document.write( "Question 1057279: Some money is invested at 8% and twice as much at 5%. If the annual interest adds up to $720, how much is invested at each rate? \n" ); document.write( "

Algebra.Com's Answer #672328 by jorel555(1290) $\"\"$ $\"About$

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Let n be the amount invested at 8%. Then the amount invested at 5% would be 2n. So:
\n" ); document.write( ".08n+2(.05)n=720
\n" ); document.write( ".18n=720
\n" ); document.write( "n=$4000 invested at 8%, $8000 invested at 5%. ☺☺☺☺ \n" ); document.write( "

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