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)\"\" \"About 
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( "
\n" );