document.write( "Question 1069927: Jim invests $30,000 in two separate accounts. One pays 4% each year simple interest the other pays 6% his investment earns 4620 over the three year period how much did he deposit in each account \n" ); document.write( "

Algebra.Com's Answer #685324 by jorel1380(3719) $\"\"$ $\"About$

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Let n be the amount invested at 6%. Then the amount invested at 4% would be 30000-n. So:
\n" ); document.write( "3*.06n+3*.04(30000-n)=4620
\n" ); document.write( ".06n+.04(3000-n)=1540
\n" ); document.write( ".02n+1200=1540
\n" ); document.write( ".02n=340
\n" ); document.write( "n=17000
\n" ); document.write( "$17000 was invested at 6%; and $13000 was invested at 4%. ☺☺☺☺ \n" ); document.write( "

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