document.write( "Question 630993: A part of $10,000 is invested at 4% and the remaining amount at 3 1/2%.The annual income both is $380.Find the amount of each investment? \n" ); document.write( "
Algebra.Com's Answer #397271 by stanbon(75887)\"\" \"About 
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A part of $10,000 is invested at 4% and the remaining amount at 3 1/2%.The annual income both is $380.Find the amount of each investment?
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\n" ); document.write( "Equations:
\n" ); document.write( "value: x + y = 10,000
\n" ); document.write( "interest: 0.04x + 0.035y = 360
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\n" ); document.write( "Modify:
\n" ); document.write( "40x + 40y = 40\"10,000
\n" ); document.write( "40x + 35y = 360,000
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\n" ); document.write( "5y = 40,000
\n" ); document.write( "y = 8000 (amt. invested at 3.5%
\n" ); document.write( "x = 2000 (amt. invested at 4%
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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