document.write( "Question 160209: A sum of $37,500 is invested, part of it at 11% annual interest and the remainder at 8% annual interest. If the yearly interest from the 11% investment is $303 less than twice the interest from the 8% investment, how much is invested at each interest rate? \n" ); document.write( "
Algebra.Com's Answer #118137 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! .11x=2[.08(37,500-x)]-303 \n" ); document.write( ".11x=2[(3000-.08x)]-303 \n" ); document.write( ".11x=6000-.16x-303 \n" ); document.write( ".11x+.16x=5697 \n" ); document.write( ".27x=5697 \n" ); document.write( "x=5697/.27 \n" ); document.write( "x=21,100 invested @ 11%. \n" ); document.write( "37,500-21,100=16,400 invested @ 8% \n" ); document.write( "Proof: \n" ); document.write( ".11*21,100=2(.08*16,400)-303 \n" ); document.write( "2,321=2*1,312-303 \n" ); document.write( "2,321=2,624-303 \n" ); document.write( "2,321=2,321 \n" ); document.write( " |