document.write( "Question 1100317: A broker invested $10000, part earning interest at 8% per annum, and the remainder at 10% per annum. If the investments earned $872 in one year, how much is invested at 10%. (Use algebra) \n" ); document.write( "
Algebra.Com's Answer #714907 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! x=8% earned \n" ); document.write( "10000-x=10% earned \n" ); document.write( ".08x+.10(10000-x)=872 \n" ); document.write( ".08x+1000-.10x=872 \n" ); document.write( "-.02x=-128 \n" ); document.write( "multiply both sides by -50 \n" ); document.write( "x=6400 @8%=$512 \n" ); document.write( "10000-x=3600@10%=$360 \n" ); document.write( " \n" ); document.write( " |