document.write( "Question 221964: part of $12,000 investment earned interest at a rate of 7% and the rest earned interest at a rate of 9%. the combined interest earned at the end of 1 year was $890. how much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #166337 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
.09x+.07(12,000-x)=890
\n" ); document.write( ".09x+840-.07x=890
\n" ); document.write( ".02x=890-840
\n" ); document.write( ".02x=50
\n" ); document.write( "x=50/.02
\n" ); document.write( "x=2,500 amount invested @ 9%.
\n" ); document.write( "12,000-2,500=9,500 invested @ 7%.
\n" ); document.write( "Proof:
\n" ); document.write( ".09*2,500+.07*9,500=890
\n" ); document.write( "225+665=890
\n" ); document.write( "890=890
\n" ); document.write( "
\n" );