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)\"\" \"About 
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( "
\n" );