document.write( "Question 995264: The annual interest on a $10,000 investment exceeds the interest earned on an $8000 investment by $100. The $10,000 is invested at a 0.5% higher rate of interest than the $8000. What is the interest rate of each investment?\r
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Algebra.Com's Answer #614146 by macston(5194)\"\" \"About 
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\n" ); document.write( "x=interest on $10000
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\n" ); document.write( "(x+0.005)$10000=(x)$8000+$100
\n" ); document.write( "$10000x+$50=$8000x+$100
\n" ); document.write( "$2000x=$50
\n" ); document.write( "x=0.025
\n" ); document.write( "ANSWER 1: The rate on the $8000 investment is 2.5%.
\n" ); document.write( "x+0.005=0.03
\n" ); document.write( "ANSWER 2: The rate on the $10000 investment is 3%.
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\n" ); document.write( "CHECK:
\n" ); document.write( "(x+0.005)$10000=(x)$8000+$100
\n" ); document.write( "0.03($10000)=0.025($8000)+$100
\n" ); document.write( "$300=$200+$100
\n" ); document.write( "$300=$300
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