A $10,000 principal is invested in two accounts, one earning 3% interest and another earning
8% interest. If the total interest for the year is $595, then how much is invested in each
account?
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10 000 = A + B
595 = A *(0.03) + B* (0.08)
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Solving the system of equations by substitution
10 000 = A + B
A = 10 000 - B
595 = (10 000 - B ) *(0.03) + B* (0.08)
595 = 300 - 0.03 B + 0.08 B
B ( 0.08 -0.03) = 595 - 300
B= 295 / 0.05
B= 5900
A = 10 000 -5900
A = 4100
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Answer
Of the 10,000, 4,100 were invested in the first account ( 3 % ) and 5900 in the second account (8%).