document.write( "Question 677729: Two investments were made totaling $10,500. Part of the $10,500 was invested at 9% and the remaining amount was invested at 12%. The annual interest from both investments was $1,140. How much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #421028 by partha_ban(41)![]() ![]() You can put this solution on YOUR website! Let, x be the amount invested at 9% interest. \n" ); document.write( "(10500 - x) be the amount invested at 12% interest. \n" ); document.write( "From I = Prt, interest from first investment = \n" ); document.write( "= \n" ); document.write( "Interest from 2nd investment = \n" ); document.write( "= \n" ); document.write( "By condition, \n" ); document.write( " \n" ); document.write( "9x + 126000 - 12x = 1140 * 100 \n" ); document.write( "9x - 12 x = 114000 - 126000 \n" ); document.write( "-3x = -12000 \n" ); document.write( " \n" ); document.write( "x = 4000 \n" ); document.write( "Therefore, $4,000 has been invested at 9% interest. \n" ); document.write( "$(10,500 - 4,000) = $6,500 has been invested at 12% interest. \n" ); document.write( " |