document.write( "Question 1108863: A total of $3600 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $385, how much was invested at each rate?
\n" ); document.write( "6 %+
\n" ); document.write( "11%=
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Algebra.Com's Answer #723861 by addingup(3677)\"\" \"About 
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Let \"part of it\" be x. So, x was invested at 6% or 0.06x
\n" ); document.write( "The remainder is invested at 11%, and that amount = 3600 - x
\n" ); document.write( "Put it all together:
\n" ); document.write( "0.06x + 0.11(3600 - x) = 385
\n" ); document.write( "0.06x + 396 - 0.11x = 385
\n" ); document.write( "-0.05x = -11 divide, and remember that -/- = + so when you divide you can ignore the - sign, just divide 11/0.05
\n" ); document.write( "x = 220 this is how much is invested at 6%
\n" ); document.write( "3600 - 220 = 3380 this is the amount invested at 11%
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\n" ); document.write( "Check:
\n" ); document.write( "220 * 0.06 = 13.2
\n" ); document.write( "3380 * 0.11 = 371.80
\n" ); document.write( "13.2 + 371.80 = 385 Correct.
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