document.write( "Question 52078This question is from textbook intermediate algebra
\n" ); document.write( ": How many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution? \n" ); document.write( "
Algebra.Com's Answer #34734 by Earlsdon(6294)\"\" \"About 
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Let x = the number of liters of the 10% alcohol solution.
\n" ); document.write( "From the problem description, you can write the equation: (Note: Change the percents to their decimal equivalents) The final solution will be (40+x) liters.
\n" ); document.write( "\"x%280.1%29+%2B+40%280.5%29+=+%2840%2Bx%29%280.4%29\" Simplify and solve for x.
\n" ); document.write( "\"0.1x+%2B+20+=+16+%2B+0.4x\" Subtract 0.1x from both sides of the equation.
\n" ); document.write( "\"20+=+16+%2B+0.3x\" Subtract 16 from both sides.
\n" ); document.write( "\"4+=+0.3x\" Finally, divide both sides by 0.3
\n" ); document.write( "\"13.33+=+x\"\r
\n" ); document.write( "\n" ); document.write( "You will need to mix 13.33...(or 13 1/3) liters of 10% alcohol solution with 40 liters of 50% alcohol solution to obtain 53.33... (or 53 1/3) liters of 40% alcohol solution.\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "13.33(0.1) + 40(0.5) = 53.33(0.4)
\n" ); document.write( "1.333 + 20 = 21.333
\n" ); document.write( "21.333 = 21.333
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