document.write( "Question 915702: Forty gallons of a 60% acid solution is obtained by mixing a 75% solution with a 50% solution. How many gallons of each solution must be used to obtain the desired mixture? \n" ); document.write( "
Algebra.Com's Answer #555733 by josgarithmetic(39618)\"\" \"About 
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The method here makes sense only if the percents are volume per volume; otherwise, other details might be necessary.\r
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\n" ); document.write( "\n" ); document.write( "u, volume of 50%
\n" ); document.write( "v, volume of 75%\r
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\n" ); document.write( "\n" ); document.write( "\"%28acid%29%2F%28mixture%29=fraction\"
\n" ); document.write( "The fraction could be used as a percentage.
\n" ); document.write( "\"%2850u%2B75v%29%2F40=60\"\r
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\n" ); document.write( "\n" ); document.write( "\"50u%2B75v=2400\"
\n" ); document.write( "\"10u%2B15v=480\"
\n" ); document.write( "\"2u%2B3v=96\"\r
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\n" ); document.write( "\n" ); document.write( "\"u%2Bv=40\", using just sum of the material volume quantities.\r
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\n" ); document.write( "\n" ); document.write( "Linear system of equations, \"highlight_green%28system%282u%2B3v=96%2Cu%2Bv=40%29%29\"
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