document.write( "Question 1115540: $500,000 was borrowed in 3 loans, at 7%, 8%, and 9%. The annual interest was $38,600. If 3 times as much was borrowed at 7% than at 9%, how much was borrowed at each rate? \n" ); document.write( "
Algebra.Com's Answer #730384 by addingup(3677)![]() ![]() You can put this solution on YOUR website! 0.07x + 0.09y + 0.08(500,000-x+y) = 38,600 \n" ); document.write( "x = 3y \n" ); document.write( "0.07(3y) + 0.09y + 0.08(500,000-4y) = 38600 \n" ); document.write( "0.21y + 0.09y + 40,000 - 0.32y = 38600 \n" ); document.write( "-0.02y = -1400 \n" ); document.write( "y = 70,000 amount at 9% \n" ); document.write( "70,000(3) = 210,000 amount at 7% \n" ); document.write( "500,000 - 280,000 = 220,000 amount at 8% \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \n" ); document.write( "Check: \n" ); document.write( "70,000(0.09) + 210,000(0.07) + 220,000(0.08) = 38,600 Correct \n" ); document.write( " \n" ); document.write( " |