document.write( "Question 1017053: The annual interest on a $5000 investment exceeds the interest earned on a $3000 investment by $185. The $5000 is invested at a 0.5% higher rate of interest than the $3000. What is the interest rate of each investment? \n" ); document.write( "
Algebra.Com's Answer #633389 by addingup(3677)![]() ![]() You can put this solution on YOUR website! 5000x+0.005 = 3000x-185 \n" ); document.write( "2000x = 184.995 \n" ); document.write( "x = 0.0925 or 9.25%. This is the rate on the $3,000. For the $5,000: \n" ); document.write( "9.25%+0.5% = 9.75% \n" ); document.write( "J \n" ); document.write( " |