document.write( "Question 211194: A man invested $6,000, part of it at 5% simple intrest and the rest at 7% simple intrest. If his annual intrest income is $372, how much did he invest in each rate? \n" ); document.write( "
Algebra.Com's Answer #159571 by math problem solving(30)\"\" \"About 
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Let x be the amount of money invested at 5%.
\n" ); document.write( "Then the amount invested at 7% is 6000 - x.
\n" ); document.write( "The interest earned from the money invested at 5% is x*5%
\n" ); document.write( "The interest earned from the money invested at 7% is (6000 - x)*7%
\n" ); document.write( "As the total interest earned from the two investments is given as $372, we have:
\n" ); document.write( "x*5% + (6000 - x)*7% = 372
\n" ); document.write( "Solving for x, we get
\n" ); document.write( "5x + 7(6000 - x) = 37200 (multiply both sides by 100)
\n" ); document.write( "5x + 42000 - 7x = 37200
\n" ); document.write( "-2x + 42000 = 37200
\n" ); document.write( "-2x = -4800
\n" ); document.write( "x = 2400
\n" ); document.write( "So $2,400 is invested at 5%, and $6,000 - $2,400 = $3,600 is invested at 7%.
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