document.write( "Question 156951: A man invested $6000 dollars and if 5% is in simple intrest and the rest is in 7% interest. if the anunal income on the interest is $372 how much money was put in each one? \n" ); document.write( "
Algebra.Com's Answer #115776 by orca(409)\"\" \"About 
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Let x be the amount invested at 5%, the the amount invested at 7% is 6000 - x.
\n" ); document.write( "The interest earned from the 5% investment is x*5%.
\n" ); document.write( "The interest earned from the 7% investment is (6000 - x)*7%.
\n" ); document.write( "Their total is 372, we have:
\n" ); document.write( "x*5% + (6000 - x)*7% = 372
\n" ); document.write( "Solving the equation for x, we have
\n" ); document.write( "5x + 7(6000 - x) = 37200 (multiply both sides by 100)
\n" ); document.write( "5x + 42000 - 7x = 37200
\n" ); document.write( "-2x = -4800
\n" ); document.write( "x = 2400
\n" ); document.write( "So
\n" ); document.write( "the amount invested at 5% is $2400.
\n" ); document.write( "the amount invested at 7% is 6000 - x = 6000 - 2400 = $3600.\r
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