been worked using two unknowns
\n" ); document.write( "Let x=amount of 30% needed
\n" ); document.write( "And let y=amount of 80% needed
\n" ); document.write( "x+y=800--------------------------------------eq1
\n" ); document.write( "0.30x+0.80y=0.60*800----------------------------eq2
\n" ); document.write( "From eq1, we see that y=800-x; substitute this into eq2 and then we will have the same equation as before\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor \n" ); document.write( "

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