document.write( "Question 1025269: A man invested part of 10,000 @ 6% and the rest @ 7%. The annual income from the 7% was 35 less than seven times the annual income from the 6% investment. How much did he invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #642643 by fractalier(6550)\"\" \"About 
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Call what he invests at 6%, x. Thus he invests 10000-x at 7%.
\n" ); document.write( "The interest he gets from each is .06x and .07(10000-x).
\n" ); document.write( "So we set this up like this
\n" ); document.write( ".07(10000-x) = 7(.06x) - 35
\n" ); document.write( "Now solve for x...
\n" ); document.write( "700 - .07x = .42x - 35
\n" ); document.write( "735 = .49x
\n" ); document.write( "x = $1500 at 6% and thus
\n" ); document.write( "10000-x = $8500 at 7%
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