document.write( "Question 358362: debra invested $18000.Part of it was invested at 3% annual simple interest, and the rest was invested at 4%. Her interest income for the first year was $650.How much did she invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #255750 by robertb(5830)![]() ![]() You can put this solution on YOUR website! x = investment at 3%, \n" ); document.write( "18,000 - x = investment at 4%. Then \n" ); document.write( "0.03x + 0.04(18,000 - x) = 650. \n" ); document.write( "0.03x + 720 - 0.04x = 650 \n" ); document.write( "-0.01x = -70 \n" ); document.write( "x = 7000, investment at 3%. and \n" ); document.write( "18,000 - x = 11,000 = investment at 4%. \n" ); document.write( " |