document.write( "Question 312958: A broker invested 10,000$. Part of the money was invested at 8% per year and the remainder at 10% per year. if the investment earned 872 in 1 year, how much was invested at each amount? \n" ); document.write( "
Algebra.Com's Answer #223761 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! .10x+.08(10,000-x)=872 \n" ); document.write( ".10x+800-.08x=872 \n" ); document.write( ".02x=872-800 \n" ); document.write( ".02x=72 \n" ); document.write( "x=72/.02 \n" ); document.write( "x=3,600 Invested @ 10% \n" ); document.write( "10,000-3,600=6,400 Invested @ 8% \n" ); document.write( "Proof: \n" ); document.write( ".10*3,600+.08*6,400=872 \n" ); document.write( "360+512=872 \n" ); document.write( "872=872 \n" ); document.write( " |