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)\"\" \"About 
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.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
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