document.write( "Question 899881: The annual interest on a $9000 investment exceeds the interest earned on a $6000 investment by $135. The $9000 is invested at a 0.5% higher rate of interest than the $6000. What is the interest rate of each investment? \n" ); document.write( "

Algebra.Com's Answer #545643 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi
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document.write( "9000r + 6000(r-.005) = 135
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document.write( "  r = (135+30)/15000)= .011  0r 1.100%, rate of $9000 investment
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document.write( "(r-.005) = 1.095%, rate of $6000 investment \n" );
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