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)![]() ![]() You can put this solution on YOUR website! 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% \n" ); document.write( " |