document.write( "Question 1013748: joe invested $5000 part at 11 annual interest and the rest at 13% annual interest if he receives $610 interest at the end of one year how much did he invest at each rate \n" ); document.write( "

Algebra.Com's Answer #630387 by fractalier(6550) $\"\"$ $\"About$

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Call what he invests at 11%, x.
\n" ); document.write( "Then what he invests at 13% must be 5000-x.
\n" ); document.write( "The set up looks like this:
\n" ); document.write( ".11x + .13(5000-x) = 610
\n" ); document.write( ".11x + 650 - .13x = 610
\n" ); document.write( "-.02x = -40
\n" ); document.write( "x = $2000 at 11%
\n" ); document.write( "5000-x = $3000 at 13% \n" ); document.write( "

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