document.write( "Question 568850: How many ounces of a solution that is 90% alcohol need to be mixed with 5 ounces of a solution that is 50% alcohol in order to obtain a solution that is 80% alcohol?\r
\n" ); document.write( "\n" ); document.write( "I have tried a formula: 5(.50) + x(.90) = .80, but the answer yields a negative amount of ounces which is not possible.
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Algebra.Com's Answer #367150 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+x+\" = ounces of 90% solution needed
\n" ); document.write( "\"+.9x+\" = ounces of alcohol in 90% solution
\n" ); document.write( "\"+.5%2A5+=+2.5+\" ounces of alcohol in 50% solution
\n" ); document.write( "given:
\n" ); document.write( "\"+%28+.9x+%2B+2.5+%29+%2F+%28+x+%2B+5+%29+=+.8+\"
\n" ); document.write( "\"+.9x+%2B+2.5+=+.8%2A%28+x+%2B+5+%29+\"
\n" ); document.write( "\"+.9x+%2B+2.5+=+.8x+%2B+4+\"
\n" ); document.write( "\"+.1x+=+1.5+\"
\n" ); document.write( "\"+x+=+15+\"
\n" ); document.write( "15 ounces of 90% solution are needed
\n" ); document.write( "check answer:
\n" ); document.write( "\"+%28+.9%2A15+%2B+2.5+%29+%2F+%28+15+%2B+5+%29+=+.8+\"
\n" ); document.write( "\"+%28+13.5+%2B+2.5+%29+%2F+20+=+.8+\"
\n" ); document.write( "\"+16+=+.8%2A20+\"
\n" ); document.write( "\"+16+=+16+\"
\n" ); document.write( "OK
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