Question 108197
shawn has an annual interest income of $3290 from two investments.  he has $7000 more invested at 8% than he has invested at 6%. find the amount invested in each rate ?
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.08x+.06y=3290
y=x+7000
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.08x+.06(x+7000)=3290
.08x+.06x+420=3290
.14x=2870
x=20500
.08(20500)+.06y=3290
1640+.06y=3290
.06y=1650
y=27500
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At 8% interest = $20500
at 6%interest = $27500
:)