document.write( "Question 190688: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #143187 by checkley75(3666)![]() ![]() ![]() You can put this solution on YOUR website! .12x+.08(7000-x)=764 \n" ); document.write( ".12x+560-.08x=764 \n" ); document.write( ".04x=764-560 \n" ); document.write( ".04x=204 \n" ); document.write( "x=204/.04 \n" ); document.write( "x=$5,100 is the amount invested @ 12%. \n" ); document.write( "7,000-5,100=$1,900 is the amounr invested @ 8%. \n" ); document.write( "Proof: \n" ); document.write( ".12*5,100+.08*1,900=764 \n" ); document.write( "612+152=764 \n" ); document.write( "764=764\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |