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)\"\" \"About 
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( "
\n" );