document.write( "Question 102266: two investments were made totaling $15,000. For a certain year these investments yielded $1432 in simple intrest. part of the $15,000 was invested at 9% and part at 10%. find the amount invested at each rate \n" ); document.write( "
| Algebra.Com's Answer #74322 by ptaylor(2198)     You can put this solution on YOUR website! \n" ); document.write( "Interest(I)=Principal(P) times Rate(R) times Time(T) or I=PRT \n" ); document.write( "Let P-amount invested at 9% \n" ); document.write( "(1) Interest accrued at 9%=P*0.09*1 or 0.09P\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then 15000-P=amount invested at 10% \n" ); document.write( "(2) Interest accrued at 10%=(15000-P)*0.10*1 or 0.10(15000-P)\r \n" ); document.write( "\n" ); document.write( "Now we are told that (1)+(2)=1432 or\r \n" ); document.write( "\n" ); document.write( "0.09P+0.10(15000-P)=1432 get rid of parens\r \n" ); document.write( "\n" ); document.write( "0.09P+1500-0.10P=1432 subtract 1500 from both sides \n" ); document.write( "0.09P+1500-1500-0.10P=1432-1500 collect like terms\r \n" ); document.write( "\n" ); document.write( "-0.01P=-68 divide both sides by -0.01\r \n" ); document.write( "\n" ); document.write( "P=$6800---------------------amount invested at 9%\r \n" ); document.write( "\n" ); document.write( "15000-P=15000-6800=$8200----------------amount invested at 10%\r \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( "6800*0.09+8200*0.10=1432 \n" ); document.write( "$612+$820=$1432 \n" ); document.write( "$1432=$1432\r \n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |